{"id":1304,"date":"2015-09-02T01:36:56","date_gmt":"2015-09-02T01:36:56","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1304"},"modified":"2024-08-23T13:04:01","modified_gmt":"2024-08-23T13:04:01","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-forca-de-atrito","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/dinamica\/forca-de-atrito\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-forca-de-atrito\/","title":{"rendered":"For\u00e7a de atrito – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre<\/b><\/span><\/span><\/span><\/p>\n

For\u00e7a de atrito<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

01<\/b><\/span><\/span>– Calculando as intensidades das for\u00e7as de atrito:<\/b><\/span><\/span><\/p>\n

est\u00e1tico\u00a0 —\u00a0 F<\/b><\/span><\/span>ate<\/b><\/span><\/span><\/sub>=\u03bc<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>.N==m<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>.P=\u03bc<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>.mg=0,5.10.10\u00a0 —\u00a0 F<\/b><\/span><\/span>atr<\/b><\/span><\/span><\/sub>=50N \u2013 ele s\u00f3 entra em movimento se F for superior a 50N\u00a0 —\u00a0\u00a0\u00a0\u00a0 din\u00e2mico\u00a0 —\u00a0 F<\/b><\/span><\/span>atd<\/b><\/span><\/span><\/sub>=\u03bc<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>.N==m<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>.P=\u03bc<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>.mg=0,3.10.10\u00a0 —\u00a0 F<\/b><\/span><\/span>atd<\/b><\/span><\/span><\/sub>=30N\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

a) Se F=0, Fat=0, ele continua em repouso\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

b) Se F=20N, F<\/b><\/span><\/span>ate<\/b><\/span><\/span><\/sub>=20N e ele continua em repouso\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

c) Se F=40N, F<\/b><\/span><\/span>ate<\/b><\/span><\/span><\/sub>=40N e ele continua em repouso\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

d) Se F=50N, F<\/b><\/span><\/span>ate<\/b><\/span><\/span><\/sub>=50N e ele continua em repouso, mas est\u00e1 na imin\u00eancia de movimento.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

e) Se F=60N, ele entra em movimento e est\u00e1 acelerando com acelera\u00e7\u00e3o a tal que F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 F \u2013 F<\/b><\/span><\/span>atd<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

60 –\u00a0\u00a0 30=10a\u00a0 —\u00a0 a=3\/10\u00a0 —\u00a0 a=3m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0.<\/b><\/span><\/span><\/p>\n

f) Como o bloco j\u00e1 est\u00e1 em movimento a for\u00e7a de atrito sobre ele \u00e9 o din\u00e2mico ou cin\u00e9tico e vale 30N.\u00a0 —\u00a0 F<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>=ma\u00a0 — 35 \u2013 30=10.a\u00a0 —\u00a0 a=5\/10\u00a0 —\u00a0 a=0,5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>. Ele acelera com a velocidade aumentando de 0,5m\/s em cada segundo.<\/b><\/span><\/span><\/p>\n

g) Como F=30N e F<\/b><\/span><\/span>atd<\/b><\/span><\/span><\/sub>=30N, a for\u00e7a resultante sobre ele \u00e9 nula e ele est\u00e1 em equil\u00edbrio din\u00e2mico, movendo-se em MRU.<\/b><\/span><\/span><\/p>\n

h) Nesse caso a for\u00e7a resultante sobre ele \u00e9 o F<\/b><\/span><\/span>atd<\/b><\/span><\/span><\/sub>=30N que o est\u00e1 freando com\u00a0 —\u00a0 F<\/b><\/span><\/span>atd<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 30=10.a\u00a0 —\u00a0 a=3m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0e sua<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

velocidade vai diminuindo de 3m\/s em cada segundo, at\u00e9 parar.\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

02<\/b><\/span><\/span>– Pelo gr\u00e1fico observamos que quando F=15N, a for\u00e7a de atrito est\u00e1tico \u00e9 m\u00e1xima e vale F<\/b><\/span><\/span>ate<\/b><\/span><\/span><\/sub>=15N. Qualquer for\u00e7a acima desse valor coloca o bloco em movimento. Quando F=30N, ele j\u00e1 entrou em movimento e o F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0sobre ele \u00e9 o din\u00e2mico e vale F<\/b><\/span><\/span>atd<\/b><\/span><\/span><\/sub>=10N (vide gr\u00e1fico)\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 30 \u2013 10=5.a\u00a0\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a=4m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u00a0\u00a0\u00a0—\u00a0\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>ate<\/b><\/span><\/span><\/sub>=\u03bc<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>N=\u03bc<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>P=\u03bc<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>mg\u00a0 —\u00a0 15=me.5.10\u00a0 —\u00a0 \u03bc<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=15\/50\u00a0 —\u00a0\u00a0\u03bc<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=0,3<\/b><\/span><\/span><\/p>\n

R- A<\/b><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span>\u00a0A \u00fanica for\u00e7a que age sobre a caixa, na dire\u00e7\u00e3o horizontal \u00e9 a for\u00e7a de atrito din\u00e2mico que \u00e9 a for\u00e7a resultante, freando-o at\u00e9 parar.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

V= V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ a.t\u00a0 —\u00a0 0 = 10 + a.5\u00a0 —\u00a0 a= -2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0(negativo, pois est\u00e1 freando)\u00a0 —\u00a0 em m\u00f3dulo a=2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 F<\/b><\/span><\/span>atd<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0\u03bcN=ma\u00a0 —\u00a0 \u03bcP =ma\u00a0 —\u00a0 \u03bcmg=ma\u00a0 —\u00a0 \u03bc=a\/g\u00a0 —\u00a0 \u03bc=2\/10\u00a0 —\u00a0\u00a0\u03bc<\/b><\/span><\/span>=0,2<\/b><\/span><\/span>\u00a0 R- B<\/b><\/span><\/span><\/p>\n

04<\/b><\/span><\/span>– As for\u00e7as verticais\u00a0\"\"<\/b><\/span><\/span>\u00a0e\u00a0\"\"se anulam, sobrando apenas a for\u00e7a de atrito din\u00e2mico que \u00e9 a for\u00e7a resultante sobre ele.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=72km\/h\/3,6=20m\/s\u00a0 —\u00a0 Torricelli\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.a.DS\u00a0 —\u00a0 0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= 20<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.a.50\u00a0 —\u00a0 a=-400\/100\u00a0 — a=-4m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 em m\u00f3dulo\u00a0 —\u00a0 a=4m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>atd<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 F<\/b><\/span><\/span>atd<\/b><\/span><\/span><\/sub>=10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.4\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>atd<\/b><\/span><\/span><\/sub>=4.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>N<\/b><\/span><\/span>\u00a0 —\u00a0 V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ a.t\u00a0 —\u00a0 0=20 + (-4).t\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>t=5s<\/b><\/span><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span>atd<\/b><\/span><\/span><\/sub>=\u03bc<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>N=m<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>P=\u03bc<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>mg — 4.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=\u03bc<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10\u00a0 —\u00a0 \u03bc<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=4.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\/10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0\u03bc<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=0,4<\/b><\/span><\/span><\/p>\n

05-\u00a0<\/b><\/span><\/span>Calculando a intensidade da for\u00e7a de atrito din\u00e2mico que age sobe cada bloco<\/b><\/span><\/span><\/p>\n

A\u00a0 —\u00a0 F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>=\u03bc<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.N<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>=\u03bc<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>=\u03bc<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.g\u00a0 —\u00a0 F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>=0,2.6.10\u00a0 —\u00a0 F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>=12N\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

B\u00a0 —\u00a0 F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=\u03bc<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.N<\/b><\/span><\/span>B\u00a0<\/b><\/span><\/span><\/sub>\u00a0—\u00a0\u00a0 F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=\u03bc<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=\u03bc<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.g\u00a0 —\u00a0 F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=0,3.4.10\u00a0 —\u00a0 F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=12N<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>\u00a0<\/b><\/span><\/span><\/p>\n

Colocando as for\u00e7as na dire\u00e7\u00e3o do movimernto que \u00e9 horizontal:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Bloco A\u00a0 —\u00a0 F<\/b><\/span><\/span>RA<\/b><\/span><\/span><\/sub>=\u03bc<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.a\u00a0 —\u00a0 F \u2013 (N + F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>)=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>a\u00a0 —\u00a0 60 \u2013 N \u2013 12=6.a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>48 \u2013 N=6.a\u00a0 I<\/b><\/span><\/span><\/p>\n

Bloco B\u00a0 —\u00a0 F<\/b><\/span><\/span>RB<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.a\u00a0 —\u00a0 N \u2013 F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>N \u2013 12=4.a\u00a0 II<\/b><\/span><\/span><\/p>\n

Somando membro a membro I com II\u00a0 —\u00a0 36=10.a\u00a0 —\u00a0 a=3,6m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 N \u2013 12=4.a\u00a0 —\u00a0 N=12 + 4.3,6\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>N=26,4N\u00a0 R- A<\/b><\/span><\/span><\/p>\n

