{"id":1315,"date":"2015-09-10T03:33:24","date_gmt":"2015-09-10T03:33:24","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1315"},"modified":"2024-08-23T13:39:22","modified_gmt":"2024-08-23T13:39:22","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-plano-inclinado-com-atrito","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/dinamica\/plano-inclinado-com-atrito\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-plano-inclinado-com-atrito\/","title":{"rendered":"Plano inclinado com atrito – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

Plano inclinado com atrito<\/b><\/span><\/span><\/span><\/p>\n

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

01-<\/b><\/span><\/span>O plano inclinado possui uma sec\u00e7\u00e3o transversal que \u00e9 um tri\u00e2ngulo ret\u00e2ngulo de hipotenusa 50 cm e cateto 30 cm. O outro cateto, por Pit\u00e1goras, deve ser de 40 cm.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

sen\u03b1=cateto oposto\/hipotenusa=30\/50\u00a0 —\u00a0 sen\u03b1=0,6\u00a0\u00a0—\u00a0\u00a0\u00a0\u00a0 cos\u03b1=cateto adjacente\/hipotenusa=40\/50\u00a0 —\u00a0 cosa=0,8<\/b><\/span><\/span><\/p>\n

Como o bloco est\u00e1 em equil\u00edbrio a for\u00e7a resultante sobre ele \u00e9 nula e assim, PP=Fat\u00a0 —\u00a0 mgsen\u03b1=\u03bcmgcos\u03b1\u00a0 —\u00a0 0,6=\u03bc.0,8\u00a0 —\u00a0 \u03bc=0,75<\/b><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span>\u00a0Como ele est\u00e1 com velocidade constante a for\u00e7a resultante sobre ele \u00e9 nula e, nesse caso Fat=PP. Seja M a massa total (caminh\u00e3o + combust\u00edvel)\u00a0 —\u00a0\u00a0\u03bcMgcos\u03b1=Mgsen\u03b1\u00a0 —\u00a0\u00a0\u03bc=sena\/cosa=tga\u00a0 —\u00a0 R- D<\/b><\/span><\/span><\/p>\n

03-\u00a0A menor for\u00e7a de atrito para mant\u00ea-lo em equil\u00edbrio vale F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=Psen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=700.0,5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=350N\u00a0 R-A<\/b><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span>\u00a0Esquema:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=Psen80<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=P.0,98 —\u00a0 no equil\u00edbrio \u2013 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=P.0,98\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\/P=0,98\u00a0 —\u00a0 0,98=98\/100\u00a0 —\u00a0 98%\u00a0 R-E<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span>\u00a0Colocando as for\u00e7as na dire\u00e7\u00e3o do movimento.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>sen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=200.0,5\u00a0 — \u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=100N<\/b><\/span><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>=\u03bcP<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>cos30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=0,25.200.0,87\u00a0 —\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>=43,5N<\/b><\/span><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=\u03bcN=mP<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=0,25.400\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=100N<\/b><\/span><\/span><\/p>\n

Na imin\u00eancia de movimento a intensidade da for\u00e7a resultante sobre cada bloco \u00e9 nula\u00a0 —\u00a0 bloco B \u2013 T=F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=100N\u00a0 —\u00a0<\/b><\/span><\/span> bloco A \u2013\u00a0 F + P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>atA<\/b><\/span><\/span><\/sub>\u00a0+ T\u00a0 —\u00a0 F + 100=43,5 + 100\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F=43,5N<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span>\u00a0a) A rea\u00e7\u00e3o do plano inclinado \u00e9 N=P<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=Pcos\u03b8=mgcos\u03b8=1.10.0,8\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=8N<\/b><\/span><\/span><\/p>\n

b) A velocidade inicial \u00e9 nula (partiu do repouso) e a final tamb\u00e9m (parou). Assim,\u00a0<\/b><\/span><\/span>\u2206V=0<\/b><\/span><\/span><\/p>\n

c) No plano inclinado AB\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=mN<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 F<\/b><\/span><\/span>ati<\/b><\/span><\/span><\/sub>=\u03bcP<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=0,2.8\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=1,6N<\/b><\/span><\/span><\/p>\n

No plano horizontal BC\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=\u03bcN<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>=mP=0,2.10\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=2N<\/b><\/span><\/span><\/p>\n

d) No plano horizontal a for\u00e7a resultante sobre ele \u00e9 a for\u00e7a de atrito horizontal que o obriga a parar em C, ou seja, F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/p>\n

