{"id":1514,"date":"2015-11-16T15:08:32","date_gmt":"2015-11-16T15:08:32","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1514"},"modified":"2024-08-23T14:16:06","modified_gmt":"2024-08-23T14:16:06","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-estatica-de-um-corpo-extenso-maquinas-simples","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/estatica\/estatica-de-um-corpo-extenso-maquinas-simples\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-estatica-de-um-corpo-extenso-maquinas-simples\/","title":{"rendered":"Est\u00e1tica de um corpo extenso \u2013 M\u00e1quinas Simples – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

Est\u00e1tica de um corpo extenso \u2013 M\u00e1quinas Simples<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

01-<\/b><\/span><\/span>\u00a0A \u00fanica for\u00e7a que n\u00e3o precisa ser decomposta no sentido de girar a barra no sentido hor\u00e1rio \u00e9 F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A\u00a0<\/b><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span>\u00a0O port\u00e3o tende a girar no sentido hor\u00e1rio\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span>\u00a0F<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=F\u00a0 —\u00a0 M<\/b><\/span><\/span>FA<\/b><\/span><\/span><\/sub>=+F.a\u00a0 —\u00a0 M<\/b><\/span><\/span>FB<\/b><\/span><\/span><\/sub>=+F.a\u00a0 —\u00a0 M<\/b><\/span><\/span>FC<\/b><\/span><\/span><\/sub>=0\u00a0 — M<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=+2Fa\u00a0 —\u00a0 gira no sentido hor\u00e1rio\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span>\u00a0a) O momento resultante \u00e9 a soma alg\u00e9brica do momento de cada for\u00e7a em rela\u00e7\u00e3o ao ponto O\u00a0 —\u00a0 M<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=+F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.d<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=+1,0.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>.100\u00a0 —\u00a0 M<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=10,0 .10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N.m\u00a0 —\u00a0 M<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=-F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=-2,0.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>.80\u00a0 —\u00a0 M<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=-16,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N.m\u00a0 —\u00a0 M<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=10,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0\u2013 16,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 M<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=-6,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N.m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>M<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=6,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N.m e tende a girar no sentido anti-hor\u00e1rio<\/b><\/span><\/span><\/p>\n

b) F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=3.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>N\u00a0 —\u00a0 \u0394t=1min=60s\u00a0 —\u00a0 I=F.\u0394t=3.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>.60\u00a0 —<\/b><\/span><\/span>\u00a0I=1,8.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>N.s<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span>\u00a0Para desatarraxar\u00a0 —\u00a0 M<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=F.d=400.15\u00a0 —\u00a0 M<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=6.000N.cm\u00a0 —\u00a0 com a extens\u00e3o\u00a0 —\u00a0 M<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=M<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 6.000=F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.75\u00a0 —\u00a0 F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=6.000\/75\u00a0 —\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=80N<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span>\u00a0Jovem\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

M<\/b><\/span><\/span>j<\/b><\/span><\/span><\/sub>=F.d=750.20\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>M<\/b><\/span><\/span>j<\/b><\/span><\/span><\/sub>=15.000N.cm<\/b><\/span><\/span>\u00a0 —\u00a0 namorada\u00a0 —\u00a0 M<\/b><\/span><\/span>n<\/b><\/span><\/span><\/sub>=510.30\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>M<\/b><\/span><\/span>n<\/b><\/span><\/span><\/sub>=15.300N.cm<\/b><\/span><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>sim, consegue<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span>\u00a0M<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>=+F<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>.d=80.10\u00a0 —\u00a0 M<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>=800N.cm\u00a0 —\u00a0 M<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=-F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>.d=-5.200\u00a0 —\u00a0 M<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=1.000N.cm\u00a0 —\u00a0 ganha o menino\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span>\u00a0As for\u00e7as que agem sobre o trampolim est\u00e3o indicadas abaixo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

09-<\/b><\/span><\/span>\u00a0Para abrir a porta\u00a0 —\u00a0 M<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=F.d=20.0,4\u00a0 —\u00a0 M<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=8N.m\u00a0 —\u00a0 esse momento \u00e9 o que o menino deve aplicar para abri a porta com for\u00e7a de 10N a uma dist\u00e2ncia d\u00a0 —\u00a0 M=F.d\u00a0 —\u00a0 8=10.d\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0d=0,8m<\/b><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span>\u00a0Fd=F\u2019d\/2\u00a0 —\u00a0 F\u2019=2F\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span>\u00a0Observe que os momentos das for\u00e7as F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0e F<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0s\u00e3o nulos, pois a linha de a\u00e7\u00e3o dessas for\u00e7as coincide com o eixo de<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0rota\u00e7\u00e3o (p\u00f3lo O), n\u00e3o fazendo, portanto a chapa girar. Quem a tende a girar \u00e9 o momento de F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0tal que M<\/b><\/span><\/span>F1<\/b><\/span><\/span><\/sub>=F.d e que deve ter o mesmo m\u00f3dulo que o momento de F<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sub>\u00a0(M<\/b><\/span><\/span>F4<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sub>.d)\u00a0 —\u00a0 M<\/b><\/span><\/span>F1<\/b><\/span><\/span><\/sub>=M<\/b><\/span><\/span>F4<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Fd=F<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sub>d\u00a0 —\u00a0 F<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sub>=F\u00a0 —\u00a0 a for\u00e7a F, aplicada em D deve impedir a tend\u00eancia de rota\u00e7\u00e3o provocada por F=1 e, assim, F em D deve ser vertical e para cima\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span>\u00a0(1) \u2013 Verdadeira\u00a0 —\u00a0 n\u00e3o existe atrito entre a escada e o ch\u00e3o, ent\u00e3o ela escorrega n\u00e3o podendo ficar em equil\u00edbrio.<\/b><\/span><\/span><\/p>\n

(2) –\u00a0 Falsa\u00a0 —\u00a0 decompondo as for\u00e7as conforme figura abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

(3) \u2013 Falsa\u00a0 —\u00a0 depende da dire\u00e7\u00e3o e sentido de cada for\u00e7a<\/b><\/span><\/span><\/p>\n

(4) \u2013 Verdadeira\u00a0 —\u00a0 quanto menor o \u00e2ngulo com o teto (\"\"), maior ser\u00e1 a for\u00e7a de tra\u00e7\u00e3o nele.\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

R- V F F V<\/b><\/span><\/span><\/p>\n

13- R- C<\/b><\/span><\/span>\u00a0 —\u00a0 veja figura abaixo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