06<\/b><\/span><\/span>–\u00a0<\/b><\/span><\/span>Colocando as for\u00e7as<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span><\/p>\n

F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>=\u03bc<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.N<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>=\u03bc<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.P\u00a0 —\u00a0 F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>=0,5.P\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=\u03bc<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.N<\/b><\/span><\/span>B\u00a0<\/b><\/span><\/span><\/sub>\u00a0—\u00a0\u00a0 F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=\u03bc<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.P\u00a0 —\u00a0 F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=0,5.P<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>\u00a0<\/b><\/span><\/span><\/p>\n

Bloco A —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 F \u2013 (N + F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>)=m.a\u00a0 —\u00a0 10 \u2013 N \u2013 0,5P=m.a I<\/b><\/span><\/span><\/p>\n

Bloco B\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 N \u2013 F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 N \u2013 0,5P=m.a II\u00a0 —\u00a0 somando I com II\u00a0 —\u00a0 10 \u2013 P=2m.a\u00a0 —\u00a0 a=(10 \u2013 P)\/2m III<\/b><\/span><\/span><\/p>\n

Substituindo III em II\u00a0 —\u00a0 N \u2013 0,5P = m(10 \u2013 P)\/2m\u00a0 —\u00a0 N=0,5P + 5 \u2013 0,5P\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>N=5N R-D<\/b><\/span><\/span><\/p>\n

07<\/b><\/span><\/span>– F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>=\u03bc.m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.g\u00a0 —\u00a0 F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>=m.2.10\u00a0 —\u00a0 F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>=20m\u00a0\u00a0 —\u00a0\u00a0 F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=\u03bcm<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>g\u00a0 —\u00a0 F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=m.1.10\u00a0 —\u00a0 F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=10m<\/b><\/span><\/span><\/p>\n

a) Colocando as for\u00e7as:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0Como os blocos est\u00e3o com velocidade constante, est\u00e3o em equil\u00edbrio din\u00e2mico e a for\u00e7a resultante sobre eles \u00e9 nula\u00a0 —bloco B\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0 —\u00a0 F \u2013 F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>\u00a0\u2013 T=0\u00a0 —\u00a0 18 – 10m – T=0\u00a0 I\u00a0\u00a0 —\u00a0\u00a0 bloco A\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 T – 20m=0\u00a0 II\u00a0 —\u00a0 somando I com II\u00a0 —\u00a0 18 – 30m=0\u00a0 —\u00a0<\/b><\/span><\/span>m=0,6<\/b><\/span><\/span><\/p>\n

b) T – 20m=0\u00a0 —\u00a0 T=20.0,6\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=12N<\/b><\/span><\/span><\/p>\n

08<\/b><\/span><\/span>– a)<\/b><\/span><\/span>\u00a0F<\/b><\/span><\/span>at<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=Fat<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=Fat<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=\u03bcMg\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

Como a velocidade \u00e9 constante o sistema est\u00e1 em equil\u00edbrio din\u00e2mico e a for\u00e7a resultante sobre cada bloco \u00e9 nula, assim,\u00a0<\/b><\/span><\/span>a=0<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

b) bloco B\u00a0 — F<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 T=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0+ T<\/b><\/span><\/span>1\u00a0<\/b><\/span><\/span><\/sub>\u00a0—\u00a0 T= \u03bcMg + T<\/b><\/span><\/span>1\u00a0<\/b><\/span><\/span><\/sub>I\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/sub>—\u00a0\u00a0 bloco A\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=Fat\u00a0 —\u00a0 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=\u03bcMg\u00a0 II\u00a0\u00a0 — substituindo II em I\u00a0 —\u00a0 T=\u03bcMg<\/b><\/span><\/span><\/p>\n

09<\/b><\/span><\/span>– a) Como o caminh\u00e3o e o caixote est\u00e3o com velocidade constante (equil\u00edbrio din\u00e2mico) a for\u00e7a resultante sobre o caixote \u00e9 nula e, se F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0, F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0tamb\u00e9m ser\u00e1 nula.<\/b><\/span><\/span><\/p>\n

b) A for\u00e7a resultante que age sobre o caixote, na dire\u00e7\u00e3o do movimento \u00e9 a for\u00e7a de atrito F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0que ele troca com a carroceria.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Sobre o caixote\u00a0 —\u00a0 F<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 \u03bcmg=m.a\u00a0 —\u00a0 0,25.10=a\u00a0 —\u00a0 a=2,5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u00a0—\u00a0 a=2,5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ a.t\u00a0 —\u00a0 0 = 20 – 2,5.t\u00a0 —\u00a0 t=20\/2,5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>t=8s<\/b><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span>\u00a0A intensidade da for\u00e7a de atrito est\u00e1tico entre as caixa e a carroceria vale F<\/b><\/span><\/span>ate<\/b><\/span><\/span><\/sub>=\u03bcN\u00a0 —\u00a0 F<\/b><\/span><\/span>ate<\/b><\/span><\/span><\/sub>=\u03bcP\u00a0 —\u00a0 F<\/b><\/span><\/span>ate<\/b><\/span><\/span><\/sub>=mg\u00a0 —\u00a0 F<\/b><\/span><\/span>ate<\/b><\/span><\/span><\/sub>=0,1.500.10\u00a0 —\u00a0 F<\/b><\/span><\/span>ate<\/b><\/span><\/span><\/sub>=500N (acima desse valor as caixas se movem).<\/b><\/span><\/span><\/p>\n

No movimento acelerado\u00a0 —\u00a0 V=V<\/b><\/span><\/span>o\u00a0<\/b><\/span><\/span><\/sub>+ a.t\u00a0 —\u00a0 30=0 + a.20\u00a0 —\u00a0 a=1,5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=500.1,5\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=750N ( como F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>>F<\/b><\/span><\/span>ate<\/b><\/span><\/span><\/sub>, as caixas se movem para tr\u00e1s e ficam juntas na parte traseira do caminh\u00e3o).<\/b><\/span><\/span><\/p>\n

No movimento uniforme, elas est\u00e3o em equil\u00edbrio din\u00e2mico e F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0. Assim, elas n\u00e3o se movem e ficam juntas na parte traseira.<\/b><\/span><\/span><\/p>\n

No movimento retardado\u00a0 —\u00a0 V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ a.t\u00a0 —\u00a0 0=30 \u2013 a.40\u00a0 —\u00a0 a=3\/4m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=500.3\/4\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=375N (menor que 500N, portanto elas n\u00e3o se movem e ficam na parte traseira do caminh\u00e3o) R- A<\/b><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span>\u00a0A for\u00e7a de atrito sobre cada bloco \u00e9 a mesma e vale f<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

figura 1\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 F \u2013 f=m.a I\u00a0\u00a0\u00a0 —\u00a0\u00a0\u00a0 figura 2\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 2F \u2013 f=m.3.a II\u00a0 —\u00a0 multiplicando I por (-1) e somando membro a membro I com II\u00a0 —\u00a0 F=2ma III\u00a0\u00a0\u00a0 —\u00a0\u00a0\u00a0 substituindo III em I\u00a0 —\u00a0 f=ma IV\u00a0\u00a0\u00a0 —\u00a0\u00a0\u00a0 dividindo membro a membro IV por III\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

f\/F = 1\/2<\/b><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span>\u00a0R- C (vide teoria)<\/b><\/span><\/span><\/p>\n

13<\/b><\/span><\/span>– a) Considerando os dois blocos como um corpo s\u00f3 de massa 3kg\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 4,5=3.a \u00a0—\u00a0 a=1,5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.<\/b><\/span><\/span><\/p>\n

b) colocando as for\u00e7as de atrito que os blocos A e B trocam entre si e que constituem par a\u00e7\u00e3o e rea\u00e7\u00e3o (mesma intensidade,<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

mesma dire\u00e7\u00e3o e sentidos opostos) e aplicando F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a apenas no bloco B\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=1,2.1,5\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=1,8N\u00a0 — \u00a0agora, aplicando F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a apenas sobre o bloco A, teremos\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 4,5 \u2013 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=1,6.1,5\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=1,8N\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=\u03bcN\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=\u03bcP\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=\u03bcmg\u00a0 —\u00a0 1,8=\u03bc.1,8.10\u00a0 —\u00a0 \u03bc=1,8\/18\u00a0 —\u00a0\u00a0\u03bc<\/b><\/span><\/span>=0,1<\/b><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span>\u00a0a) F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=(2,0 + 3,0).a\u00a0 —\u00a0 2=5.a\u00a0 —\u00a0 a=2\/5\u00a0 —\u00a0 a=0,4m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.<\/b><\/span><\/span><\/p>\n

b) colocando as for\u00e7as de atrito que os blocos trocam entre si<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Bloco de baixo\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 2 \u2013 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=3.0,4\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=0,8N<\/b><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span>\u00a0A for\u00e7a de atrito \u00e9 a mesma nos dois casos:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