\"\"<\/p>\n

F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span><\/sub>=ma=F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 2=1a\u00a0 — \u00a0<\/b><\/span><\/span><\/span>a=2m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/p>\n

e) Ele parte do repouso em A, desce acelerando e atinge a velocidade m\u00e1xima em B (V\u00ad<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span><\/p>\n

Torricelli\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/span><\/sup>=V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/span><\/sup>\u00a0+ 2.a.DS\u00a0 —\u00a0 0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>=V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/span><\/sup>\u00a0+ 2.(-2).4\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/span><\/sub>=4m\/s<\/b><\/span><\/span><\/span><\/p>\n

f) Colocando as for\u00e7as que agem sobre o corpo, na dire\u00e7\u00e3o do movimento, em sua descida pelo plano inclinado<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=1,6N<\/b><\/span><\/span>\u00a0 —\u00a0 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=Psen\u03b8=10.0,6\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=6N<\/b><\/span><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>\u00a0\u2013 F<\/b><\/span><\/span>ati<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 6 \u2013 1,6=1.a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a=4,4m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

Torricelli\u00a0 —\u00a0 V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>=V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.a.\u2206S\u00a0 —\u00a0 4<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.4,4.\u2206S\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u2206S\u00bb1,8m<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span>\u00a0a) Como no plano inclinado n\u00e3o existe atrito a for\u00e7a resultante sobre o corpo \u00e9 o P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>= P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 \u03bcgsen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=ma\u00a0 —\u00a0 10.10.0,5=10.a\u00a0 —\u00a0 a=5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0\u2206S=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.t + a.t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0\u00a0\u2206S=0.2 + 5.2<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0<\/b><\/span><\/span>\u2206S=10m<\/b><\/span><\/span><\/p>\n

b) C\u00e1lculo da velocidade V<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0com que o corpo chega \u00e0 base do plano inclinado\u00a0 —\u00a0 V<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ at\u00a0 —\u00a0 V<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=0 + 5.2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=10m\/s<\/b><\/span><\/span><\/p>\n

No plano horizontal existe atrito e a for\u00e7a resultante \u00e9 o pr\u00f3prio F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span><\/sub>=F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u03bc<\/b><\/span><\/span>N=m.a\u00a0 —\u00a0\u00a0\u03bcP=ma\u00a0 —\u00a0 0,2.100=10.a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>a=2m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/p>\n

Torricelli\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>\u00a0= V<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/span><\/sup>\u00a0+ 2.a.\u2206S\u00a0 —\u00a0 0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>\u00a0= 10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>\u00a0+ 2.(-2).\u2206S\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u2206S=25m<\/b><\/span><\/span><\/span><\/p>\n

08<\/b><\/span><\/span>– For\u00e7as que agem sobre o bloco na dire\u00e7\u00e3o do movimento:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=\u03bcmgcosa=0,5.4.10.0,8\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=16N<\/b><\/span><\/span>\u00a0 —\u00a0 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=mgsena=4.10.0,6\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=24N<\/b><\/span><\/span>\u00a0 —\u00a0 no equil\u00edbrio din\u00e2mico\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0F=P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>\u00a0+ F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 F=24 + 16\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F=40N<\/b><\/span><\/span><\/p>\n

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

09-<\/b><\/span><\/span>\u00a0Como g \u00e9 o mesmo (mesmo focal), se P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=10P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=10m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

Como est\u00e1 com velocidade constante est\u00e1 em equil\u00edbrio din\u00e2mico e a for\u00e7a resultante sobre cada bloco \u00e9 nula\u00a0 —\u00a0 bloco A\u00a0 —\u00a0 T=P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>g\u00a0<\/b><\/span><\/span>\u00a0\u00a0—\u00a0 bloco B\u00a0 —\u00a0 F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>PB<\/b><\/span><\/span><\/sub>\u00a0+ T\u00a0 —\u00a0\u00a0\u03bcm<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>gcos\u03b1\u00a0= m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>gsen\u03b1\u00a0+ m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>g\u00a0 —\u00a0\u03bc10m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>gcos\u03b1\u00a0=10m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>gsen\u03b1\u00a0+ m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>g\u00a0 —\u00a0\u00a0\u03bc.10.0,8=10.0,6 + 1\u00a0 —\u00a0 8\u03bc=1,6\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u03bc=0,875<\/b><\/span><\/span><\/p>\n