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

\"\"<\/p>\n

b) Sim, desde que ela n\u00e3o escorregue em C, onde deve ter atrito para manter o equil\u00edbrio\u00a0 —\u00a0 a for\u00e7a normal\u00a0\"\"\u00a0tem a mesma intensidade que o peso\u00a0\"\"\u00a0\u00a0—\u00a0\u00a0 N=P=mg=40.10\u00a0 —\u00a0 N=400N\u00a0 —\u00a0 tg\u03b2=4\/3\u00a0 —\u00a0 tg\u03b2=N\/Fat\u00a0 —\u00a0 4\/3=400\/Fat\u00a0 —\u00a0 Fat=1.200\/4\u00a0 —\u00a0 Fat=300N<\/b><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span>\u00a0a)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

b) Decompondo a for\u00e7a\u00a0\"\"\u00a0que o ch\u00e3o troca com a barra<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

Equil\u00edbrio na horizontal\u00a0 —\u00a0 N<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 equil\u00edbrio na vertical\u00a0 —\u00a0 N<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=P\u00a0 —\u00a0 colocando o p\u00f3lo em Q\u00a0 —\u00a0 a soma alg\u00e9brica dos momento s\u00e9 igual a zero\u00a0 —\u00a0 M<\/b><\/span><\/span>N1<\/b><\/span><\/span><\/sub>=N<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.PQsen\u03b1\u00a0 —\u00a0 M<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=P.(PQ)\/2.cos\u03b1\u00a0 —\u00a0 M<\/b><\/span><\/span>Fat<\/b><\/span><\/span><\/sub>=M<\/b><\/span><\/span>N2<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 +F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>.PQ.sen\u03b1 \u2013 P.(PQ)\/2.cos\u03b1=0\u00a0 —\u00a0 2.F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>.sen\u03b1=P.cos\u03b1\u00a0 —\u00a0 tg\u03b1=P\/2F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=\u03bcP\u00a0 —\u00a0 tg\u03b1=P\/2\u03bcP\u00a0 —\u00a0 tg\u03b1=1\/2.0,25\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0tg\u03b1=2<\/b><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span>\u00a0Colocando as for\u00e7as que atuam na r\u00e9gua e colocando o p\u00f3lo em P\u00a0 —\u00a0 P<\/b><\/span><\/span>corpo<\/b><\/span><\/span><\/sub>=mg=9.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.10\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

M<\/b><\/span><\/span>Pr\u00e9gua<\/b><\/span><\/span><\/sub>=- P<\/b><\/span><\/span>r\u00e9gua<\/b><\/span><\/span><\/sub>.d= – P<\/b><\/span><\/span>r\u00e9gua<\/b><\/span><\/span><\/sub>.0,05\u00a0 —\u00a0 M<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 M<\/b><\/span><\/span>Pcorpo<\/b><\/span><\/span><\/sub>= + P<\/b><\/span><\/span>corpo<\/b><\/span><\/span><\/sub>.d=+9,0.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>.0,45=4,05.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>N.m\u00a0 —\u00a0 a soma dos momentos deve ser nula\u00a0 —\u00a0 – 0,05.P<\/b><\/span><\/span>r\u00e9gua<\/b><\/span><\/span><\/sub>\u00a0+ 4,05.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>=0\u00a0 —\u00a0 P<\/b><\/span><\/span>r\u00e9gua<\/b><\/span><\/span><\/sub>=4,05.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\/5.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>=0,81N\u00a0 —\u00a0 m<\/b><\/span><\/span>r\u00e9gua<\/b><\/span><\/span><\/sub>=0,81\/10=0,081kg=81g\u00a0 —\u00a0<\/b><\/span><\/span>m<\/b><\/span><\/span>r\u00e9gua<\/b><\/span><\/span><\/sub>=81g<\/b><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span>\u00a0Para que a prancha esteja na imin\u00eancia de tombar, a for\u00e7a de rea\u00e7\u00e3o normal N<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>, no primeiro apoio deve ser nula (N<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=0)\u00a0 — \u00a0\u00a0o cilindro deve estar a uma dist\u00e2ncia x do segundo apoio\u00a0 —\u00a0 a soma dos momentos de cada for\u00e7a deve ser nula\u00a0 —<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

–\u00a0 colocando o p\u00f3lo O no ponto de aplica\u00e7\u00e3o de N<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0(segundo apoio) —\u00a0 0.2d \u2013 2mg.d + N<\/b><\/span><\/span>2.<\/b><\/span><\/span><\/sub>0 + mg.x=0\u00a0 —\u00a0 2mgd=mgx\u00a0 —\u00a0 x=2d\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span>\u00a0a) Observe a figura abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Colocando o p\u00f3lo em N<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 a soma dos momentos em rela\u00e7\u00e3o ao p\u00f3lo deve ser nula\u00a0 —\u00a0 M<\/b><\/span><\/span>N2<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 M<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=+P.4\u00a0 —\u00a0 M<\/b><\/span><\/span>N1<\/b><\/span><\/span><\/sub>=-N<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.6\u00a0 — M<\/b><\/span><\/span>N1<\/b><\/span><\/span><\/sub>=-6.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N.m\u00a0 —\u00a0 +4P \u2013 6.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>=0\u00a0 —\u00a0 P=6.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\/4\u00a0 —\u00a0 P=1,5.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N<\/b><\/span><\/span><\/p>\n

b) N<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ N<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=1,5.105\u00a0 —\u00a0 1,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0+ N<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=1,5.1<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>5\u00a0 —\u00a0 N<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=0,5.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0N<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=5,0.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>N<\/b><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span>\u00a0I \u2013 Correta\u00a0 —\u00a0 veja teoria\u00a0 —\u00a0 II \u2013 Correta\u00a0 —\u00a0 dentro da \u00e1gua o corpo E fica mais leve devido ao empuxo, vertical e para cima e fora da \u00e1gua ele tender\u00e1 a descer, fazendo a barra girar em torno de C no sentido hor\u00e1rio\u00a0 —\u00a0 III \u2013 Falsa\u00a0 —\u00a0 veja II\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span>\u00a0a) colocando o p\u00f3lo em C\u00a0 —\u00a0 -F.d + P.a=0\u00a0 —\u00a0 -F.0,04 + 20.0,3=0\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0F=150N<\/b><\/span><\/span><\/p>\n