O F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0em cada situa\u00e7\u00e3o \u00e9 o mesmo, pois F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=mP e m e P s\u00e3o os mesmos.<\/b><\/span><\/span><\/p>\n

situa\u00e7\u00e3o 1\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 — F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0\u2013 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0+ ma I\u00a0\u00a0\u00a0 —-\u00a0\u00a0\u00a0 situa\u00e7\u00e3o 2\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u2013 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=ma II\u00a0 —\u00a0 Observe em I e em II que F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>, portanto F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=1 R- B<\/b><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span>\u00a0C\u00e1lculo da acelera\u00e7\u00e3o do sistema\u00a0 —\u00a0\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 12=(1 + 2,5 + 0,5).a\u00a0 —\u00a0 a=3m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0(acelera\u00e7\u00e3o dos tr\u00eas blocos).<\/b><\/span><\/span><\/p>\n

For\u00e7as que atuam no corpo C:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Peso de C, aplicado pela Terra, de intensidade p=mg\u00a0 —\u00a0 P=0,5.10com m\u00f3dulo 5,0 N.<\/b><\/span><\/span><\/p>\n

For\u00e7a normal para cima, aplicada por B sobre C, de intensidade N=5,0N (C est\u00e1 em equil\u00edbrio vertical).<\/b><\/span><\/span><\/p>\n

For\u00e7a de atrito que B aplica sobre C\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=0,5.3\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=1,5N<\/b><\/span><\/span><\/p>\n

Observa\u00e7\u00e3o: As for\u00e7as\u00a0\"\"\u00a0e\u00a0\"\"podem ser somadas vetorialmente e se tornarem uma \u00fanica for\u00e7a de B sobre C que chamamos de<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0for\u00e7a de contato\u00a0\"\"\u00a0e que tem intensidade (F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=(1,5)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ (5)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=5,2N<\/b><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span>\u00a0a) Quando o cart\u00e3o \u00e9 puxado pela for\u00e7a\u00a0\"\", horizontal e para a direita, ele aplica sobre a moeda uma for\u00e7a de atrito cin\u00e9tico (din\u00e2mico)\u00a0\"\", puxado-a tamb\u00e9m para a direita.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Sobre a moeda, al\u00e9m de\"\"\u00a0\u00b4(horizontal e para a direita) agem tamb\u00e9m as for\u00e7a verticais:\u00a0\"\"\u00a0(peso da moeda) e\u00a0\"\"(for\u00e7a normal exercida pelo cart\u00e3o).<\/b><\/span><\/span><\/p>\n

b) O Princ\u00edpio da In\u00e9rcia afirma que todo corpo em repouso, por in\u00e9rcia, tende a permanecer em repouso, mas somente se a for\u00e7a resultante sobre ele for nula e, no caso da moeda esse princ\u00edpio n\u00e3o \u00e9 v\u00e1lido, pois existe atrito (\"\").<\/b><\/span><\/span><\/p>\n

Vamos calcular a for\u00e7a de atrito est\u00e1tico\u00a0 —\u00a0 F<\/b><\/span><\/span>ate<\/b><\/span><\/span><\/sub>=\u03bcN\u00a0 —\u00a0 F<\/b><\/span><\/span>ate<\/b><\/span><\/span><\/sub>=\u03bcP\u00a0 —\u00a0 F<\/b><\/span><\/span>ate<\/b><\/span><\/span><\/sub>=mmg\u00a0 —\u00a0 F<\/b><\/span><\/span>ate<\/b><\/span><\/span><\/sub>=0,15.0,01.10\u00a0 —\u00a0 F<\/b><\/span><\/span>ate<\/b><\/span><\/span><\/sub>=0,15N (para qualquer valor de F maior que 0,15N, a moeda entra em movimento, ou seja, destaca-se do cart\u00e3o.<\/b><\/span><\/span><\/p>\n

Assim, se F for maior que 0,15N a moeda poder\u00e1 cair dentro do copo e se F for menor ou igual a 0,15N, a moeda acompanhar\u00e1 o cart\u00e3o e n\u00e3o se destacar\u00e1 dele.<\/b><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span>\u00a0Como entre o bloco A e o bloco B n\u00e3o h\u00e1 atrito, a for\u00e7a resultante, na dire\u00e7\u00e3o do movimento, sobre o bloco A ser\u00e1\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>= M<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.a<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 24=1.a<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0a<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=24m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

\"\"<\/p>\n

Entre o bloco B e o solo existe atrito que vale F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=\u03bcN\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=\u03bc (P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0+ P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=0,2.50\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=10N\u00a0 — bloco B\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.a<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 F<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0\u2013 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=4.a<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 12 \u2013 10=4.a<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span>a<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=0,5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

Assim, a acelera\u00e7\u00e3o relativa entre A e B \u00e9 de a=24 \u2013 0,5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a=23,5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.<\/b><\/span><\/span><\/p>\n

19<\/b><\/span><\/span>-Colocando todas as for\u00e7as que agem sobre o bloco e sobre a parede<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0 equil\u00edbrio na vertical\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=P=mg=3.10\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=30N\u00a0 —\u00a0 equil\u00edbrio na horizontal\u00a0 —\u00a0 F=N\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=\u03bcN\u00a0 —\u00a0 30=0,2.F\u00a0 —\u00a0 F=30\/0,2\u00a0 —\u00a0 F=150N\u00a0 R- D<\/b><\/span><\/span><\/p>\n

 <\/p>\n

20<\/b><\/span><\/span>-No equil\u00edbrio est\u00e1tico na imin\u00eancia de movimento, a for\u00e7a de atrito est\u00e1tica \u00e9 dada por \u03bceN, e se equilibra com o peso P. Na situa\u00e7\u00e3o do exerc\u00edcio, a for\u00e7a normal N \u00e9 tal que N = F. Assim, \u03bceN = P implica em \u03bce = P\/F. Se \u03bce < 1, logo 0 < P < F.\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

21<\/b><\/span><\/span>– Colocando todas as for\u00e7as que se relacionam com a pessoa, a parede e o solo.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0e F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 for\u00e7as trocadas entre a pessoa e a parede de intensidade 120N\u00a0 —\u00a0\u00a0 F<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0e F<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0 for\u00e7as de atrito trocadas entre os p\u00e9s da pessoa e o solo\u00a0 —\u00a0 F<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 peso da pessoa trocada com o centro da Terra e F<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sub>, for\u00e7a normal que a pessoa troca com o solo.<\/b><\/span><\/span><\/p>\n

Como a pessoa est\u00e1 em repouso, a for\u00e7a resultante \u00e9 nula e as for\u00e7as verticais e horizontais se anulam\u00a0 R- D<\/b><\/span><\/span><\/p>\n

22-\u00a0<\/b><\/span><\/span>Calculando a intensidade da for\u00e7a de atrito\u00a0\"\"\u00a0que age sobre a pessoa\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=\u03bcN\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=\u03bcP\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=\u03bcmg\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=\u03bc80.10\u00a0 —<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=800\u03bc\u00a0 —\u00a0 mas, F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=F=120N (equil\u00edbrio horizontal)\u00a0 —\u00a0 120=800\u03bc\u00a0 —\u00a0 \u03bc=120\/800\u00a0 —\u00a0\u00a0\u03bc<\/b><\/span><\/span>=0,15\u00a0 R- A<\/b><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span>\u00a0Colocando as for\u00e7as que agem sobre cada bloco na dire\u00e7\u00e3o do movimento<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe que, como os blocos est\u00e3o em equil\u00edbrio\u00a0 —\u00a0 P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=T e T=F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>. Assim, F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>A\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/sub>R- A<\/b><\/span><\/span><\/p>\n

24- Colocando as for\u00e7as sobre cada bloco na dire\u00e7\u00e3o do movimento<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>=\u03bcm<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>g\u00a0 —\u00a0 F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>=0,2.3.10\u00a0 —\u00a0 F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>=6N\u00a0 —\u00a0 F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=\u03bcm<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>g\u00a0 —\u00a0 F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=\u03bcm<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>g\u00a0 —\u00a0 F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=0,2.2.10\u00a0 —\u00a0 F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>=4N\u00a0 —\u00a0 P<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>.g\u00a0 — P<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=7.10\u00a0 —\u00a0 P<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=70N<\/b><\/span><\/span><\/p>\n

bloco C\u00a0 —\u00a0 desce\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>a\u00a0 —\u00a0 P<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0\u2013 T=m<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>.a\u00a0 —\u00a0 70 \u2013 T=7.a\u00a0 I\u00a0 —\u00a0 bloco B\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.a\u00a0 —\u00a0 N \u2013 4=2.a\u00a0 II\u00a0 —\u00a0 bloco A\u00a0\u00a0 —\u00a0F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.a\u00a0 —\u00a0 T \u2013 N \u2013 6=3.a\u00a0 III\u00a0\u00a0 —\u00a0\u00a0 somando membro a membro I, II e II\u00a0 —\u00a0 70 \u2013 10=12.a\u00a0 —\u00a0 a=5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