R- C<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

sen\u03b1=0,6\/1\u00a0 —\u00a0 sen\u03b1=0,6\u00a0 —\u00a0 cos\u03b1=0,8\/1\u00a0 —\u00a0 cos\u03b1=0,8\u00a0 —\u00a0 como ele est\u00e1 em repouso a for\u00e7a resultante sobre ele \u00e9 nula, ent\u00e3o\u00a0 —\u00a0\u00a0 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 mgsen\u03b1\u00a0=\u00a0\u03bcN\u00a0 —\u00a0 mgsen\u03b1\u00a0=\u03bc(P<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>\u00a0+ F)\u00a0 —\u00a0 mgsen\u03b1\u00a0=\u00a0\u03bc(mgcosa\u00a0+ F)\u00a0 —\u00a0 2.9,8.0,6 = 0,4.(2.9,8.0,8 + F)\u00a0 —\u00a0 11,76 = 6,272 + 0,4F\u00a0 —\u00a0 F=5,488\/0,4\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F=13,72N\u00a0 R- A<\/b><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span>\u00a0P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>\u00a0e F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0s\u00e3o os mesmos tanto na figura 1 como na figura 2<\/b><\/span><\/span><\/p>\n

a) figura 1\u00a0 —\u00a0 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=mgsen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=5.10.0,5\u00a0 —\u00a0 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=25N\u00a0 —\u00a0 comparando V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ a.t com V=2,0.t, conclu\u00edmos que V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0 e a=2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

\"\"<\/p>\n

F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/sub>\u00a0\u2013 F \u2013F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 25 \u2013 F \u2013 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub>=5.2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>F + F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub>=15 I<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Figura 2\u00a0 —\u00a0 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=25N\u00a0 —\u00a0 comparando V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ a\u2019.t com V=3,0.t, conclu\u00edmos que V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0 e a\u2019=3m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span><\/sub>=ma\u2019\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/sub>\u00a0+ F \u2013 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub>=ma\u2019\u00a0 —\u00a0 25 + F \u2013 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub>=5.3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>F=F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub>\u00a0\u2013 10 II<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 II em I\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub>\u2013 10 + F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub>=15\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub>=12,5N<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 F + F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub>=15\u00a0 —\u00a0 F + 12,5 = 15\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>F=2,5N<\/b><\/span><\/span><\/span><\/p>\n

b) F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub>=<\/b><\/span><\/span><\/span>\u03bc<\/b><\/span><\/span>mgcos30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0 12,5=<\/b><\/span><\/span><\/span>\u03bc<\/b><\/span><\/span>.5.10.\u221a3\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u03bc<\/b><\/span><\/span>=12,5\/25\u221a3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u03bc<\/b><\/span><\/span>\u00bb0,3<\/b><\/span><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span>\u00a0a)C\u00e1lculo da acelera\u00e7\u00e3o pelo gr\u00e1fico\u00a0 —\u00a0 a=\u2206V\/\u2206t=2,0\/1,0\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a=2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

b) F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span><\/sub>=ma —\u00a0 P<\/b><\/span><\/span><\/span>P\u00a0<\/b><\/span><\/span><\/span><\/sub>\u2013 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 mgse30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sup>\u00a0\u2013 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub>\u00a0= m.a\u00a0 —\u00a0 0,2.10.0,5 \u2013 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub>=0,2.2\u00a0 —\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub>=0,6N<\/b><\/span><\/span><\/span><\/p>\n

13-\u00a0<\/b><\/span><\/span>\u00a0I- Correta, pois P<\/b><\/span><\/span>PB<\/b><\/span><\/span><\/sub>\u00a0\u00e9 sempre menor que P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.\u00a0\u00a0 \u00a0\u00a0\u00a0II- Falsa, vide I.\u00a0\u00a0\u00a0\u00a0\u00a0 III- N\u00e3o necessariamente, pois F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>\u00a0de dos valores do coeficiente de atrito\u00a0m\u00a0e do \u00e2ngulo de inclina\u00e7\u00e3o do plano.\u00a0\u00a0\u00a0\u00a0\u00a0 IV- Correta, vide I.\u00a0\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

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

a)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

b) Na dire\u00e7\u00e3o do plano de apoio as for\u00e7as que agem sobre a caixa s\u00e3o:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como a caixa est\u00e1 em repouso, a for\u00e7a resultante sobre ela \u00e9 nula\u00a0 —\u00a0\u00a0\"\"<\/b><\/span><\/span><\/p>\n

