{"id":1484,"date":"2015-11-02T21:56:04","date_gmt":"2015-11-02T21:56:04","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1484"},"modified":"2024-08-23T14:18:00","modified_gmt":"2024-08-23T14:18:00","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-estatica-de-um-ponto-material","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/estatica\/estatica-de-um-ponto-material\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-estatica-de-um-ponto-material\/","title":{"rendered":"Est\u00e1tica de um Ponto Material – 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 Ponto Material<\/b><\/span><\/span><\/span><\/p>\n

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

01-<\/b><\/span><\/span><\/span>\u00a0Sobre o homem atuam tr\u00eas for\u00e7as, seu peso (vertical e para baixo) e as duas for\u00e7as de tra\u00e7\u00e3o aplicadas pelo teto atrav\u00e9s dos fios, agindo sobre suas m\u00e3os \u00a0—<\/b><\/span><\/span>\u00a0 R- A<\/b><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span>\u00a0A parede da direita aplica em B, sobre a esfera 2 uma for\u00e7a horizontal e para a esquerda, analogamente,\u00a0 parede da esquerda aplica em A, sobre a esfera 2 uma for\u00e7a horizontal e para a direita e a esfera 1 comprime-a em A\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

04-<\/b><\/span><\/span><\/span>\u00a0Quando elas tiverem mesma dire\u00e7\u00e3o e mesmo sentido\u00a0 —\u00a0 valor m\u00e1ximo da terceira for\u00e7a que deve anula-las F<\/b><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub>=20 + 5=25N\u00a0 —\u00a0 quando elas tiverem mesma dire\u00e7\u00e3o e sentidos opostos\u00a0 —\u00a0 valor m\u00ednimo da terceira que deve anula-las F<\/b><\/span><\/span>m\u00edn<\/b><\/span><\/span><\/sub>=20 – 5=15N\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0Observe atentamente a figura abaixo onde todas as for\u00e7as foram colocadas<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

06-<\/b><\/span><\/span><\/span>\u00a0Colocando as for\u00e7as que agem nos im\u00e3s\u00a0 —\u00a0 F<\/b><\/span><\/span>m\u00a0<\/b><\/span><\/span><\/sub>\u2013 for\u00e7a magn\u00e9tica que deve ser de atra\u00e7\u00e3o, caso contr\u00e1rio n\u00e3o haveria equil\u00edbrio\u00a0 —\u00a0 peso P\u00a0 —\u00a0 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0\u2013 for\u00e7a de tra\u00e7\u00e3o no fio de cima\u00a0 —\u00a0 T=2 \u2013 for\u00e7a de tra\u00e7\u00e3o no fio de baixo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

07-<\/b><\/span><\/span><\/span><\/span>\u00a0(1)\u00a0 —\u00a0 2T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>=P\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>=P\/2\u00a0 —\u00a0 (2) 2T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>sen60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sup>=P\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>=P\/2sen60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>=0,58P\u00a0 —\u00a0 (3) 2T<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sub>sen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sup>=P\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sub>=P\/2sen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sub>=P\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>\/m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>=3\/2\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>=3\/2m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub>=<\/b><\/span><\/span><\/span>\u03bc<\/b><\/span><\/span>N=<\/b><\/span><\/span><\/span>\u03bc<\/b><\/span><\/span>P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>=\u03bcm<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>g\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/span><\/sub>=3\/2\u03bcm<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>g\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>=m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>g\u00a0 —\u00a0 colocando as forces:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

O bloco m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0est\u00e1 em equil\u00edbrio\u00a0 —\u00a0 T=P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 T=m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>g\u00a0 —\u00a0 o bloco m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0tamb\u00e9m est\u00e1 em equil\u00edbrio\u00a0 —\u00a0 equil\u00edbrio horizontal\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=Tcos\u03b8\u00a0 — 3\/2<\/b><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>g =Tcos30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 3\/2<\/b><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>g=m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>g.\u221a3\/2\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0\u03bc=\u221a3\/3<\/b><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Colocando as for\u00e7as<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Equil\u00edbrio na horizontal\u00a0 —\u00a0 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>cos60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>cos30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.1\/2=T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.\u221a3\/2\u00a0 —\u00a0 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=\u221a3.T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 equil\u00edbrio na vertical\u00a0 —\u00a0 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>sen60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0+ T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>sen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=P\u00a0 —\u00a0 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.\u221a3\/2 + T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.1\/2=6.10\u00a0 —\u00a0 (\u221a3.T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.\u221a3)\/2 + T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/2=60\u00a0 —\u00a0 3T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=120\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=30N\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=\u221a3.T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=\u221a3.30\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=30\u221a3N<\/b><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0colocando as for\u00e7as<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

sen\u03b2=L\/(L\u221a2)\u00a0 —\u00a0 sen\u03b2=\u221a2\/2\u00a0 —\u00a0 cos\u03b2= L\/(L\u221a2)\u00a0 —\u00a0 cos\u03b2=\u221a2\/2\u00a0 —\u00a0 equil\u00edbrio vertical\u00a0 —\u00a0 P=(\u221a2\/2)T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=P\u221a2\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Observe a figura abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