A for\u00e7a de contato entre A e B \u00e9 N \u00a0—\u00a0 substituindo a=5m\/s<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>\u00a0em II\u00a0 —\u00a0 N \u2013 4=2.5\u00a0 —\u00a0 N=14N\u00a0 R- B<\/b><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span>\u00a0Colocando o bloco de massa m sobre o bloco de massa 10kg e todas as for\u00e7as que influem na tend\u00eancia de movimento:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como o sistema est\u00e1 em repouso, a for\u00e7a resultante sobre cada bloco \u00e9 nula\u00a0 —\u00a0 bloco de massa 5kg\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 P=T=5.10\u00a0 —\u00a0 T=50N\u00a0 —\u00a0 blocos apoiados na superf\u00edcie horizontal\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=T\u00a0 —\u00a0 \u03bcN=50\u00a0 —\u00a0 m(P<\/b><\/span><\/span>10kg<\/b><\/span><\/span><\/sub>\u00a0+ P<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>)=50\u00a0 —\u00a0\u00a0 0,2.(10 + m).10=50\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

20 + 2m=50\u00a0 —\u00a0 m=30\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>m=<\/b><\/span><\/span>15kg\u00a0\u00a0\u00a0 (para qualquer m<15kg colocada sobre 10kg, o sistema se mover\u00e1 no sentido hor\u00e1rio).<\/b><\/span><\/span><\/p>\n

26-\u00a0\"\"<\/b><\/span><\/span>\u00a0\u00a0—\u00a0 for\u00e7a que o bloco A troca com o bloco B\u00a0 —\u00a0\u00a0\"\"\u00a0– peso do bloco B\u00a0 —\u00a0\u00a0\"\"\u00a0– for\u00e7a de atrito que o bloco A aplica no bloco B, impedindo-o de escorregar.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

bloco B\u00a0 —\u00a0 equil\u00edbrio vertical\u00a0 —\u00a0 F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=P=mg=0,4.10\u00a0 —F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=4N\u00a0 —\u00a0 F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=\u03bcN\u00a0 —\u00a0 4=0,9N\u00a0 —\u00a0 N=4,4N<\/b><\/span><\/span><\/p>\n

c\u00e1lculo da acelera\u00e7\u00e3o de B em movimento horizontal para a direita onde a for\u00e7a resultante \u00e9\u00a0\"\"\u00a0—\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.a<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0 N=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.a<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 — 4,4=0,4.a<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 a<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=11m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 mas, para o bloco B n\u00e3o cair e acompanhar o bloco A eles devem ter a mesma acelera\u00e7\u00e3o, ou seja,<\/b><\/span><\/span><\/p>\n

a<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=11m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 bloco A\u00a0 —\u00a0 F<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.a<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 F \u2013 N=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.a<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 F \u2013 4,4=5.11\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F=59,4N<\/b><\/span><\/span>(para qualquer F>59,4N, o bloco B n\u00e3o cair\u00e1).<\/b><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span>\u00a0R \u2013D Como o movimento \u00e9 retil\u00edneo uniforme, a for\u00e7a resultante sobre o caixote na dire\u00e7\u00e3o do movimento (horizontal) \u00e9 nula e , na vertical, a for\u00e7a peso e a rea\u00e7\u00e3o normal da esteira se anulam.\u00a0<\/b><\/span><\/span><\/p>\n

28-<\/b><\/span><\/span>\u00a0a)\u00a0\"\"– peso da caixa\u00a0 —\u00a0\u00a0\"\"– for\u00e7a normal que a ca\u00e7amba aplica sobre a caixa\u00a0 —\u00a0\u00a0\"\"\u00a0– for\u00e7a de atrito trocada entre a ca\u00e7amba e a caixa.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

b) Essa for\u00e7a pedida \u00e9 a for\u00e7a de atrito F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0 C\u00e1lculo da acelera\u00e7\u00e3o do caminh\u00e3o por Torricelli\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.a.DS\u00a0 —\u00a0 0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=(20)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 2.a.100\u00a0 —\u00a0 a=2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

Se a caixa n\u00e3o se mover em rela\u00e7\u00e3o ao caminh\u00e3o eles ter\u00e3o a mesma acelera\u00e7\u00e3o de 2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>e a for\u00e7a resultante\u00a0 na dire\u00e7\u00e3o horizontal vale\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=30.2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=60N<\/b><\/span><\/span><\/p>\n

29-<\/b><\/span><\/span>\u00a0Colocando as for\u00e7as<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

 <\/p>\n

m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 sobe com a=2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.a\u00a0 —\u00a0 T \u2013 50=5.2\u00a0 —\u00a0 T=60N\u00a0 —\u00a0 bloco C se move para a esquerda com acelera\u00e7\u00e3o de 2m\/s<\/b><\/span><\/span>2\u00a0\u00a0<\/b><\/span><\/span><\/sup>—\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>.a\u00a0 —F \u2013 T=m<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>.a\u00a0 —\u00a0 F \u2013 60=10.2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F=80N<\/b><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span>\u00a0Colocando as for\u00e7as e decompondo\u00a0\"\"em\u00a0\"\"\u00a0e em\u00a0\"\"\u00a0:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

sen\u03b1=1,2\/1,5=0,8\u00a0 —\u00a0\u00a0 cos\u03b1=0,9\/1,5=0,6\u00a0 —\u00a0 F<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>=Fcos\u03b1=F.0,6\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>=0,6F<\/b><\/span><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub>=Fsen\u03b1=F.0,8\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub>=0,8F<\/b><\/span><\/span>\u00a0 —\u00a0 P=mg=0,4.10\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P=4N<\/b><\/span><\/span>\u00a0 —\u00a0 estando o sistema em equil\u00edbrio est\u00e1tico, F<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0e N=P + F<\/b><\/span><\/span>Y\u00a0\u00a0<\/b><\/span><\/span><\/sub>—\u00a0 F<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 0,6F=\u03bcN\u00a0 — 0,6F=\u03bc (P + F<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 0,6F=1\/8.(4 + 0,8F)\u00a0 —\u00a0 0,6F=4\/8 + 0,8F\/8\u00a0 —\u00a0 0,6F=0,5 + 0,1F\u00a0 —\u00a0 0,5F=0,5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F=1N\u00a0<\/b><\/span><\/span>(para qualquer valor de F<1N, a vassoura n\u00e3o se deslocar\u00e1).<\/b><\/span><\/span><\/p>\n

31-<\/b><\/span><\/span>\u00a0Como F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=\u03bcN, e como m \u00e9 constante, das alternativas, a \u00fanica for\u00e7a capaz de diminuir a for\u00e7a de com press\u00e3o N da caixa com o ch\u00e3o \u00e9 a C<\/b><\/span><\/span><\/p>\n

32-<\/b><\/span><\/span>\u00a0Colocando as for\u00e7as e decompondo\u00a0\"\"em\u00a0\"\"\u00a0e\u00a0\"\"<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

P=mg=42.10\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P=420N<\/b><\/span><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>=Fcos60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=F.0,5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>=0,5F<\/b><\/span><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub>=Fsen60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=F.\u221a3\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub>=0,85F<\/b><\/span><\/span><\/p>\n

No equil\u00edbrio est\u00e1tico temos, na horizontal F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>\u00a0e na vertical, P=N + F<\/b><\/span><\/span>Y\u00a0\u00a0<\/b><\/span><\/span><\/sub>—\u00a0 N=P – F<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>X\u00a0\u00a0<\/b><\/span><\/span><\/sub>—\u00a0 \u03bcN=0,5F\u00a0 —\u00a0 \u03bc (P – F<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub>)= 0,5F\u00a0 —\u00a0 0,4(420 – 0,85F)=0,5F\u00a0 —\u00a0 168 – 0,34F=0,5F\u00a0 —\u00a0 0,84F=168\u00a0 —\u00a0 F=200N (qualquer valor acima de 200N desloca o bloco) R- C<\/b><\/span><\/span><\/p>\n

33-<\/b><\/span><\/span>Colocando as for\u00e7as apenas na dire\u00e7\u00e3o do movimento (horizontal):<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como o movimento \u00e9 na horizontal, vamos decompor\u00a0\"\"\u00a0apenas em\u00a0\"\"\u00a0\u00a0— \u00a0F<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>=Fcos37<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>=0,8F.<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

Na horizontal\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 F<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>\u00a0\u2013 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 0,8F \u2013 0,6=2,0.0,1\u00a0 —\u00a0 F=0,8\/0,8\u00a0 —\u00a0<\/b><\/span><\/span>F=1,0N\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