b) colocando o p\u00f3lo em F\u00a0 — -C.d + P.(a \u2013 d)=0\u00a0 —\u00a0 -C.0.04 + 20.(0,3 \u2013 0,04)=0\u00a0 —\u00a0<\/b><\/span><\/span>C=130N<\/b><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span>\u00a0Colocando o p\u00f3lo em F<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 600.30 \u2013 F<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>.120=0\u00a0 —\u00a0 F<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=150N\u00a0 —\u00a0 colocando o p\u00f3lo em F<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 -600.90 + F<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>.120=0\u00a0 —\u00a0 F<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>=150N\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span>\u00a0A soma dos momentos de todas as for\u00e7as em rela\u00e7\u00e3o ao p\u00f3lo 0 deve ser nula\u00a0 —\u00a0 -F.4 +23.20 + 100.35=0\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F=990N<\/b><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span>\u00a0A dist\u00e2ncia m\u00e1xima (x) que o homem pode se deslocar sobre o trilho a partir de P, ocorre quando o trilho estiver na<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

imin\u00eancia de girar e, nessas condi\u00e7\u00f5es N<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 com o p\u00f3lo em P\u00a0 —\u00a0 + (3.500).1 \u2013 1000.(4 \u2013 x)=0\u00a0 —\u00a0 x=0,5m\u00a0 —\u00a0 d=4 \u2013 x\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0d=3,5m<\/b><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span>\u00a0Observe o comprimento das hastes em rela\u00e7\u00e3o ao ponto de apoio e verifique que o lado de comprimento 5cm deve equilibrar o dobro da massa que o lado de comprimento 10cm (figura abaixo)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

25-<\/b><\/span><\/span>\u00a0Colocando o p\u00f3lo em N<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/p>\n

\"\"<\/p>\n

+700.1,2 \u2013 2N2=0\u00a0 —\u00a0 N<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=420N\u00a0 —<\/b><\/span><\/span>\u00a0 R- C<\/b><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span>\u00a0P=90.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10\u00a0 —\u00a0 P=9.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\u00a0 —\u00a0 polo em D\u00a0 —\u00a0 -9.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>.16 + 2N<\/b><\/span><\/span>T<\/b><\/span><\/span><\/sub>.20=0\u00a0 —\u00a0 N<\/b><\/span><\/span>T<\/b><\/span><\/span><\/sub>=36.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\u00a0 —\u00a0 M<\/b><\/span><\/span>T<\/b><\/span><\/span><\/sub>=36<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

toneladas\u00a0 —\u00a090=36 + M<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 M<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>=18 toneladas\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span>\u00a0Primeira situa\u00e7\u00e3o\u00a0 —\u00a0 colocando o p\u00f3lo em M\u00a0 —\u00a0 a soma dos momentos de cada for\u00e7a em rela\u00e7\u00e3o ao p\u00f3lo deve ser<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

nula\u00a0 —\u00a0 – 100.50 + 50P<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\u00a0+ 60(150 \u2013 x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)=0\u00a0 —\u00a0 -5.000 + 50P<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\u00a0\u2013 9.000 \u2013 60x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>5P<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\u2013 6x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>= – 400 (I)<\/b><\/span><\/span><\/p>\n

Segunda situa\u00e7\u00e3o\u00a0 —\u00a0\u00a0 colocando o p\u00f3lo em M\u00a0 —\u00a0 a soma dos momentos de cada for\u00e7a em rela\u00e7\u00e3o ao p\u00f3lo deve ser<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

nula\u00a0 —\u00a0 -60.50 + 50P<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\u00a0+ 100(150 \u2013 1,6x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)=0\u00a0 —\u00a0 -3.000 + 50P<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\u00a0+ 15.000 \u2013 160x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0<\/b><\/span><\/span>5P<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\u00a0\u2013 16x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>= -1.200 (II)<\/b><\/span><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

resolvendo o sistema composto por I e II\u00a0 —\u00a0 x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=80cm\u00a0 —\u00a0 P=16N\u00a0 —\u00a0 m=16\/10\u00a0 —<\/b><\/span><\/span>\u00a0m=1,6kg<\/b><\/span><\/span><\/p>\n

28-<\/b><\/span><\/span>\u00a0P\u00f3lo em N<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

-200.2 + F.0,5 + 350.1=0\u00a0 —\u00a0 -400 + 0,5F + 350=0\u00a0 —\u00a0 F=100N\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

29-<\/b><\/span><\/span>\u00a0p\u00f3lo no apoio\u00a0 —\u00a0 -25.140 \u2013 30.100 \u2013 50.60 + 40.60 + m.100 + 30.140=0\u00a0 —\u00a0 m=2.900\/100 \u00a0—\u00a0 m=29kg\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span>\u00a0colocando as for\u00e7as e o p\u00f3lo na posi\u00e7\u00e3o indicada e lembrando que no equil\u00edbrio de transla\u00e7\u00e3o\u00a0 —\u00a0 T=P=3.000N<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

+T.10 -3.000.20 + S.30=0\u00a0 —\u00a0 3.000×10 \u2013 60.000 + 30S=0\u00a0 —\u00a0 S=1.000N\u00a0 —\u00a0 T\/S=3.000\/1.000\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T\/S=3<\/b><\/span><\/span><\/p>\n

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

 <\/p>\n

\"\"<\/p>\n

E=dvg=10<\/b><\/span><\/span>3.<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.10\u00a0 —\u00a0 E=10N\u00a0 —\u00a0 E.d<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>\u00a0\u2013 F.d<\/b><\/span><\/span>BC<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 10.5 \u2013 F.1=0\u00a0 —\u00a0 F=50N\u00a0 —<\/b><\/span><\/span>\u00a0 R- A<\/b><\/span><\/span><\/p>\n

32-<\/b><\/span><\/span>\u00a0Observe a figura abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

+100.0,5 \u2013 0,4P=0\u00a0 —\u00a0 P=125N\u00a0 —\u00a0 m=125\/10\u00a0 —\u00a0 m=12,5kg\u00a0 —\u00a0 a maior massa a partir dessa \u00e9 10kg\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

\"\"<\/p>\n

34- a) Interpotente\u00a0 —\u00a0 a for\u00e7a aplicada pelo operador\u00a0\"\"\u00a0(pot\u00eancia) localiza-se entre o p\u00f3lo e a for\u00e7a transmitida (resist\u00eancia)\u00a0\"\".<\/b><\/span><\/span><\/p>\n

b) Observe a figura\u00a0 —\u00a0\u00a0 a soma dos momentos de cada for\u00e7a em rela\u00e7\u00e3o ao p\u00f3lo deve ser nula\u00a0 —\u00a0 -F<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>.30 + 20.70 +<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