F<\/b><\/span><\/span>RAB<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0 F<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>=(160)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ (120)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F<\/b><\/span><\/span>RAB<\/b><\/span><\/span><\/sub>=200N=F<\/b><\/span><\/span>C\u00ad<\/b><\/span><\/span><\/sub>\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 R- E<\/b><\/span><\/span><\/p>\n

12<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span> a) Observe na figura abaixo que 2T=P=mg=60.10\u00a0 —\u00a0 T=600\/2\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0T=300N<\/b><\/span><\/span><\/p>\n

b) Prolongando as duas cordas obtemos os dois tri\u00e2ngulos da figura abaixo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Pela semelhan\u00e7a entre os tri\u00e2ngulos de alturas c e (3 + c)\u00a0 —\u00a0 0,25\/c=0,75\/(3 + c)\u00a0 —\u00a0 c=1,5m\u00a0 —\u00a0 equil\u00edbrio na vertical\u00a0 —\u00a0 2Tcos\u03b8=P\u00a0 —\u00a0 Tcos\u03b8=300\u00a0\u00a0 —\u00a0 T=300\/cos\u03b8 (I)\u00a0 —\u00a0 na horizontal, a componente horizontal de T \u00e9\u00a0 —\u00a0 T<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>=Tsen\u03b8\u00a0 —\u00a0 T=T<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>\/sen\u03b8 (II)\u00a0 —\u00a0 igualando (I) com (II)\u00a0 —\u00a0 300\/cos\u03b8=T<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>\/sen\u03b8\u00a0 —\u00a0 sen\u03b8\/cos\u03b8=T<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>\/300\u00a0 —\u00a0 tg\u03b8=T<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>\/300\u00a0 —\u00a0 0,75\/4,5=T<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>\/300\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0T<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>=50N<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0Para que as placas fiquem em equil\u00edbrio est\u00e1tico, resultante das for\u00e7a que agem sobre elas devem ter componentes na horizontal e na vertical para que possam se anular\u00a0 —\u00a0 observe que a \u00fanica alternativa que n\u00e3o tem componentes que possam se anular na horizontal \u00e9 a V\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0Para mov\u00ea-lo, a for\u00e7a de tra\u00e7\u00e3o deve valer T=500N, conforme a figura abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Equil\u00edbrio na horizontal\u00a0 —\u00a0 Tcos45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=500\u00a0 —\u00a0 T\u221a2\/2=500\u00a0 —\u00a0 T=500\u221a2N\u00a0 —\u00a0 equil\u00edbrio na vertical\u00a0 —\u00a0 Tsen45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>= P\u00a0 —\u00a0 500\u221a2.(\u221a2\/2)=P\u00a0 —\u00a0 P=500N\u00a0 —\u00a0 para mov\u00ea-lo \u00e9 necess\u00e1rio um peso de 500N, ou seja, 500\/150=3,3 blocos\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0a) Somente quando \u03b1=\u03b2<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

Equil\u00edbrio na horizontal\u00a0 —\u00a0 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>cos\u03b1=T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>cos\u03b2\u00a0 —\u00a0 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u221a3\/2=T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.1\/2\u00a0 —\u00a0 T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=\u221a3.T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 equil\u00edbrio na vertical\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=P=100N\u00a0 –<\/b><\/span><\/span>—\u00a0 100=T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>sen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0+ T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>sen60<\/b><\/span><\/span>o\u00a0\u00a0<\/b><\/span><\/span><\/sup>—\u00a0 100=T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.1\/2 + (\u221a3T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>).\u221a3\/2\u00a0 —\u00a0 100=T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/2 + 3T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=50N\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=\u221a350\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=50\u221a3N<\/b><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0\u00c2ngulo de 60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=2Tcos60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=2T1\/2=T\u00a0 —\u00a0 \u00e2ngulo de 45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=2Tcos45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=2T\u221a2\/2=\u221a2T\u00a0 —\u00a0 F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=\u221a2F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

equil\u00edbrio na vertical\u00a0 —\u00a0 2.60.sen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=P\u00a0 —\u00a0 2.60.1\/2=P\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0P=60kgf<\/b><\/span><\/span><\/p>\n

b) analogamente\u00a0 —\u00a0 2.130.sen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=P\u00a0 —\u00a0 P=130kgf\u00a0 —\u00a0 130\/30=4,33\u00a0 —\u00a0<\/b><\/span><\/span>\u00a04 crian\u00e7as<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