34-<\/b><\/span><\/span>\u00a0a) como os corpos est\u00e3o em movimento e como m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>>m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>, o sistema tem que se mover no sentido hor\u00e1rio (A sobe e B desce). Colocando as for\u00e7as:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"– for\u00e7a horizontal que cada um dos mancais exerce sobre o corpo A\u00a0 —\u00a0\u00a0\"\"\u00a0– for\u00e7a de atrito que cada um dos mancais exerce sobre o corpo A tentando impedi-lo de subir e F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=mN\u00a0 —\u00a0 corpo A \u2013 sobe\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.a\u00a0 —\u00a0 T \u2013 P<\/b><\/span><\/span>a\u00ad\u00a0 \u2013<\/b><\/span><\/span><\/sub>\u00a02F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.a\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 T \u2013 m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>g – 2<\/b><\/span><\/span>\u03bc<\/b><\/span><\/span>N=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.a I<\/b><\/span><\/span>\u00a0 —\u00a0 corpo B \u2013 desce\u00a0 —\u00a0 F<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.a\u00a0 — P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0\u2013 T=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>g \u2013 T=mB.a II<\/b><\/span><\/span>\u00a0 —\u00a0 somando membro a membro I com II\u00a0 —<\/b><\/span><\/span><\/p>\n

m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>g –\u00a0 m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>g – 2mN=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.a + m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.a\u00a0 — ( m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0\u2013 m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>)g – 2\u03bcN=(m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>)a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a=(m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0\u2013 m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>)g – 2<\/b><\/span><\/span>\u03bc<\/b><\/span><\/span>N \/( m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span><\/p>\n

b) Se a velocidade \u00e9 constante os corpos A e B est\u00e3o em equil\u00edbrio din\u00e2mico e a for\u00e7a resultante sobre eles \u00e9 nula.<\/b><\/span><\/span><\/p>\n

corpo A\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 T = P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0+ 2F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>g + 2<\/b><\/span><\/span>\u03bc<\/b><\/span><\/span>N I<\/b><\/span><\/span>\u00a0 —\u00a0 corpo B\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 T=P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>g II\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 substituindo II em I\u00a0 —\u00a0 m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>g =m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>g + 2\u03bcN\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>N=(m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0– m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>)g \/2<\/b><\/span><\/span>\u03bc<\/b><\/span><\/span><\/p>\n

35-<\/b><\/span><\/span>\u00a0a) continua em repouso<\/b><\/span><\/span><\/p>\n

b) Sobre o bloco B n\u00e3o existe nenhuma for\u00e7a horizontal<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

36-<\/b><\/span><\/span>\u00a0Colocando as for\u00e7as sobre cada bloco<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>=\u03bc<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.N<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>=\u03bc<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>=0,25.100\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>=25N<\/b><\/span><\/span>\u00a0\u00a0 —\u00a0\u00a0 como o sistema est\u00e1 repouso\u00a0 —\u00a0 bloco A\u00a0 —\u00a0\u00a0 F<\/b><\/span><\/span>RA<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>=25N<\/b><\/span><\/span>\u00a0\u00a0 —\u00a0\u00a0 bloco B\u00a0 —\u00a0 F=F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>\u00a0+ F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 125=F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>\u00a0+ 25\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=100N\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=\u03bc (P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0+ P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 100=\u03bc.300\u00a0 —\u00a0\u00a0\u03bc<\/b><\/span><\/span>=1\/3<\/b><\/span><\/span><\/p>\n

37-<\/b><\/span><\/span>\u00a0Como o coeficiente de atrito \u00e9 o mesmo e o peso de cada bloco \u00e9 tamb\u00e9m o mesmo a for\u00e7a de atrito entre A e B vale F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>=\u03bcN\u00a0 —\u00a0 F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>=\u03bcP=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0e entre B e o ch\u00e3o vale F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=\u03bcN=m(P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0+ P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>)= \u03bc2P\u00a0 —\u00a0 F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=2\u03bcP\u00a0 —\u00a0 F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=2F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>.<\/b><\/span><\/span><\/p>\n

Colocando todas as for\u00e7as, em cada caso:<\/b><\/span><\/span><\/p>\n

\"\"\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\"\"\u00a0\u00a0\u00a0\u00a0\"\"<\/p>\n

Situa\u00e7\u00e3o I\u00a0 —\u00a0 A- nenhuma for\u00e7a e B \u2013\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=2F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 situa\u00e7\u00e3o II\u00a0 —\u00a0 A \u2013 T=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0e B \u2013<\/b><\/span><\/span>F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=3F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 situa\u00e7\u00e3o III\u00a0 —\u00a0 A \u2013 T=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0e<\/b><\/span><\/span><\/p>\n

B \u2013\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sub>=4F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>\/F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>=3F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub>\/2F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>\/F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>=1,5<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sub>\/F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>=4F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub>\/2F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sub>\/F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>=2<\/b><\/span><\/span><\/span><\/p>\n

38-<\/b><\/span><\/span>\u00a0Vamos chamar de L\u2019, comprimento total do pano e de L\u2019 \u2013 L o comprimento da parte pendente. Como o pano \u00e9 de constitui\u00e7\u00e3o homog\u00eanea, sua densidade \u00e9 a mesma.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

d=m<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 d (densidade) \u2013 m<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>\u00a0(massa total) \u2013 V<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>\u00a0(volume total)\u00a0 —\u00a0 m<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=d.V<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 m<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=d.(A.L\u2019), sendo A (\u00e1rea se se\u00e7\u00e3o transversal, espessura do pano) e L\u2019 (comprimento total do pano)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

massa apoiada \u2013 m<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=d.A.L e massa pendente \u2013 m<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=d.A.(L\u2019 \u2013 L)<\/b><\/span><\/span><\/p>\n

\"\"– peso da parte pendente\u00a0 —\u00a0 P=m<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>g (m<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>, massa da parte pendente)\u00a0 —\u00a0 a for\u00e7a de atrito entre o pano de comprimento L e a mesa est\u00e1 segurando o peso da parte pendente de comprimento (L\u2019 \u2013 L), ou seja, F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=P\u00a0 —\u00a0 \u03bcm<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>g= m<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>g\u00a0 —\u00a0 \u03bc.d.A.L.g=d.A.(L\u2019 \u2013 L).g\u00a0 —\u00a0\u00a0 0,5.L=(60 \u2013 L)\u00a0 —\u00a0 1,5L=60\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>L=40cm R- A<\/b><\/span><\/span><\/p>\n

39-<\/b><\/span><\/span>\u00a0Colocando as for\u00e7as<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe na figura que a tend\u00eancia de movimento \u00e9 no sentido hor\u00e1rio e, no bloco de massa 10kg a for\u00e7a de atrito est\u00e1tico \u00e9 para a esquerda (contr\u00e1rio \u00e0 tend\u00eancia ao movimento).<\/b><\/span><\/span><\/p>\n

Todos os blocos est\u00e3o em repouso\u00a0 —\u00a0 bloco de massa 4kg\u00a0 —\u00a0 T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=40N\u00a0 —\u00a0 bloco de massa 6kg\u00a0 —\u00a0 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=60N\u00a0 —bloco de massa 10kg\u00a0 —\u00a0 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 60=40 + F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=20N<\/b><\/span><\/span><\/p>\n

40-<\/b><\/span><\/span>\u00a0Nas rodas com tra\u00e7\u00e3o a for\u00e7a de atrito \u00e9 a favor do movimento e nas rodas sem tra\u00e7\u00e3o, contr\u00e1ria ao movimento. R- C<\/b><\/span><\/span><\/p>\n

41-<\/b><\/span><\/span>\u00a0Nas rodas com tra\u00e7\u00e3o a for\u00e7a de atrito \u00e9 a favor do movimento e nas rodas sem tra\u00e7\u00e3o, contr\u00e1ria ao movimento. R- B<\/b><\/span><\/span><\/p>\n

42-<\/b><\/span><\/span>\u00a0I \u2013 A\u00a0\u00a0\u00a0 II \u2013 C\u00a0\u00a0\u00a0 III \u2013 D\u00a0\u00a0\u00a0\u00a0 IV \u2013 B\u00a0\u00a0 (vide resolu\u00e7\u00e3o do exerc\u00edcio anterior)<\/b><\/span><\/span><\/p>\n

43-<\/b><\/span><\/span>\u00a0\u00a0a) Fat=F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma=3.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.3\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=9.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>N\u00a0\u00a0 b) Para o bloco n\u00e3o escorregar F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 ma=\u03bcmg\u00a0 —\u00a0 a=mg=0,4.10\u00a0 —\u00a0 a=4,0m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

44-<\/b><\/span><\/span>\u00a0a)Colocando as for\u00e7as no im\u00e3-bonequinha:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"\u00a0– for\u00e7a peso, vertical e para baixo aplicada pela Terra\u00a0 —\u00a0\u00a0\"\"– for\u00e7a magn\u00e9tica aplicada pelo metal da parede da geladeira, atraindo o im\u00e3 para ela\u00a0 —\u00a0\u00a0\"\"\u00a0– for\u00e7a normal de compress\u00e3o trocada entre a geladeira e o im\u00e3\u00a0 —\u00a0\u00a0\"\"\u00a0– for\u00e7a de atrito entre a geladeira e o im\u00e3<\/b><\/span><\/span><\/p>\n

b) equil\u00edbrio vertical\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=P\u00a0 —\u00a0 \u03bcN=mg\u00a0 —\u00a0 0,5.N=0,02.10\u00a0 —\u00a0 N=0,2\/0,5\u00a0 —\u00a0 N=0,4N\u00a0 —\u00a0 equil\u00edbrio horizontal\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>mag<\/b><\/span><\/span><\/sub>=N=0,4N<\/b><\/span><\/span><\/p>\n