c) P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 mgsen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=\u03bcmgcos30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 0,5=\u03bc0,87\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u03bc\u00bb0,57<\/b><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span>\u00a0C\u00e1lculo da acelera\u00e7\u00e3o\u00a0 —\u00a0 Torricelli\u00a0 —\u00a0 V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>=V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.a.\u2206S\u00a0 —\u00a0 9=4 +2.a.1\u00a0 —\u00a0<\/b><\/span><\/span>a=2,5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>\u00a0\u2013 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0mgsen60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0–\u00a0\u03bcmgcos60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=m.a\u00a0 —\u00a0 10.\u221a3\/2 –\u00a0m.10.0,5=2,5\u00a0 —\u00a0 5\u221a3 – 5m=2,5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>m=\u221a3 \u2013 1\/2\u00a0\u00a0 R- E<\/b><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span>\u00a0As for\u00e7as que agem sobre a telha s\u00e3o P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>\u00a0e F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Se a hipotenusa \u00e9 10 e um cateto \u00e9 6, aplicando Pit\u00e1goras, o outro cateto \u00e9 8\u00a0 —\u00a0 sen\u03b1=6\/10=0,6\u00a0 —\u00a0 cos\u03b1=8\/10=0,8\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>\u00a0\u2013 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 mgsena\u00a0– \u03bcmgcosa=ma\u00a0 —\u00a0 10.0,6 \u2013 0.2.10.0,8=a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a=4,4m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

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

\"\"<\/p>\n

Como o bloco A est\u00e1 descendo, B est\u00e1 subindo e o F<\/b><\/span><\/span>at\u00a0<\/b><\/span><\/span><\/sub>nele \u00e9 para baixo\u00a0 —\u00a0 bloco A \u2013 n\u00e3o tem atrito\u00a0 \u2013 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 P<\/b><\/span><\/span>PA<\/b><\/span><\/span><\/sub>\u00a0\u2013 T=ma\u00a0 —\u00a0 mgsen60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0– T=ma\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u221a3\/2mg \u2013 T=ma I\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 bloco B \u2013 tem atrito\u00a0 —\u00a0 T \u2013 P<\/b><\/span><\/span>PB<\/b><\/span><\/span><\/sub>\u00a0\u2013 F<\/b><\/span><\/span>atB<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 T \u2013 mgsen60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0–\u00a0\u03bcmgcos60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=ma\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T – mg\u221a3\/2 \u2013 0,2.m.g.0,5=ma II\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>somando membro a membro I com II\u00a0\u00a0 —\u00a0 -0,1m=2ma\u00a0 —\u00a0 a=-0,5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V=Vo<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>+a.t\u00a0 —\u00a0 V=3 \u2013 0,5.2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V=2m\/s<\/b><\/span><\/span><\/p>\n

 <\/p>\n

18-<\/b><\/span><\/span>Colocando as for\u00e7as<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Sobre o homem agem as for\u00e7as\u00a0 —\u00a0 F<\/b><\/span><\/span>ate<\/b><\/span><\/span><\/sub>=\u03bcN\u00a0 —\u00a0 F<\/b><\/span><\/span>ate<\/b><\/span><\/span><\/sub>=\u03bcP<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>ate<\/b><\/span><\/span><\/sub>=\u03bcmgcosq\u00a0<\/b><\/span><\/span>\u00a0– for\u00e7a de atrito est\u00e1tico m\u00e1ximo que ele troca com o plano inclinado e que \u00e9 respons\u00e1vel por ele subir sem escorregar\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>PC<\/b><\/span><\/span><\/sub>=Mgsen\u03b8\u00a0–\u00a0<\/b><\/span><\/span>for\u00e7a com que ele puxa o carrinho de massa M atrav\u00e9s do fio —\u00a0\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>PH<\/b><\/span><\/span><\/sub>=mgsenq<\/b><\/span><\/span>\u03b8\u00a0 —\u00a0 parcela de seu peso de massa m, que ele deve vencer F<\/b><\/span><\/span>ate<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>Pc<\/b><\/span><\/span><\/sub>\u00a0+ P<\/b><\/span><\/span>PH\u00a0<\/b><\/span><\/span><\/sub>\u00a0—\u00a0\u00a0\u03bcmgcos\u03b8=Mgcos\u03b8\u00a0+ mgsen\u03b8\u00a0 —\u00a0 M=m(mcosq\u00a0– senq)\/senq\u00a0 —\u00a0 M=m(mcosq\/senq\u00a0– senq\/senq)\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