Equil\u00edbrio vertical no bloco P\u00a0 —\u00a0 150=N + 100sen\u03b8\u00a0 —\u00a0 150=N + 100.0,87\u00a0 —\u00a0 N=63N\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

Equil\u00edbrio na vertical\u00a0 —\u00a0 Fcos60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=P + 8,8.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F.1\/2=8,8.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 + 200 —\u00a0 F=(8.800 + 200).2\u00a0 —\u00a0 F=18.000\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F=18.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>N<\/b><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Observe a figura abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Equil\u00edbrio na horizontal\u00a0 —\u00a0 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>cos45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>cos37<\/b><\/span><\/span>0<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.0,7=T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.0,8\u00a0 —\u00a0 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=8T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/7\u00a0 —\u00a0 equil\u00edbrio na vertical\u00a0 —\u00a0 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>sen45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0+ T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>sen37<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=P\u00a0 —\u00a0 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.0,7 + T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.0,6=280\u00a0 —\u00a0 (8T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/7).0,7 + 0,6T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=280\u00a0 —\u00a0 0,8T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ 0,6T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=280\u00a0 —\u00a0 T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=280\/1,4\u00a0 —\u00a0 T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=200N\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span>\u00a0Se os tratores e o barco se movem em trajet\u00f3ria retil\u00ednea com velocidade constante eles est\u00e3o em equil\u00edbrio din\u00e2mico e a resultante das for\u00e7as que agem sobre eles \u00e9 nula\u00a0 —\u00a0 F<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=F\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=2Fsen\u03b8=2F.0,8\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=1,6F\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0 —1,6F=1,92.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F=1,20.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>N\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span>
\n<\/b><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0Equil\u00edbrio na vertical —\u00a0 Tsen45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=P\u00a0 —\u00a0 T\u221a2\/2=55\u00a0 —\u00a0 T=55\u221a2N\u00a0 —equil\u00edbrio na horizontal\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

N=Tcos45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —N=55\u221a2.\u221a2\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>N=55N\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0Se voc\u00ea pendurar um quadro na parede por meio de dois fios, quanto menor for o \u00e2ngulo formado com o teto, ou o suporte, maior ser\u00e1 a for\u00e7a de tra\u00e7\u00e3o (tens\u00e3o) no fio.Exemplo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

R-\u00a0 T1 > T2 > T3<\/b><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span>\u00a0 (veja resolu\u00e7\u00e3o do exerc\u00edcio anterior)<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

sen\u03b8=0,2\/5\u00a0 —\u00a0 sen\u03b8=0,04\u00a0 —\u00a0 T<\/b><\/span><\/span>y<\/b><\/span><\/span><\/sub>=Tsen\u03b8=0,04T\u00a0\u00a0 —\u00a0 2T<\/b><\/span><\/span>y<\/b><\/span><\/span><\/sub>=P\u00a0 —\u00a0 2.0,04T=20\u00a0 —\u00a0 T=2540N\u00a0 —<\/b><\/span><\/span>\u00a0 R- D<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>sen\u03b2=30\/50\u00a0 —\u00a0 sen\u03b2=0,6\u00a0 —\u00a0 P=2Tsen\u03b2\u00a0 —\u00a0 36=2T.0,6\u00a0 —\u00a0 T=36\/1,2\u00a0 —\u00a0 T=30N\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Nas figuras abaixo est\u00e3o colocadas as for\u00e7as que\u00a0 agem sobre a esfera\u00a0 —\u00a0 como a esfera est\u00e1 em equil\u00edbrio, a resultante das<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

for\u00e7as \u00e9 nula\u00a0 —\u00a0 sen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=T<\/b><\/span><\/span>din<\/b><\/span><\/span><\/sub>\/P\u00a0 —\u00a0 1\/2=10\/P\u00a0 —\u00a0 P=20N\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

28-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) Observe o diagrama abaixo que mostra as for\u00e7as atuantes no terceiro elo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

b) Dados\u00a0 —\u00a0 \u00a0m = 200 g = 0,2 kg\u00a0 —\u00a0\u00a0 g = 10 m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 P = m g = 0,2(10)\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0P = 2 N<\/b><\/span><\/span>\u00a0 —\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>43<\/b><\/span><\/span><\/sub>\u00a0= 2 P = 4 N<\/b><\/span><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>23<\/b><\/span><\/span><\/sub>\u00a0= 3 P = 6 N<\/b><\/span><\/span><\/p>\n