5.120=0\u00a0 —\u00a0 F<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=690\/30\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=23N<\/b><\/span><\/span><\/p>\n

c) \u03b7=F<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>\/F<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=5\/23\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u03b7=22%<\/b><\/span><\/span><\/p>\n

35-<\/b><\/span><\/span>\u00a0Observe a rela\u00e7\u00e3o F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=d<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 se d<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0> d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u2013 F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0> F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

36-<\/b><\/span><\/span>\u00a0a) Press\u00e3o=for\u00e7a\/\u00e1rea\u00a0 —\u00a0 P=F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\/S\u00a0 —\u00a0 1,3.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>=F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\/0,1.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0 — F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=1,3.10<\/b><\/span><\/span>9<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>-7<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=1,3.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=130N<\/b><\/span><\/span><\/p>\n

b) d<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=2.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>m\u00a0 — F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=130N\u00a0 — a soma dos momentos das for\u00e7as em rela\u00e7\u00e3o ao eixo de rota\u00e7\u00e3o deve ser nula\u00a0 —\u00a0\u00a0 -F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>.d<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0+ F<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>.d<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 130.2.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>=F<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>.10.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=260\/10\u00a0 —<\/b><\/span><\/span>\u00a0F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=26N<\/b><\/span><\/span><\/p>\n

37-<\/b><\/span><\/span>\u00a0Interpotente\u00a0 —\u00a0 a for\u00e7a aplicada pelo operador\u00a0\"\"\u00a0(pot\u00eancia) localiza-se entre o p\u00f3lo e a for\u00e7a transmitida (resist\u00eancia)\u00a0\"\".<\/b><\/span><\/span><\/p>\n

b) -F<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>.d<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>+Mg.d<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ mg(d<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)=0\u00a0 —\u00a0 F<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=(Mg.d<\/b><\/span><\/span>1\u00a0<\/b><\/span><\/span><\/sub>+ mg(d<\/b><\/span><\/span>1\u00a0<\/b><\/span><\/span><\/sub>+ d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>))\u00a0 —\u00a0 F<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=Mg + (mg(d<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>))\/d<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0F<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=\u00a0\"\"<\/b><\/span><\/span><\/p>\n

38-\u00a0<\/b><\/span><\/span>1- interpotente\u00a0 —\u00a0 2- interpotente\u00a0 —\u00a0 3- interpotente\u00a0 —\u00a0 4- interfixa\u00a0 —\u00a0 5- interpotente\u00a0 —\u00a0 6- interfixa\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

7- interfixa\u00a0 —\u00a0\u00a0 8- interpotente\u00a0 —\u00a0 9- inter-resistente<\/b><\/span><\/span><\/p>\n

39-<\/b><\/span><\/span>\u00a0Observe a figura abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Havendo equil\u00edbrio\u00a0 —\u00a0 -F<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>.dr + F<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>.d<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=0\u00a0 — \u00a0F<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>\/F<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=d\/a\u00a0 —\u00a0 como a>d\u00a0 — F<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0tem que ser maior que F<\/b><\/span><\/span>r\u00a0\u00a0<\/b><\/span><\/span><\/sub>para que o equil\u00edbrio seja mantido\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

40-<\/b><\/span><\/span>\u00a0Trata-se de uma alavanca interfixa\u00a0 — supondo que a for\u00e7a exercida por Arquimedes seja seu pr\u00f3prio peso, por<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0exemplo de massa 75kg\u00a0 —\u00a0 p=75g\u00a0 —\u00a0\u00a0 -F<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>.d<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0+ F<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>.d<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 75g.d<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=6.10<\/b><\/span><\/span>24<\/b><\/span><\/span><\/sup>g.d<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 d<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\/d<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>=6.10<\/b><\/span><\/span>24<\/b><\/span><\/span><\/sup>\/75\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

d<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\/d<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>=8.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>24<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>d<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\/d<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>=8.10<\/b><\/span><\/span>22<\/b><\/span><\/span><\/sup>\u00a0<\/b><\/span><\/span>(d<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0deve ser 8.10<\/b><\/span><\/span>22<\/b><\/span><\/span><\/sup>\u00a0vezes maior que d<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>).<\/b><\/span><\/span><\/p>\n

41-<\/b><\/span><\/span>\u00a0a) interpotente<\/b><\/span><\/span><\/p>\n

b) F<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>.d<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0\u2013 F<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>.d<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 1.200.2=F<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>.8\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>=300N<\/b><\/span><\/span><\/p>\n

42-<\/b><\/span><\/span>\u00a04.OA \u2013 F.4OA=0\u00a0 —\u00a0<\/b><\/span><\/span>F=1N\u00a0 —\u00a0 alavanca interpotente<\/b><\/span><\/span><\/p>\n

43-<\/b><\/span><\/span>\u00a0a) \u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

F<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0= for\u00e7a exercida pela pessoa que opera o abridor\u00a0 —\u00a0 F<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0= for\u00e7a de rea\u00e7\u00e3o que a tampinha exerce no abridor na regi\u00e3o da borda da tampinha\u00a0 —\u00a0 F<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0= for\u00e7a de rea\u00e7\u00e3o que a tampinha exerce no apoio<\/b><\/span><\/span><\/p>\n

b) F<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>.8,4 \u2013 F<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>.1,4=0\u00a0 —<\/b><\/span><\/span>\u00a0 F<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\/F<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=0,17<\/b><\/span><\/span><\/p>\n

 <\/p>\n

44-\u00a0<\/b><\/span><\/span>a) Momento de uma for\u00e7a \u00e9 a grandeza vetorial que mede o poder de uma for\u00e7a provocar rota\u00e7\u00e3o\u00a0 —\u00a0 depende da intensidade da for\u00e7a<\/b><\/span><\/span>\u00a0F<\/b><\/span><\/span>\u00a0e do m\u00f3dulo da dist\u00e2ncia\u00a0<\/b><\/span><\/span>d\u00a0<\/b><\/span><\/span>da linha de a\u00e7\u00e3o da for\u00e7a at\u00e9 o eixo de rota\u00e7\u00e3o, denominada bra\u00e7o da alavanca\u00a0 —\u00a0 express\u00e3o matem\u00e1tica\u00a0 —\u00a0 M<\/b><\/span><\/span>F<\/b><\/span><\/span><\/sub>=F.d.sen\u03b1, onde \u03b1 \u00e9 o \u00e2ngulo entre\u00a0F\u00a0e\u00a0d.<\/b><\/span><\/span><\/p>\n