M=\u03bc(mcos\u03b8\/sen\u03b8\u00a0– 1) R- E<\/b><\/span><\/span><\/p>\n

19<\/b><\/span><\/span>-C\u00e1lculo da velocidade com que o corpo chega ao ponto Q. No trecho PQ as for\u00e7as que agem sobre o corpo na subida s\u00e3o P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>\u00a0e F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=mgsen\u03b8=m.10.0,8\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=8m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=\u03bcmgcos\u03b8=1\/3.m.10.0,6\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=2m\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>\u00a0+ F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 8m + 2m=ma\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a=10m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 c\u00e1lculo da velocidade com que chega em Q\u00a0 —\u00a0 Torricelli\u00a0 —\u00a0 V<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0= V<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.a.\u2206S\u00a0 —\u00a0 V<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>=25 + 2.(-10).0,8\u00a0 —\u00a0 <\/b><\/span><\/span>V<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>=3m\/s\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 c\u00e1lculo do tempo que demora para chegar em Q\u00a0 —\u00a0 V<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>\u00a0+ a.t\u00a0 —\u00a0 3=5 + (-10).t\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>t=0,2s<\/b><\/span><\/span><\/p>\n

A partir de Q, aonde chega com velocidade de 3m\/s, ocorre um lan\u00e7amento obl\u00edquo com \u00e2ngulo de lan\u00e7amento\u00a0q.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como o exerc\u00edcio quer o instante em que a componente vertical da velocidade se anula, vamos trabalhar apenas com ela (eixo Y).<\/b><\/span><\/span><\/p>\n

Decompondo V<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>\u00a0no eixo vertical Y\u00a0 —\u00a0 V<\/b><\/span><\/span>QY<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>Q<\/b><\/span><\/span><\/sub>sen\u03b8\u00a0\u00a0 —\u00a0 V<\/b><\/span><\/span>QY<\/b><\/span><\/span><\/sub>=3.0,68\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>QY<\/b><\/span><\/span><\/sub>=2,4m\/s \u00a0\u00a0<\/b><\/span><\/span>—\u00a0 quando a componente vertical da velocidade se anula, V<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 V<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ g.t\u00a0 —\u00a0 0=2,4 + (-10).t\u00a0 —\u00a0 t=0,24s\u00a0 —\u00a0 t<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=0,2 + 0,24\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>t<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=0,44s<\/b><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span>a) As for\u00e7as que agem sobre a caixa s\u00e3o:<\/b><\/span><\/span><\/p>\n

 <\/p>\n

 <\/p>\n

\"\"<\/p>\n

figura I\u00a0 —\u00a0\u00a0\"\"– peso, vertical e para baixo exercida pela Terra\u00a0 —\u00a0\u00a0\"\"\u00a0– for\u00e7a de tra\u00e7\u00e3o, exercida pelo fio\u00a0 —\u00a0\u00a0\"\"\u00a0– for\u00e7a exercida pelo plano.<\/b><\/span><\/span><\/p>\n

Figura II \u2013 a for\u00e7a exercida pelo plano\u00a0\"\"\u00a0\u00e9 a soma vetorial da for\u00e7a normal\u00a0\"\"com a for\u00e7a de atrito\u00a0\"\", ou seja,\u00a0\"\"\u00a0=\u00a0\"\"\u00a0+\u00a0\"\".<\/b><\/span><\/span><\/p>\n

 <\/p>\n

b) Decompondo as for\u00e7as nos eixos X e Y<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

T<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>=Tcos30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>=T\u221a3\/2<\/b><\/span><\/span>\u00a0 —\u00a0 T<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub>=Tsen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub>=T\/2<\/b><\/span><\/span>\u00a0 —\u00a0 P<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>=mgsen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>=mg\/2<\/b><\/span><\/span>\u00a0 —\u00a0 P<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub>=mgcos30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>P<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub>=mg\u221a3\/2<\/b><\/span><\/span> <\/b>F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=\u03bcN<\/b><\/span><\/span><\/p>\n

Equil\u00edbrio na dire\u00e7\u00e3o X\u00a0 —\u00a0 T<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>X<\/b><\/span><\/span><\/sub>\u00a0+ F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T\u221a3\/2=mg\/2 +\u00a0\u03bcN\u00a0 I\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>—-\u00a0\u00a0\u00a0 equil\u00edbrio na dire\u00e7\u00e3o Y\u00a0 —\u00a0 N + T<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 <\/b><\/span><\/span>N=mg\u221a3\/2 \u2013 T\/2 II<\/b><\/span><\/span>\u00a0 —\u00a0 II em I\u00a0 —\u00a0 T\u221a3\/2=mg\/2 +\u00a0m(mg\u221a3\/2 \u2013 T\/2)\u00a0 —\u00a0 T(\u221a3 +m)=mg(1 +\u00a0m\u221a3)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T= mg(1 +\u00a0m\u221a3)\/ (\u221a3 +\u00a0m)<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