45- C\u00e1lculo da intensidade for\u00e7a de atrito F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0em cada bloco:<\/b><\/span><\/span><\/p>\n

F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>=\u03bcm<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>g=0,1.4.10\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>=4,0N<\/b><\/span><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=\u03bcm<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>g=0,1.2.10\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=2,0N<\/b><\/span><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span>atC<\/b><\/span><\/span><\/sub>=\u03bcm<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>g=0,1m<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>10\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>atC<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\u00a0—\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>=4,0N<\/b><\/span><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=\u03bcm<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>g=0,1.2.10\u00a0 —\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=2,0N<\/b><\/span><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span>atC<\/b><\/span><\/span><\/sub>=\u03bcm<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>g=0,1m<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>10\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>atC<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/p>\n

a=2,0m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

Colocando as for\u00e7as na dire\u00e7\u00e3o do movimento:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

bloco A\u00a0 —\u00a0\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>a\u00a0 —\u00a0 30 \u2013 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0\u2013 4=4.2\u00a0 — \u00a0<\/b><\/span><\/span>T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=18N\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 bloco B\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>a\u00a0 —\u00a0 18 \u2013 2 – T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=2.2\u00a0 —\u00a0 T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=12N\u00a0 —\u00a0 bloco C\u00a0 —<\/b><\/span><\/span><\/p>\n

F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>a\u00a0 —\u00a0 12= m<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>.2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>m<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=6kg<\/b><\/span><\/span>\u00a0\u00a0\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>RB<\/b><\/span><\/span><\/sub>=18 \u2013 14=4N<\/b><\/span><\/span><\/p>\n

1 \u2013 F; 2 \u2013 V; 3 \u2013 V; 4 \u2013 F; 5 – V<\/b><\/span><\/span><\/p>\n

46-<\/b><\/span><\/span>\u00a0F=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=\u03bcN\u00a0 —\u00a0 360=0,2.N\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>N=1,8.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>N<\/b><\/span><\/span><\/p>\n

47-<\/b><\/span><\/span>\u00a0Sobre o bloco B, na dire\u00e7\u00e3o do movimento agem as for\u00e7as F<\/b><\/span><\/span>at1<\/b><\/span><\/span><\/sub>=\u03bcN\u00a0 —\u00a0 F<\/b><\/span><\/span>at1<\/b><\/span><\/span><\/sub>=\u03bcP<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 F<\/b><\/span><\/span>at1<\/b><\/span><\/span><\/sub>=\u03bcm<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>g\u00a0 —\u00a0 F<\/b><\/span><\/span>at1<\/b><\/span><\/span><\/sub>=0,3.7.10\u00a0 —\u00a0 F<\/b><\/span><\/span>at1<\/b><\/span><\/span><\/sub>=21N, que \u00e9 a a\u00e7\u00e3o de A sobre B e F<\/b><\/span><\/span>at2<\/b><\/span><\/span><\/sub>=\u03bcN\u00a0 —\u00a0 F<\/b><\/span><\/span>at2<\/b><\/span><\/span><\/sub>=m(P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0+ P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 F<\/b><\/span><\/span>at2<\/b><\/span><\/span><\/sub>=0,3.(70 + 70)\u00a0 —\u00a0 F<\/b><\/span><\/span>at2<\/b><\/span><\/span><\/sub>=42N que \u00e9 a rea\u00e7\u00e3o de C sobre B.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como o bloco B se move com velocidade constante, F<\/b><\/span><\/span>RB<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 F=F<\/b><\/span><\/span>at1<\/b><\/span><\/span><\/sub>\u00a0+ F<\/b><\/span><\/span>at2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 F=21 + 42\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F=63N<\/b><\/span><\/span><\/p>\n

48-<\/b><\/span><\/span><\/p>\n

48-<\/b><\/span><\/span>a) Dist\u00e2ncia D, percorrida com velocidade constante V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/p>\n

\"\"<\/p>\n

V=DS\/Dt\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=D\/2<\/b><\/span><\/span><\/p>\n

Dist\u00e2ncia D, percorrida com velocidade variando de V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0para 0, com desacelera\u00e7\u00e3o -a, no tempo t.\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Torricelli\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.a.\u2206S\u00a0 —\u00a0 0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.(-a).D\u00a0 —\u00a0 0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=(D\/2)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 2.a.D\u00a0 —\u00a0<\/b><\/span><\/span>a=D\/8\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ a.t\u00a0 —\u00a0 0=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ (-a).t — 0=D\/2 \u2013 (D\/8).t\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>t=4s<\/b><\/span><\/span><\/p>\n

b) Rela\u00e7\u00e3o entre a e V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 a=D\/8\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=D\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=4a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>Trata-se de uma fun\u00e7\u00e3o do 1<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0grau e o gr\u00e1fico \u00e9 uma reta, que \u00e9 definida por apenas dois pontos, como por exemplo:<\/b><\/span><\/span><\/p>\n

Quando a=0, V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0\u00a0<\/b><\/span><\/span>(0,0)<\/b><\/span><\/span>\u00a0e quando a=10m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>, V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=4.a\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=4.10\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=40m\/s (<\/b><\/span><\/span>40,10).<\/b><\/span><\/span><\/p>\n

Colocando no gr\u00e1fico:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

c) F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 ma=\u03bcmg\u00a0 —\u00a0 a=0,6.10\u00a0 —\u00a0 a=6m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 quando a=6m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>, V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>\u00a0e , pelo gr\u00e1fico\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>\u00bb23,5m\/s<\/b><\/span><\/span><\/p>\n

 <\/p>\n

49-\u00a0<\/b><\/span><\/span>No Bloco A na dire\u00e7\u00e3o horizontal e sentido da for\u00e7a F \u00e9 verdadeiro escrever\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>\u00a0= m.a\u00a0 —\u00a0 F \u2013 F(el\u00e1stica) \u2013 F(atrito) = m.a\u00a0 —\u00a0 F \u2013 k.x – \u03bc.m.g = m.a\u00a0 —\u00a0 60 \u2013 800.x \u2013 0,4.6.10 = 6.a\u00a0 —\u00a0 60 \u2013 800.x \u2013 24 = 6.a\u00a0 —\u00a0 36 \u2013 800.x = 6.a\u00a0 —\u00a0 bloco B nas mesmas condi\u00e7\u00f5es j\u00e1 citadas\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>\u00a0= m.a\u00a0 —\u00a0 F(el\u00e1stica) \u2013 F(atrito) = m.a\u00a0 —\u00a0 k.x – \u03bc.m.g = m.a\u00a0 —\u00a0 800.x \u2013 0,4.4.10 = 4.a\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

800.x \u2013 16 = 4.a\u00a0 —\u00a0 resolvendo, por adi\u00e7\u00e3o, o sistema formado pelas duas equa\u00e7\u00f5es\u00a0 —\u00a0 36 \u2013 800.x = 6.a\u00a0 —\u00a0 800.x \u2013 16 = 4.a\u00a0 —\u00a0 36 \u2013 16 = 10.a\u00a0 —\u00a0 10.a = 20\u00a0 —\u00a0 a = 20\/10= 2 m\/s<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>\u00a0—\u00a0 800.x \u2013 16 = 4.a\u00a0 —\u00a0 \u00a0800.x = 16 + 4.2 = 16 + 8 = 24\u00a0 —\u00a0 \u00a0x =24\/800\u00a0 —\u00a0 x= 0,03 m = 3 cm<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sup>\u00a0—\u00a0 R- A<\/b><\/span><\/span><\/p>\n

50-\u00a0<\/b><\/span><\/span>A m\u00e1xima dist\u00e2ncia (d) do bloco \u00e0 parede ocorre quando a for\u00e7a de atrito m\u00e1xima (F<\/b><\/span><\/span>atmax<\/b><\/span><\/span><\/sub>) tem a mesma intensidade da for\u00e7a<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

el\u00e1stica (F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>). F<\/b><\/span><\/span>atmax<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=\u03bcP=Kx\u00a0 —\u00a0 x=<\/b><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>P\/K=0,4.30\/20\u00a0 —\u00a0 x=0,6m\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

51-\u00a0<\/b><\/span><\/span>As ranhuras tornam o piso mais \u00e1spero, aumentando o coeficiente de atrito.\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

R- A<\/b><\/span><\/span><\/p>\n

52-\u00a0<\/b><\/span><\/span>Se F=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>, a resultante das for\u00e7a \u00e9 nula e ao caixa se move em movimento retil\u00edneo e uniforme\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