Aplica\u00e7\u00f5es pr\u00e1ticas:A chave de roda para se trocar um pneu, o martelo, o alicate, a ma\u00e7aneta da porta e o pr\u00f3prio abrir e fechar da<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

porta.<\/b><\/span><\/span><\/p>\n

b) Para arrastar objetos pesados torna-se menos dificultoso faz\u00ea-lo em etapas, apoiando uma extremidade e girando a outra, alternadamente.<\/b><\/span><\/span><\/p>\n

Esse truque \u00e9 muito usado pelos oper\u00e1rios de empresas que fazem mudan\u00e7as. Ao transportar m\u00f3veis (geladeira, fog\u00e3o, guarda-roupas etc.) em vez de levantar os objetos, um funcion\u00e1rio apoia uma das extremidades, enquanto outro d\u00e1 um pequeno giro no m\u00f3vel, aplicando for\u00e7a na outra extremidade. A seguir, invertem-se as opera\u00e7\u00f5es. Prosseguindo essa altern\u00e2ncia, o m\u00f3vel vai avan\u00e7ando.\u00a0<\/b><\/span><\/span><\/p>\n

45-\u00a0<\/b><\/span><\/span>Observe a figura abaixo\u00a0 —\u00a0 equil\u00edbrio de transla\u00e7\u00e3o\u00a0 —\u00a0 a resultante das for\u00e7as na vertical deve<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

ser nula\u00a0 —\u00a0 T<\/b><\/span><\/span>E<\/b><\/span><\/span><\/sub>\u00a0+ T<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>=P\u00a0 —\u00a0equil\u00edbrio de rota\u00e7\u00e3o\u00a0 —\u00a0 a soma dos momentos de cada for\u00e7a deve ser nula\u00a0 —\u00a0 p\u00f3lo em P\u00a0 —\u00a0 +T<\/b><\/span><\/span>E<\/b><\/span><\/span><\/sub>.(x) \u2013 T<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>.(y)=0\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

T<\/b><\/span><\/span>E<\/b><\/span><\/span><\/sub>.(x)=T<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>.(y)\u00a0 —\u00a0 observe que, como x>y, T<\/b><\/span><\/span>E<\/b><\/span><\/span><\/sub>\u00a0deve ser menor que T\u00ad<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 T<\/b><\/span><\/span>E<\/b><\/span><\/span><\/sub><T<\/b><\/span><\/span>D<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

46-\u00a0<\/b><\/span><\/span>Como a alavanca est\u00e1 em equil\u00edbrio de rota\u00e7\u00e3o, o somat\u00f3rio dos momentos hor\u00e1rios \u00e9 igual ao somat\u00f3rio dos momentos anti-hor\u00e1rios\u00a0 —\u00a0 com o p\u00f3lo (eixo de rota\u00e7\u00e3o) indicado na figura\u00a0 —\u00a0 Q.(x) = P.(y)\u00a0 —\u00a0 200x=600y\u00a0 —\u00a0 x=3y\u00a0 —\u00a0 R- C<\/b><\/span><\/span><\/p>\n

47-\u00a0<\/b><\/span><\/span>Na situa\u00e7\u00e3o 1\u00a0 —\u00a0 F=Q\u00a0 —\u00a0 Q\/F=1\u00a0 —\u00a0 na situa\u00e7\u00e3o 2\u00a0 —\u00a0 F.3=Q.1\u00a0 —\u00a0 Q\/F=3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

48-\u00a0<\/b><\/span><\/span>Observe na figura que, se o comprimento da barra for L, cada divis\u00e3o corresponde a L\/6\u00a0 —\u00a0 como a barra est\u00e1 em equil\u00edbrio,<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

soma dos momentos hor\u00e1rios \u00e9 igual \u00e0 soma dos momentos anti-hor\u00e1rios\u00a0 —\u00a0 chamando de P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0o peso do corpo A, de P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0o peso do corpo B e de P o peso da barra, colocando-os conforme\u00a0 a figura voc\u00ea ter\u00e1 matematicamente, com o p\u00f3lo no ponto de apoio (cunha)\u00a0 —\u00a0 M.P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=M.P + M<\/b><\/span><\/span>.<\/b><\/span><\/span><\/sub>P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 P<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.(L\/6) = P.((L\/6) + P<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.d\u00a0 —\u00a0 4.M.g.L\/6 = M.g.L\/6 + M.g.d\u00a0 —\u00a0 d=3L\/6 (\u00e0 direita da cunha)\u00a0 a\u00a0 —<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

49-\u00a0<\/b><\/span><\/span>Observe as for\u00e7as colocadas na figura abaixo\u00a0 —\u00a0 como n\u00e3o h\u00e1 rota\u00e7\u00e3o, o somat\u00f3rio dos momentos em rela\u00e7\u00e3o ao eixo de rota\u00e7\u00e3o \u00e9 nulo\u00a0 —\u00a0 matematicamente\u00a0 —\u00a0 F.(8) = F\u2019.(4)\u00a0 —\u00a0 2.(8) =4.F\u2019\u00a0 —\u00a0 F\u2019 = 4 N\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

50-\u00a0<\/b><\/span><\/span>Como \u00e9 uma situa\u00e7\u00e3o de equil\u00edbrio de um corpo extenso, temos que considerar equil\u00edbrio de transla\u00e7\u00e3o (a resultante das for\u00e7as deve ser nula) e equil\u00edbrio de rota\u00e7\u00e3o (o momento resultante deve ser nulo). Analisando cada uma das op\u00e7\u00f5es:<\/b><\/span><\/span><\/p>\n

a) Falsa. A resultante das for\u00e7as na dire\u00e7\u00e3o horizontal \u00e9 n\u00e3o nula.<\/b><\/span><\/span><\/p>\n

b) Falsa. A resultante das for\u00e7as na dire\u00e7\u00e3o vertical \u00e9 n\u00e3o nula.<\/b><\/span><\/span><\/p>\n

c) Correta.<\/b><\/span><\/span><\/p>\n

d) Falsa. O momento resultante \u00e9 n\u00e3o nulo, provocando rota\u00e7\u00e3o no sentido hor\u00e1rio.\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