Substituindo os valores\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T\u00bb14N<\/b><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span>\u00a0(a) Com rodas n\u00e3o existe atrito e a for\u00e7a resultante sobre o carrinho \u00e9 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 mgsena=ma\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a=gsena<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u2206S=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.t + a.t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u2206S=V<\/b><\/span><\/span>0<\/b><\/span><\/span><\/sub>.t + gsen\u03b1.t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2<\/b><\/span><\/span>\u00a0I\u00a0 —\u00a0 sen\u03b1=h\/\u2206S\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u2206S=h\/sen\u03b1\u00a0II\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 II em I\u00a0 —\u00a0 h\/sen\u03b1=gsen\u03b1t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=h\/sen\u03b1X2\/gsena\u00a0 —\u00a0 t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=2h\/g(sen\u03b1)<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>\u00a0\u00a0—\u00a0\u00a0<\/b><\/span><\/span>t=(\u221a2h\/g)\/sen\u03b1<\/b><\/span><\/span><\/p>\n

b) Sem rodas, existe for\u00e7a de atrito de intensidade\u00a0 (F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>). Temo de descida agora \u00e9 t<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=2t\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>t<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=2(\u221a2h\/g)\/sen\u03b1<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u2206S=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.t + a.t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 h\/sen\u03b1=0.t + a.t<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 h\/sen\u03b1=a.4.(2h\/g)\/(sen\u03b1)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 a=2h.(sen\u03b1)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.g\/4sen\u03b1.2h\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a=g.sena\/4 II<\/b><\/span><\/span><\/p>\n

F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>\u00a0\u2013 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 mgsen\u03b1\u00a0–\u00a0\u03bcmgcos\u03b1=m.g(sen\u03b1)\/4\u00a0 —\u00a0 sen\u03b1\u00a0–\u00a0mcos\u03b1=(sen\u03b1)\/4\u00a0 —\u00a0\u00a0mcos\u03b1=sen\u03b1\u00a0– (sen\u03b1)\/4\u00a0 —\u00a0mcos\u03b1=(3sen\u03b1)\/4\u00a0 —\u00a0\u00a0m=3sen\u03b1\/4cos\u03b1\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>m=(3tg\u03b1)\/4<\/b><\/span><\/span><\/p>\n

 <\/p>\n

22-<\/b><\/span><\/span>\u00a0A for\u00e7a de atrito que atua sobre o corpo \u00e9 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0= \u03bc.m.g.cos\u03bc = 8m.\u03bc\u00a0 —\u00a0 \u00a0for\u00e7a resultante na dire\u00e7\u00e3o do movimento\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>= m.g.sen \u03b2 – 8m.\u03bc = m.a \u00a0—\u00a0 6m – 8m.\u03bc = m.a\u00a0 —\u00a0 \u00a06 – 8\u03bc. = a\u00a0 —\u00a0 a fun\u00e7\u00e3o hor\u00e1ria do movimento \u00e9 S = S0 + v0.t + a.t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 S<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0= 0 e v<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0= 0\u00a0 —\u00a0 S = a.t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 para t = 1 s \u00a0—\u00a0 \u00a0S = 1 m \u00a0—\u00a0 \u00a01 = a.1<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 \u00a0—\u00a0 a = 2 m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 ent\u00e3o: 6 – 8m. = a \u00a0—\u00a0 6 – 8m. = 2 \u00a0— 8m. = 6 \u2013 2 \u00a0—\u00a0 \u00a08m. = 4 \u00a0—\u00a0\u00a0<\/b><\/span><\/span>\u00a0m = 0,5<\/b><\/span><\/span>\u00a0 —\u00a0 para t = 1 + 3 = 4 s \u00a0—\u00a0 S = x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u00a0—\u00a0 \u00a0x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= 2. 4<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=16m<\/b><\/span><\/span><\/p>\n

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

23-\u00a0<\/b><\/span><\/span>Observe ao figura abaixo onde est\u00e3o colocadas todas as for\u00e7as que agem sobre o bloco\u00a0 —\u00a0 \u00a0P<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>\u00a0 = P sen 45\u00b0 = m g sen 45\u00b0\u00a0 —<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

–\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0\u00a0 N = P<\/b><\/span><\/span>n<\/b><\/span><\/span><\/sub>\u00a0= P cos 45\u00b0 = m g cos 45\u00b0\u00a0 —\u00a0 g\u00a0 = 10 m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 a = 5 m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u00a0\u03b8 = 45\u00b0\u00a0 —\u00a0 \u00a0princ\u00edpio fundamental da din\u00e2mica\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 P<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>\u00a0 \u2013 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0= m a\u00a0 —\u00a0 mgsen45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=\u03bcmgcos45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=ma \u00a0—\u00a0 10\u221a2\/2 \u2013 \u03bc.10.\u221a2\/2=5\u00a0 —\u00a0 m = (5\u221a2 \u2013 5)\/5\u221a2= (\u221a2 \u2013 1)\/\u221a2=(1,4 \u2013 1)\/1,4\u00a0 —\u00a0 \u03bc\u22480,3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