53-\u00a0<\/b><\/span><\/span>Dados: m = 120 kg; g = 10 m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>; \u03bc = 0,8; F\/A= 30 N\/cm<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>; N = P = 1.200 N\u00a0 —\u00a0 para colocar a caixa em movimento o operador tem que vencer a for\u00e7a de atrito\u00a0 —\u00a0 F<\/b><\/span><\/span>op<\/b><\/span><\/span><\/sub>\u00a0= F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0= m N\u00a0 —\u00a0 0,8 (1.200) = 960 N\u00a0 —\u00a0 como ele est\u00e1 usando os dois bra\u00e7os, cada bra\u00e7o aplica ent\u00e3o\u00a0 —\u00a0 \u00a0F = F<\/b><\/span><\/span>op<\/b><\/span><\/span><\/sub>\/2=480 N\u00a0 —\u00a0 F\/A=30\u00a0 —\u00a0 480\/A=30\u00a0 —\u00a0 A=16cm<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>\u00a0—\u00a0 uma das prote\u00ednas respons\u00e1veis pela contra\u00e7\u00e3o das miofibrilas \u00e9 a actina\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

54-\u00a0<\/b><\/span><\/span>Como o movimento \u00e9 uniforme, a resultante tangencial \u00e9 nula\u00a0 —\u00a0 assim, a componente de atrito tem a mesma intensidade da for\u00e7a motriz; e como o movimento \u00e9 horizontal, a componente normal tem a mesma intensidade do peso\u00a0 —\u00a0 dados: F<\/b><\/span><\/span>motriz<\/b><\/span><\/span><\/sub>\u00a0= 2.000 N\u00a0 —\u00a0 \u00a0m = 200 kg\u00a0 —\u00a0 \u00a0g = 10 m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F<\/b><\/span><\/span>atc<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>motriz<\/b><\/span><\/span><\/sub>=2.000N\u00a0 —\u00a0 N=P=mg=2.000N\u00a0 — \u00a0F<\/b><\/span><\/span>atc<\/b><\/span><\/span><\/sub>=2.000=\u03bc<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>N\u00a0 —\u00a0 \u03bc<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>.2.000=2.000\u00a0 —<\/b><\/span><\/span><\/p>\n

\u03bc<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=1\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

55-\u00a0<\/b><\/span><\/span>Dados\u00a0 —\u00a0\u00a0 x<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= 2 cm; x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= 10 cm\u00a0 —\u00a0 observe as figuras abaixo onde foram colocadas as for\u00e7as nas duas situa\u00e7\u00f5es\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Na Figura 1, o bloco est\u00e1 na imin\u00eancia de escorregar. A componente de atrito\u00a0\"\"\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>\u00e9 m\u00e1xima e, como o bloco ainda est\u00e1 em repouso, ela tem a mesma intensidade da for\u00e7a el\u00e1stica\u00a0\"\". \u00a0Pela mesma raz\u00e3o, a componente normal\u00a0\"\"\u00a0tem a mesma intensidade que o peso\"\"\u00a0do bloco\u00a0 —\u00a0 sendo k a constante el\u00e1stica da mola, m a massa do bloco e g a intensidade do campo gravitacional, temos\u00a0 —\u00a0 N = P = m g\u00a0 (I)\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0= F\u00a0 —\u00a0 \u03bc N = k x<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 (II)\u00a0 —\u00a0 substituindo (I) em (II)\u00a0 —\u00a0 \u03bc m g = k x<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0\u00a0 (III).<\/b><\/span><\/span><\/p>\n

Na Figura 2, o bloco tamb\u00e9m est\u00e1 em repouso\u00a0 —\u00a0 assim, a nova for\u00e7a el\u00e1stica\u00a0\"\"\u00a0equilibra o peso\u00a0 —\u00a0 F<\/b><\/span><\/span>el<\/b><\/span><\/span><\/sub>\u00a0= P\u00a0 —\u00a0 k x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= m g (IV)\u00a0 —\u00a0 substituindo (IV) em (III)\u00a0 —\u00a0 m k \u00adx<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= k x<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00de\u00a0 m =x<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=2\/10\u00a0 —\u00a0 m = 0,2\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

56-\u00a0<\/b><\/span><\/span>Dados\u00a0 —\u00a0 g = 10 m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u00a0\u03bc<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0= 0,60\u00a0 —\u00a0 \u00a0\u03bc<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0= 0,80\u00a0 —\u00a0 \u00a0m = 1.200 kg\u00a0 —\u00a0 \u00a0for\u00e7a que a pista exerce no ve\u00edculo tem duas componentes\u00a0 —\u00a0 normal\u00a0\"\"\u00a0e \u00a0atrito\u00a0\"\"\u00a0\u00a0—\u00a0 supondo que a frenagem ocorra em pista horizontal, a componente normal\u00a0\"\"\u00a0\u00a0da for\u00e7a que a pista aplica no ve\u00edculo tem intensidade igual \u00e0 do seu peso\u00a0\"\"\u00a0\u00a0—\u00a0 N = P = m g = 12.000 N\u00a0 —\u00a0 \u00a0componente de atrito est\u00e1tico m\u00e1xima\u00a0 —\u00a0 \u00a0F<\/b><\/span><\/span>atmax<\/b><\/span><\/span><\/sub>= \u03bc<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0N = 0,8 (12.000)\u00a0 —\u00a0 \u00a0F<\/b><\/span><\/span>atmax<\/b><\/span><\/span><\/sub>\u00a0= 9.600 N\u00a0 —\u00a0 a componente de atrito cin\u00e9tico: F<\/b><\/span><\/span>atc<\/b><\/span><\/span><\/sub>\u00a0 = \u03bc<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0N = 0,6. (12.000)\u00a0 —\u00a0 \u00a0F<\/b><\/span><\/span>atc<\/b><\/span><\/span><\/sub>\u00a0= 7.200 N\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span>\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0<\/b><\/span><\/span><\/p>\n

57-\u00a0<\/b><\/span><\/span>Na imin\u00eancia de a caixa entrar em movimento, tem-se que as componentes horizontal e vertical da for\u00e7a resultante agindo sobre ela s\u00e3o nulas\u00a0 —\u00a0 assim, para a componente horizontal, Fcos(\u03b8) = F<\/b><\/span><\/span>ate max<\/b><\/span><\/span><\/sub>=\u03bc<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>N\u00a0 —\u00a0 N e F<\/b><\/span><\/span>ate max<\/b><\/span><\/span><\/sub>\u00a0denotam, respectivamente, os m\u00f3dulos da for\u00e7a normal e de atrito est\u00e1tico m\u00e1xima nessa situa\u00e7\u00e3o\u00a0 —\u00a0 componente vertical\u00a0 —\u00a0 \u00a0N + Fsen(\u03b8) = P\u00a0 —\u00a0 combinando essas duas equa\u00e7\u00f5es, obt\u00e9m-se que F = \u03bc<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>P\/[cos(\u03b8) + \u03bc<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0sen(\u03b8)]\u00a0 —<\/b><\/span><\/span>\uf020<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

58-\u00a0<\/b><\/span><\/span>Na figura est\u00e3o colocadas as for\u00e7as que atuam sobre cada bloco\u00a0 —\u00a0 o bloco P est\u00e1 em equil\u00edbrio na vertical (n\u00e3o cai)\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=P\u00a0 —\u00a0 \u03bcN=mg\u00a0 —\u00a0 N=mg\/\u03bc (I)\u00a0 —\u00a0 para que o bloco P n\u00e3o caia, ele deve se mover na horizontal com a mesma acelera\u00e7\u00e3o que a do bloco Q\u00a0 —\u00a0 bloco P\u00a0 —\u00a0 F \u2013 N=ma (II)\u00a0 —\u00a0\u00a0 bloco Q\u00a0 —\u00a0 N=Ma (III)\u00a0 —\u00a0 (II) com (III)\u00a0 —\u00a0 F=(M + m).a (IV)\u00a0 —\u00a0 (I) em (III)\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

Mg\/\u03bc=Ma\u00a0 —\u00a0 a=mg\/\u03bcM (V)\u00a0 —\u00a0 (V) em (IV)\u00a0 —\u00a0 F=(M + m).mg\/\u03bcM\u00a0 —\u00a0 F=mg\/\u03bcM.(M + m)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

59-\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=\u03bcN= \u03bcP= \u03bcmg=0,02.20.10\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=4N\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 4=20a\u00a0 —\u00a0 a=0,2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 Torricelli\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.a.<\/b><\/span><\/span>\u0394<\/b><\/span><\/span><\/span>S\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.(-0,2).45\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=\u221a18\u22484,2m\/s\u00a0 —\u00a0 V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ at\u00a0 —\u00a0 0=4,2 \u2013 0,2t\u00a0 —\u00a0 t\u224821s\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