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

51-\u00a0<\/b><\/span><\/span>Observe na figura abaixo as for\u00e7as que agem na gangorra\u00a0 —\u00a0 a soma dos momentos das for\u00e7as no sentido hor\u00e1rio devem se<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

igualar com a soma dos momentos das for\u00e7as no sentido anti-hor\u00e1rio\u00a0 —\u00a0 440x=400×1,7 + 420×1\u00a0 — 440x=1100\u00a0 —\u00a0 x=2,5m\u00a0 —<\/b><\/span><\/span><\/p>\n

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

52-\u00a0<\/b><\/span><\/span>A figura a seguir mostra as for\u00e7as atuantes nas polias, bem como as for\u00e7as atuantes na barra, sendo P<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\u00a0o peso da barra, aplicado<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

53-\u00a0<\/b><\/span><\/span>Na figura abaixo est\u00e3o colocadas as for\u00e7as que agem no prisma que est\u00e1 em equil\u00edbrio\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

colocando o p\u00f3lo em O\u00a0 —\u00a0F.x=P.2\u00a0 —\u00a0 10.x=15.2\u00a0 —\u00a0 x=3m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

55-\u00a0<\/b><\/span><\/span>Observe na figura abaixo as for\u00e7as que agem sobre o bra\u00e7o que est\u00e1 na horizontal\u00a0 —\u00a0 peso da<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

massa M\u00a0 —\u00a0 P=mg=430.10\u00a0 —\u00a0P=4300N\u00a0 —\u00a0 P=4,3kN\u00a0 —\u00a0 para que o bra\u00e7o esteja em equil\u00edbrio, o somat\u00f3rio dos momentos (torque) nos sentidos hor\u00e1rio e anti-hor\u00e1rio deve ser nulo\u00a0 — polo em O\u00a0 —\u00a0\u00a0 F<\/b><\/span><\/span>y<\/b><\/span><\/span><\/sub>.0,6 \u2013 P.2,4=0\u00a0 —\u00a0 F<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub>=17,2kN\u00a0 — a for\u00e7a\u00a0\"\"\u00a0\u00e9 a soma vetorial de\u00a0\"\"\u00a0com\u00a0\"\"\u00a0\u00a0— veja<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

esquema\u00a0 —\u00a0 cos30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=F<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>\/F\u00a0 —\u00a0 F<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>= Fcos30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F=F<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>\/cos30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=17,2kN\/0,86\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F=20kN<\/b><\/span><\/span><\/p>\n

b) Press\u00e3o=for\u00e7a\/\u00e1rea\u00a0 —\u00a0 P<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>=4,5N\/3.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>.0,2.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>=4,5\/6.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0P<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>=7,5.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0(Pa)<\/b><\/span><\/span><\/p>\n

56-\"\"<\/b><\/span><\/span><\/p>\n

 <\/p>\n

57- Observe a figura abaixo com o p\u00f3lo P no eixo de rota\u00e7\u00e3o\u00a0 —\u00a0 c\u00e1lculo do momento de cada for\u00e7a<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

em rela\u00e7\u00e3o ao p\u00f3lo estabelecendo como positivo o sentido hor\u00e1rio de rota\u00e7\u00e3o\u00a0 —\u00a0 M<\/b><\/span><\/span>10N<\/b><\/span><\/span><\/sub>\u00a0= -10.2=-20N.m\u00a0 —\u00a0 M<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=+N.5\u00a0 —\u00a0 a somat\u00f3ria de todos os momentos deve ser nula para que haja equil\u00edbrio de rota\u00e7\u00e3o\u00a0 —\u00a0 M<\/b><\/span><\/span>10N<\/b><\/span><\/span><\/sub>\u00a0+ M<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>\u00a0= 0\u00a0 —\u00a0 – 20 + 5N = 0\u00a0 —\u00a0 N=4N\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

58- No equil\u00edbrio antes de colocar o recipiente\u00a0 —\u00a0 colocando as for\u00e7as\u00a0\u00a0 —\u00a0 equil\u00edbrio de transla\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 R<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0+ R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=P (I)\u00a0 —\u00a0equil\u00edbrio de rota\u00e7\u00e3o com o p\u00f3lo em A\u00a0 —\u00a0 M<\/b><\/span><\/span>RA<\/b><\/span><\/span><\/sub>=R<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.d=R<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.0=0\u00a0 —\u00a0 M<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=+ p.L\/2\u00a0 —\u00a0 M<\/b><\/span><\/span>RB<\/b><\/span><\/span><\/sub>= – R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.L\u00a0 —\u00a0 a somados momentos de cada for\u00e7a \u00e9 nula\u00a0 —\u00a0\u00a0 +P.L\/2 \u2013 R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.L=0\u00a0 —\u00a0 R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=P\/2 (II)\u00a0 —\u00a0 R<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=P\/2\u00a0\u00a0 —\u00a0 ap\u00f3s a inclus\u00e3o do recipiente, quando o mesmo estiver totalmente cheio de l\u00edquido de peso P<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 equil\u00edbrio de transla\u00e7\u00e3o\u00a0 —\u00a0 R<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0+ R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=P + P<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 equil\u00edbrio de rota\u00e7\u00e3o\u00a0 —\u00a0 M<\/b><\/span><\/span>RA<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 M<\/b><\/span><\/span>Pl\u00edquido<\/b><\/span><\/span><\/sub>= –<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

P<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>.x\u00a0 —M<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>= – P.L\/2\u00a0 —\u00a0 M<\/b><\/span><\/span>RB<\/b><\/span><\/span><\/sub>=R<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>.L\u00a0 —\u00a0 0 – P<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>.x \u2013 P.L\/2 + R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.L=0\u00a0 —\u00a0 R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=(P<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>.x + P.L\/2)\/L\u00a0 —\u00a0 R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=(2 P<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>.x + P.L)\/2L\u00a0 — \u00e0 medida que o l\u00edquido vai preenchendo o recipiente seu volume V\u2019 e consequentemente seu peso (P<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>) v\u00e3o aumentando provocando um aumento de R<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0e como o exerc\u00edcio pede, de R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 observe tamb\u00e9m que a rela\u00e7\u00e3o entre R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0e P<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>\u00a0e consequentemente V\u2019 do l\u00edquido \u00e9 uma fun\u00e7\u00e3o do primeiro grau, ou seja, o comportamento de R<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0\u00e9 linear\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