24-\u00a0<\/b><\/span><\/span>O corpo est\u00e1 em equil\u00edbrio din\u00e2mico (F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0)\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=Psen\u03b8\u00a0 —\u00a0 sen\u03b8=FC\/FB=6\/10=0,6\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=mgsen\u03b8=20.10.0,6\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=120N\u00a0 —<\/b><\/span><\/span>\u00a0 R- C<\/b><\/span><\/span><\/p>\n

25-\u00a0<\/b><\/span><\/span>Na imin\u00eancia de escorregar, ele est\u00e1 em equil\u00edbrio e, nessas condi\u00e7\u00f5es a parcela do peso (P<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=Psen\u03b2) que tenta desloca-lo \u00e9 equilibrada pela for\u00e7a de atrito (F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 mgsen\u03b2=\u03bcmgcos\u03b2\u00a0 —\u00a0 \u03bc=sen\u03b2\/cos\u03b2\u00a0 —\u00a0 \u03bc=tg\u03b2=0,40\u00a0 —\u00a0 tg\u03b2=cat. oposto\/cat. adjacente\u00a0 —\u00a0 0,40=10\/x\u00a0 —\u00a0 x=10\/0,4\u00a0 —\u00a0 x=25cm\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

26-\u00a0<\/b><\/span><\/span>Colocando as for\u00e7as que agem sobre o bloco\u00a0 —\u00a0 decompondo-as na vertical e na horizontal\u00a0 —\u00a0 na vertical a for\u00e7a<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

resultante \u00e9 nula\u00a0 — P=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>sen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0+ Ncos30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 20=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\/2 + N\u221a3\/2 (I)\u00a0 —\u00a0 na horizontal existe uma for\u00e7a resultante para a direita onde o bloco se move com acelera\u00e7\u00e3o a=2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0e de intensidade\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=2.2\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=4N\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=N\/2 \u2013 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>.\u221a3\/2\u00a0 —<\/b><\/span><\/span><\/p>\n

4=N\/2 \u2013 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>.\u221a3\/2 (II)\u00a0 —\u00a0 resolvendo o sistema composto por I e II e substituindo \u221a3 por 1,7, voc\u00ea obter\u00e1\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=6,6N<\/b><\/span><\/span>\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

27-\u00a0<\/b><\/span><\/span>Na figura est\u00e3o colocadas as for\u00e7as que agem sobre cada bloco\u00a0 —\u00a0 c\u00e1lculo da acelera\u00e7\u00e3o do sistema considerando os dois corpos como um \u00fanico de massa (m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)=4kg\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

28-\u00a0<\/b><\/span><\/span>Trecho CD\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=\u03bcN=\u03bcP=\u03bcmg\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=10<\/b><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>m\u00a0 —\u00a0 a intensidade da for\u00e7a resultante \u00e9 a pr\u00f3pria for\u00e7a de atrito\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 10\u03bcm=ma\u00a0 —\u00a0 a=10\u03bc\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.\u0394S\u00a0 —\u00a0 0=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.(-a).20\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>=2.10\u03bc.20\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>=400\u03bc\u00a0 —\u00a0 \u00a0trecho AC, sem atrito, desce com acelera\u00e7\u00e3o a\u2019, sujeito \u00e0 P<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=mgsen\u03b8=m.10.12\/AC=120m\/AC\u00a0 —\u00a0 P<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u2019\u00a0 —\u00a0 120m\/AC=ma\u2019\u00a0 —\u00a0 a\u2019=120\/AC\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.\u0394S\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>=0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.a\u2019.AC\u00a0 —\u00a0 400\u03bc=2.120\/AC.AC\u00a0 —\u00a0 \u03bc=240\/400\u00a0 —\u00a0 \u03bc=0,6\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

As for\u00e7as que agem sobre o bloco na dire\u00e7\u00e3o do movimento s\u00e3o\u00a0 —\u00a0 P<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0(parcela do peso respons\u00e1vel pela descida<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

do bloco)\u00a0 —\u00a0 P<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=Psen45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=mgsen45=m.10.\u221a2\/2=m.10.1,4\/2=7m\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0(for\u00e7a de atrito, sempre contr\u00e1ria ao movimento)\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=\u03bc.P.sen45<\/b><\/span><\/span>0<\/b><\/span><\/span><\/sup>=\u03bc.m.g.\u221a2\/2=\u03bc.m.10.1,4\/2=7\u03bcm\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 P<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0\u2013 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 7m – 7\u03bcm=m.2\u00a0 —\u00a0 \u03bc=5\/7\u22480,714\u00a0 —<\/b><\/span><\/span>R- A<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