 <\/p>\n

60-<\/b><\/span><\/span><\/p>\n

01) Correta<\/b><\/span><\/span><\/p>\n

02) Correta<\/b><\/span><\/span><\/p>\n

04) Falsa\u00a0 —\u00a0 constata-se experimentalmente que F<\/b><\/span><\/span>atemax<\/b><\/span><\/span><\/sub>>F<\/b><\/span><\/span>atc<\/b><\/span><\/span><\/sub>\u00a0e que \u03bc<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>>\u03bc<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>, pois as reentr\u00e2ncias e sali\u00eancias das superf\u00edcies em contato est\u00e3o mais encaixadas com o corpo em repouso, dificultando mais a movimenta\u00e7\u00e3o.<\/b><\/span><\/span><\/p>\n

08. Falsa\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=\u03bc.N\u00a0 —\u00a0 \u03bc=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0(for\u00e7a-medida em N)\/N (tamb\u00e9m for\u00e7a-medida\u00a0 em N)\u00a0 —\u00a0 o coeficiente de atrito \u00e9 uma grandeza adimensional, ou seja, n\u00e3o tem unidades.<\/b><\/span><\/span><\/p>\n

16. Falsa\u00a0 —\u00a0 \u00e9 a for\u00e7a de atrito est\u00e1tico m\u00e1xima que \u00e9 sempre maior que a for\u00e7a de atrito cin\u00e9tico.<\/b><\/span><\/span><\/p>\n

Corretas: 01 e 02\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>soma=03<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

61-<\/b><\/span><\/span><\/p>\n

Os freios antitravamento (ABS,\u00a0anti-lock braking system) ajudam a parar melhor\u00a0 —\u00a0 eles previnem o travamento das rodas e proporcionam uma dist\u00e2ncia de frenagem mais curta em superf\u00edcies escorregadias, evitando o\u00a0descontrole do ve\u00edculo\u00a0 —\u00a0 ele mant\u00e9m as rodas sempre na imin\u00eancia de deslizar, aproveitando melhor o atrito est\u00e1tico m\u00e1ximo, que \u00e9 maior que o atrito cin\u00e9tico (de deslizamento)\u00a0 —\u00a0R- B.<\/b><\/span><\/span><\/p>\n

62<\/b><\/span><\/span>–<\/b><\/span><\/span><\/p>\n

C\u00e1lculo da acelera\u00e7\u00e3o de retardamento do carro\u00a0\u00a0\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=54km\/h\/3,6=15m\/s\u00a0 —\u00a0 V=0 (p\u00e1ra)\u00a0 —\u00a0 t=5s\u00a0 —\u00a0 V= V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ a.t\u00a0 —\u00a0 0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

= 15 + a.5\u00a0 —\u00a0 a= – 3m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0(o sinal negativo de a significa que o carro est\u00e1 freando)\u00a0 —\u00a0 em m\u00f3dulo\u00a0 —\u00a0 a=3m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 depois que ele acionou os freios a intensidade da for\u00e7a resultante sobre o carro \u00e9 a pr\u00f3pria for\u00e7a de atrito\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 m.a = \u03bc.N\u00a0 —\u00a0 m.a = \u03bc.m.g\u00a0 —\u00a0 3= \u03bc.10\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\u03bc = 0,3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

63-<\/b><\/span><\/span><\/p>\n

Decompondo a for\u00e7a\u00a0\"\"\u00a0em sua componente horizontal\u00a0\"\"\u00a0e vertical\u00a0\u00a0\"\"\u00a0\u00a0—\u00a0 F<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>=Fcos37<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=50.0,8\u00a0 —\u00a0 F<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>=40N\u00a0 —\u00a0 F<\/b><\/span><\/span>y<\/b><\/span><\/span><\/sub>=Fsen37<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=50.0,6\u00a0 —\u00a0 F<\/b><\/span><\/span>y<\/b><\/span><\/span><\/sub>=30N\u00a0 —\u00a0 colocando as for\u00e7as peso\u00a0\"\"\u00a0(vertical e para baixo), a for\u00e7a normal\u00a0\"\"(vertical e para cima) e<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

a for\u00e7a de atrito\u00a0\"\"\u00a0(contr\u00e1ria ao movimento ou \u00e0 sua tend\u00eancia)\u00a0 —\u00a0 equil\u00edbrio na vertical\u00a0 —\u00a0 P = F<\/b><\/span><\/span>y<\/b><\/span><\/span><\/sub>\u00a0+ F<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 50=30 + F<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 F<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=20N\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=\u03bcF<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=0,6.20\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=12N\u00a0 —\u00a0 na horizontal existe for\u00e7a resultante, pois F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>\u00a0\u2013 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>= 40 \u2013 12\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=28N\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 28=5.a\u00a0 —\u00a0 a=5,6m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

64-<\/b><\/span><\/span><\/p>\n

Colocando todas as for\u00e7as (figura) que influem no movimento ou na sua tend\u00eancia\u00a0 — for\u00e7a de atrito\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=\u03bcN=\u03bcM<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.g\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=\u03bc.10.10\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=100\u03bc\u00a0 —\u00a0 como o sistema est\u00e1 em equil\u00edbrio est\u00e1tico (repouso) a for\u00e7a resultante sobre cada corpo \u00e9 nula\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=T\u00a0 —\u00a0 100\u03bc=T (I)\u00a0 —\u00a0 como o bloco M<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0tamb\u00e9m est\u00e1 em repouso\u00a0 —\u00a0 T=P=M<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.g=5.10\u00a0 —\u00a0 T=50N II)\u00a0 —\u00a0 substituindo (II) em (I)\u00a0 —\u00a0 100\u03bc=50\u00a0 —\u00a0 \u03bc=0,5\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

65-<\/b><\/span><\/span><\/p>\n

a) As for\u00e7as que agem sobre o conjunto im\u00e3-grampo s\u00e3o\u00a0 —\u00a0\u00a0\"\"\u00a0\u2013 for\u00e7a peso, vertical e para baixo, aplicada sobre o conjunto<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

pela Terra\u00a0 —\u00a0\u00a0\"\"\u00a0– for\u00e7a de tra\u00e7\u00e3o no fio, vertical e para baixo, aplicada sobre o conjunto pela massa M pendurada\u00a0 —\u00a0\"\"\u00a0– for\u00e7a de atrito, trocada entre o conjunto\u00a0 e a parede da geladeira, vertical e para cima, contr\u00e1ria ao movimento ou \u00e0 sua tend\u00eancia\u00a0 —\u00a0\u00a0\"\"\u00a0–\u00a0 for\u00e7a magn\u00e9tica, horizontal e que atrai o conjunto para a esquerda, \u00e9 de atra\u00e7\u00e3o magn\u00e9tica entre o im\u00e3 e a parede da geladeira\u00a0 —\u00a0\u00a0\"\"\u00a0\u2013 for\u00e7a normal, horizontal e para a direita, rea\u00e7\u00e3o da parede da geladeira sobre o conjunto.<\/b><\/span><\/span><\/p>\n

b) Se o conjunto n\u00e3o deve cair a for\u00e7a resultante sobre ele tanto na vertical como na horizontal deve ser nula\u00a0 —\u00a0 equil\u00edbrio na vertical\u00a0 —<\/b><\/span><\/span><\/p>\n

F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=P + T\u00a0 —\u00a0 P=m<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>g\u00a0 —\u00a0 T=Mg\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=<\/b><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub>N\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub>N = m<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>g + Mg (I)\u00a0 —\u00a0 equil\u00edbrio na horizontal\u00a0 —\u00a0 F<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>=N (II)\u00a0 —\u00a0 (II0 em (I)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub>F<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>\u00a0= m<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>g + Mg\u00a0 —\u00a0 Mg =\u00a0<\/b><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub>F<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>\u00a0\u2013 m<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>g\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0M= (<\/b><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub>F<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>\u00a0\u2013 m<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>g)\/g ou M=<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub>F<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>\/g – m<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0—\u00a0 para esse valor de M o conjunto est\u00e1 na imin\u00eancia de cair\u00a0 —\u00a0 para qualquer valor de M menor que esse, o conjunto n\u00e3o cai.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

Voltar para os exerc\u00edcios<\/span><\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre For\u00e7a de atrito \u00a0 01– Calculando as intensidades das for\u00e7as de atrito: est\u00e1tico\u00a0 —\u00a0 Fate=\u03bce.N==me.P=\u03bce.mg=0,5.10.10\u00a0 —\u00a0 Fatr=50N \u2013 ele s\u00f3 entra em movimento se F for superior a 50N\u00a0 —\u00a0\u00a0\u00a0\u00a0 din\u00e2mico\u00a0 —\u00a0 Fatd=\u03bcd.N==me.P=\u03bcd.mg=0,3.10.10\u00a0 —\u00a0 Fatd=30N\u00a0\u00a0\u00a0\u00a0 a) Se F=0, Fat=0, ele continua em repouso\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 b) Se F=20N, Fate=20N e ele continua em repouso\u00a0\u00a0\u00a0\u00a0 c)<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1300,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-1304","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1304","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/comments?post=1304"}],"version-history":[{"count":4,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1304\/revisions"}],"predecessor-version":[{"id":10824,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1304\/revisions\/10824"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1300"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1304"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}