59-\u00a0Primeira situa\u00e7\u00e3o\u00a0 —\u00a0 equil\u00edbrio de rota\u00e7\u00e3o\u00a0 —\u00a0 a soma dos momentos de cada for\u00e7a com o p\u00f3lo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0em N (N=0) \u00e9 nula\u00a0 —\u00a0 M<\/b><\/span><\/span>pc<\/b><\/span><\/span><\/sub>=-50x\u00a0 —\u00a0 M<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 M<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>=+15P\u00a0 —\u00a0 -50x + 15P=0\u00a0 —\u00a0 x=15P\/50\u00a0 —\u00a0 x=3P\/10<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

(constante)\u00a0 —\u00a0 segunda situa\u00e7\u00e3o\u00a0 —\u00a0 P<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=80N\u00a0 —\u00a0d\u2019=?\u00a0 —\u00a0 a soma dos momentos com o p\u00f3lo em N (N=0) deve ser nula\u00a0 —\u00a0 -80x + Pd\u2019= 0\u00a0 —\u00a0 -80.(3P\/10) = Pd\u2019\u00a0 —\u00a0 d\u2019 = 24 cm\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

60- Equil\u00edbrio de transla\u00e7\u00e3o\u00a0 —\u00a0 c\u00e1lculo da intensidade da for\u00e7a de tra\u00e7\u00e3o\u00a0 —\u00a0 T=30 + 6=36N\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

equil\u00edbrio de rota\u00e7\u00e3o\u00a0 com o p\u00f3lo (eixo de rota\u00e7\u00e3o em T)\u00a0 —\u00a0 a doma dos momentos de cada for\u00e7a em rela\u00e7\u00e3o ao p\u00f3lo deve ser nula\u00a0 —\u00a0 M<\/b><\/span><\/span>30N<\/b><\/span><\/span><\/sub>= – 30.8 = – 240N.cm\u00a0 —\u00a0 M<\/b><\/span><\/span>T<\/b><\/span><\/span><\/sub>=T.d=T.0=0\u00a0 —\u00a0 M<\/b><\/span><\/span>6N<\/b><\/span><\/span><\/sub>= + 6.d\u00a0 —\u00a0 \u03a3M=0\u00a0 —\u00a0 -240 + 0 + 6d = 0\u00a0 —\u00a0 d= 240\/6\u00a0 —\u00a0 d= 40cm\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span><\/p>\n

61- Seja F a for\u00e7a aplicada pelo motorista no pedal\u00a0 —\u00a0 pelo enunciado a alavanca tem a capacidade\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

de amplia\u00e7\u00e3o da for\u00e7a aplicada por um fator igual \u00e0 raz\u00e3o direta de seus bra\u00e7os, que \u00e9 de 40\/10=4 vezes\u00a0 —\u00a0 f=4F\u00a0 —\u00a0 a prensa hidr\u00e1ulica amplia a for\u00e7a f na raz\u00e3o direta de suas \u00e1reas, ou seja, de 8 vezes\u00a0 —\u00a0 F\u2019=8f\u00a0 —\u00a0 F\u2019=8.4F\u00a0 —\u00a0 F\u2019=32F\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A.<\/b><\/span><\/span><\/p>\n

62- \u00a0<\/b><\/span><\/span>–<\/b><\/span><\/span>01. Correta\u00a0 —\u00a0 a for\u00e7a\"\"m\u00ednima necess\u00e1ria, aplicada no bra\u00e7o de alavanca para manter suspensas as 5 barras de ferro, de peso P=5.mg=5.5.10=250N ocorre quando o sistema est\u00e1 em equil\u00edbrio\u00a0 —\u00a0\u00a0 como as duas polias s\u00e3o fixas elas transmitem integralmente a intensidade da tra\u00e7\u00e3o T=250N atrav\u00e9s do fio at\u00e9 a extremidade direita do disco\u00a0 —\u00a0 equil\u00edbrio de rota\u00e7\u00e3o com o p\u00f3lo (eixo de rota\u00e7\u00e3o) no centro do disco\u00a0 —\u00a0 momento de cada for\u00e7a\u00a0 —\u00a0 M<\/b><\/span><\/span>F<\/b><\/span><\/span><\/sub>=+F.d=F.1\u00a0 —<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

M<\/b><\/span><\/span>F<\/b><\/span><\/span><\/sub>= F\u00a0 —\u00a0 momento de T=250N\u00a0 —\u00a0 M<\/b><\/span><\/span>T<\/b><\/span><\/span><\/sub>= – T.d= – 250.0,5= – 125N.m\u00a0 —\u00a0 a soma dos momentos de cada\u00a0 for\u00e7a deve ser nulo\u00a0 —\u00a0 F \u2013 125=0\u00a0 —\u00a0 F=125N.<\/b><\/span><\/span><\/p>\n

02. Correta\u00a0 —\u00a0 a velocidade de qualquer ponto do fio (2m\/s) \u00e9 a mesma que de qualquer ponto da periferia do disco de raio r=0,5m que tamb\u00e9m \u00e9 de v=2m\/s\u00a0 —\u00a0 velocidade angular do disco que \u00e9 constante para qualquer ponto do mesmo e vale\u00a0 —\u00a0 W=v\/r=2\/0,5=4rad\/s\u00a0 —\u00a0 o bra\u00e7o de alavanca gira em torno de uma circunfer\u00eancia de raio R=1m e sua velocidade v vale\u00a0 —\u00a0 W=v\/R\u00a0 —\u00a0 4=v\/1\u00a0 —\u00a0 v=4m\/s.<\/b><\/span><\/span><\/p>\n

04. Falsa\u00a0 —\u00a0 c\u00e1lculo da tra\u00e7\u00e3o T no fio com F=250N\u00a0 —\u00a0 M<\/b><\/span><\/span>250N<\/b><\/span><\/span><\/sub>=+250.1=250N.m\u00a0 —\u00a0 M<\/b><\/span><\/span>T<\/b><\/span><\/span><\/sub>= – T.0,5\u00a0 —\u00a0 250 \u2013 0,5T=0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0—\u00a0 T=500N\u00a0 —\u00a0 essa for\u00e7a \u00e9 totalmente transmitida pelas polias fixas at\u00e9 o peso de P=250N\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 T \u2013 P=m.a\u00a0 —\u00a0 500 \u2013 250=25.a\u00a0 —\u00a0 a=250\/25\u00a0 —\u00a0 a=10m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.<\/b><\/span><\/span><\/p>\n

08. Correta\u00a0 —\u00a0 Alavanca interfixa\u00a0\u2013 o apoio est\u00e1 entre a for\u00e7a potente e a for\u00e7a resistente<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