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

Durante a descida a intensidade da for\u00e7a resultante que age sobre o bloco \u00e9 igual \u00e0 intensidade da for\u00e7a de atrito<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>\u00a0—\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=15m\/s\u00a0 —\u00a0 \u2206S=L=5m\u00a0 —\u00a0 para (V=0)\u00a0 —\u00a0 c\u00e1lculo do m\u00f3dulo da acelera\u00e7\u00e3o a pela equa\u00e7\u00e3o de 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>\u00a0= (15)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 2.a.5\u00a0 —\u00a0 a=22510\u00a0 —\u00a0 a=22,5ms<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 lei fundamental da din\u00e2mica\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a=4.22,5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=90N<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

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

Trata-se de um plano inclinado onde o peso do corpo, vertical e para baixo, \u00e9 decomposto em duas parcelas\u00a0 —\u00a0\u00a0\"\"\u00a0– parcela do peso, no plano inclinado, paralela ao plano e para baixo, respons\u00e1vel pela descida ou tentativa de descida do bloco, de intensidade\u00a0 —<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

de intensidade\u00a0 —\u00a0 P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=P.sen\u03b1\u00a0ou\u00a0P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=mgsen \u03b1—\u00a0\u00a0 a energia \u00e9 dissipada pela for\u00e7a de atrito\u00a0\"\"\u00a0 que \u00e9 sempre contr\u00e1ria ao movimento ou \u00e0 sua tend\u00eancia, de intensidade\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=\u03bcPcos45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0ou\u00a0\u00a0F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=\u03bcmgcos45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=0,5.20.10.\u221a2\/2\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=50\u221a2N\u00a0 —\u00a0 a for\u00e7a resultante, respons\u00e1vel pela descida do menino tem intensidade\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0\u2013 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=100\u221a2 – 50\u221a2=50\u221a2N\u00a0 —\u00a0 dist\u00e2ncia d percorrida pelo menino ao se deslocar de A para B\u00a0 —\u00a0 sen45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=2,8\/d\u00a0 —\u00a0 \u221a2\/2=2,8\/d\u00a0 —\u00a0 d=2,8\u221a2m\u00a0 —\u00a0 a energia dissipada na descida de A para B corresponde ao trabalho da for\u00e7a de atrito nesse trecho\u00a0 —\u00a0 W<\/b><\/span><\/span>Fat<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>.d.cos180<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0(180<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0\u00e9 o \u00e2ngulo entre o F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0e o deslocamento)\u00a0 —\u00a0 W<\/b><\/span><\/span>Fat<\/b><\/span><\/span><\/sub>=50.\u221a2.2.8.\u221a2.(-1)\u00a0 —\u00a0 W<\/b><\/span><\/span>Fat<\/b><\/span><\/span><\/sub>= – 280J (o sinal negativo significa que essa energia foi dissipada)\u00a0 —\u00a0 ingest\u00e3o de um sorvete\u00a0 —\u00a0 W<\/b><\/span><\/span>sorvete<\/b><\/span><\/span><\/sub>=112.00J \u00a0\u00a0—\u00a0 n\u00famero de vezes que a crian\u00e7a dever\u00e1 escorregar\u00a0 —\u00a0 n=112.000\/280=400 vezes\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

 <\/p>\n

Voltar para os Exerc\u00edcios<\/a><\/span><\/h3>\n

\u00a0<\/span><\/p>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Plano inclinado com atrito \u00a0 01-O plano inclinado possui uma sec\u00e7\u00e3o transversal que \u00e9 um tri\u00e2ngulo ret\u00e2ngulo de hipotenusa 50 cm e cateto 30 cm. O outro cateto, por Pit\u00e1goras, deve ser de 40 cm. sen\u03b1=cateto oposto\/hipotenusa=30\/50\u00a0 —\u00a0 sen\u03b1=0,6\u00a0\u00a0—\u00a0\u00a0\u00a0\u00a0 cos\u03b1=cateto adjacente\/hipotenusa=40\/50\u00a0 —\u00a0 cosa=0,8 Como o bloco est\u00e1 em equil\u00edbrio a for\u00e7a resultante sobre<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1311,"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-1315","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1315","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=1315"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1315\/revisions"}],"predecessor-version":[{"id":10849,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1315\/revisions\/10849"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1311"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1315"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}