16. Falsa\u00a0 —\u00a0 Se F for inclinada em rela\u00e7\u00e3o \u00e0 d, voc\u00ea pode calcular o momento da for\u00e7a decompondo-a em suas componentes horizontal x e vertical y.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Fx\u00a0n\u00e3o produz rota\u00e7\u00e3o e seu momento \u00e9 nulo\u00a0 —\u00a0 quem produz rota\u00e7\u00e3o \u00e9 apenas Fy=Fsen\u03b2\u00a0 —\u00a0 M=Fy.d\u00a0 —\u00a0 M=Fsen\u03b2.d.<\/b><\/span><\/span><\/p>\n

Corretas: 01, 02 e 08\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>Soma=11<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

63- Colocando as for\u00e7as que agem sobre a trave\u00a0 —\u00a0 peso\u00a0\"\"\u00a0da trave, vertical e para baixo e de intensidade\u00a0 —\u00a0 P=m.g=4,6.10\u00a0 —\u00a0 P=46N\u00a0 —\u00a0 for\u00e7a de tra\u00e7\u00e3o\"\"\u00a0aplicada pelo cabo, sustentando-a\u00a0 —\u00a0 decompondo a for\u00e7a em suas componentes horizontal e vertical\u00a0 —\u00a0 horizontal \u2013 T<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>=T.sen60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=T.\u221a3\/2\u00a0 —\u00a0 vertical \u2013 T<\/b><\/span><\/span>v<\/b><\/span><\/span><\/sub>=T.cos60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=T.1\/2=T\/2\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe a seq\u00fc\u00eancia da figura acima\u00a0 —\u00a0 N \u00e9 a intensidade da for\u00e7a horizontal que a parede aplica na trave\u00a0 —\u00a0 equil\u00edbrio na horizontal\u00a0 —\u00a0 N=T.\u221a3\/2\u00a0 —\u00a0 equil\u00edbrio na vertical\u00a0 —\u00a0 T\/2=P\u00a0 —\u00a0 T\/2=46\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=92N\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 se fosse pedida a intensidade da for\u00e7a que a trave troca com a parede voc\u00ea teria\u00a0 —\u00a0 N=92.\u221a3\/2=46\u221a3N.<\/b><\/span><\/span><\/p>\n

64-\u00a0\u00a0Observe na figura abaixo\u00a0 —\u00a0 equil\u00edbrio de transla\u00e7\u00e3o (n\u00e3o sobe nem desce)\u00a0 —\u00a0 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>= P<\/b><\/span><\/span>pi<\/b><\/span><\/span><\/sub>\u00a0+<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

P<\/b><\/span><\/span>pl<\/b><\/span><\/span><\/sub>\u00a0(I)\u00a0 — equil\u00edbrio de rota\u00e7\u00e3o (n\u00e3o gira)\u00a0 —\u00a0 colocando o p\u00f3lo (eixo de rota\u00e7\u00e3o) em\u00a0\"\"\u00a0e calculando o momento de cada for\u00e7a em rela\u00e7\u00e3o ao p\u00f3lo\u00a0 —\u00a0 M<\/b><\/span><\/span>T1<\/b><\/span><\/span><\/sub>=T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.0=0\u00a0 —\u00a0 M<\/b><\/span><\/span>pl<\/b><\/span><\/span><\/sub>=+P<\/b><\/span><\/span>pl<\/b><\/span><\/span><\/sub>.d=600.2,5=1500N.m\u00a0 —\u00a0 M<\/b><\/span><\/span>pi<\/b><\/span><\/span><\/sub>=+P<\/b><\/span><\/span>pi<\/b><\/span><\/span><\/sub>.d=700.3=2100N.m\u00a0 —\u00a0 M<\/b><\/span><\/span>T2<\/b><\/span><\/span><\/sub>= – T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.d= – 5T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 a soma dos momentos de cada for\u00e7a deve ser nula\u00a0 —\u00a0 0 + 1500 + 2100 \u2013 5T<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sub>= 0\u00a0 —\u00a0 T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=720N\u00a0\u00a0 —\u00a0 substituindo T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=720N em (I)\u00a0 —\u00a0 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ 720 = 1300\u00a0 —\u00a0 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=580N\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

65- Observe a figura abaixo onde est\u00e3o localizadas todas as for\u00e7as que agem sobre a barra\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

c\u00e1lculo do momento de cada for\u00e7a com o p\u00f3lo em O\u2019\u00a0 —\u00a0 M<\/b><\/span><\/span>P1<\/b><\/span><\/span><\/sub>= – P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.40= – 40P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 M<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>=N.0=0\u00a0 —\u00a0 M<\/b><\/span><\/span>4P1<\/b><\/span><\/span><\/sub>= + 4P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.6= + 24P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 M<\/b><\/span><\/span>2P1<\/b><\/span><\/span><\/sub>= + 2P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.(d + 6)\u00a0 —\u00a0 a condi\u00e7\u00e3o de equil\u00edbrio de rota\u00e7\u00e3o \u00e9 que a soma dos momentos de todas as for\u00e7as for nula\u00a0 —\u00a0 – 40P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ 0 + 24P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+<\/b><\/span><\/span><\/p>\n

2P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.(d + 6) = 0\u00a0 —\u00a0 – 16P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= – 2d \u2013 12\u00a0 —\u00a0 d=2cm\u00a0 —\u00a0 a dist\u00e2ncia pedida entre os pontos O\u2019 e C<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0vale\u00a0 —\u00a0 d\u2019 = 6 + 2= 8cm\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span>.<\/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 Est\u00e1tica de um corpo extenso \u2013 M\u00e1quinas Simples   01-\u00a0A \u00fanica for\u00e7a que n\u00e3o precisa ser decomposta no sentido de girar a barra no sentido hor\u00e1rio \u00e9 F2\u00a0 —\u00a0\u00a0R- A\u00a0 02-\u00a0O port\u00e3o tende a girar no sentido hor\u00e1rio\u00a0 —\u00a0\u00a0R- A 03-\u00a0FA=FB=F\u00a0 —\u00a0 MFA=+F.a\u00a0 —\u00a0 MFB=+F.a\u00a0 —\u00a0 MFC=0\u00a0 — MR=+2Fa\u00a0 —\u00a0 gira no sentido<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1511,"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-1514","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1514","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=1514"}],"version-history":[{"count":4,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1514\/revisions"}],"predecessor-version":[{"id":10873,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1514\/revisions\/10873"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1511"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1514"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}