29-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Observe nas figuras abaixo onde cada for\u00e7a de tra\u00e7\u00e3o \u00e9 decomposta na vertical (T<\/b><\/span><\/span>V<\/b><\/span><\/span><\/sub>=Tsen60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>) e na horizontal (T<\/b><\/span><\/span>H<\/b><\/span><\/span><\/sub>=Tcos60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>)\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como o sistema est\u00e1 em equil\u00edbrio, as componentes horizontais se anulam e na vertical voc\u00ea ter\u00e1\u00a0 —\u00a0 2T\u00ad<\/b><\/span><\/span>V<\/b><\/span><\/span><\/sub>=P\u00a0 —\u00a0 2Tsen60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=P\u00a0 —\u00a0 2T.\u221a3\/2=50\u00a0 —\u00a0 T=50\/\u221a3N\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>m = 10 kg\u00a0 —\u00a0\u00a0 x<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0= 5 cm\u00a0 —\u00a0\u00a0 sen \u03b8 = 0,6\u00a0 —\u00a0\u00a0 cos \u03b8 = 0,8\u00a0 —\u00a0 g=10m\/s<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>\u00a0—\u00a0 observe as figuras a seguir onde o corpo est\u00e1 em equil\u00edbrio nas duas situa\u00e7\u00f5es e a resultante das for\u00e7as deve ser nula em<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0cada caso\u00a0 —\u00a0 figura A\u00a0 —\u00a0 F<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=P\u00a0 —\u00a0 kx<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=mg\u00a0 —k.5=10.10\u00a0 —\u00a0 k=20N\/cm\u00a0 —\u00a0 figura B\u00a0 —\u00a0\u00a0\"\"\u00a0e\u00a0\"\"\u00a0se anulam\u00a0 —\u00a0 F<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=P<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 kx<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=mgsen\u03b8\u00a0 —\u00a0 20.x<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=10.10.0,6\u00a0 —\u00a0 x<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=3cm\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

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

31-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Dado: P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= 500 N.<\/b><\/span><\/span><\/p>\n

Como \u00e9 uma situa\u00e7\u00e3o de equil\u00edbrio, a resultante em cada um dos n\u00f3s R e S \u00e9 nula\u00a0 —\u00a0 aplicando, ent\u00e3o, a regra da poligonal em<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

cada um dos n\u00f3s\u00a0 —\u00a0 figura I\u00a0 —\u00a0 sen45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u221a2\/2=500\/T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=500\u221a2N\u00a0 —\u00a0 tg45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=P<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 1=500\/T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=500N\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

Figura II\u00a0 —\u00a0 cos30<\/b><\/span><\/span>0<\/b><\/span><\/span><\/sup>=T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u221a3\/2=500\/T<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 T<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=1000\/\u221a3N\u00a0 —\u00a0 tg30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u221a3\/3=P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/500\u00a0 —\u00a0 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=500\/\u221a3N\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

 <\/p>\n

32- <\/b><\/span><\/span><\/span>Observe na figura 1 abaixo no tri\u00e2ngulo hachurado a determina\u00e7\u00e3o do \u00e2ngulo \u03b8 entre a horizontal e a dire\u00e7\u00e3o da for\u00e7a normal,\u00a0\"\"\u00a0de\u00a0contato entre a quina do degrau e a roda\u00a0 —\u00a0 sen\u03b8=(R\/2)\/R\u00a0 —\u00a0 sen\u03b8=1\/2\u00a0 —\u00a0 \u03b8=30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 n a figura 2, todas as for\u00e7as que agem sobre a roda, normal\u00a0\"\"\u00a0de contato entre o piso e a roda, peso\u00a0\"\",\u00a0\"\"\u00a0e\u00a0\"\", foram colocadas\u00a0no centro (eixo) da roda\u00a0 —\u00a0\u00a0\u00a0 na figura 3,\u00a0\"\"\u00a0foi decomposta em sua componente horizontal N<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.cos30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=N<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u221a3\/2 e vertical N<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>sen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=N<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/2\u00a0 —\u00a0 \u00e0 medida que\u00a0\"\"\u00a0for aumentando de intensidade para erguer a roda, a intensidade da for\u00e7a normal\u00a0\"\"\u00a0vai diminuindo, at\u00e9 que, quando a roda perder contato com o piso (N=0), ela come\u00e7a a subir (figura 4)\u00a0 — no instante em que ela perde contato com o piso ela est\u00e1 em equil\u00edbrio e a resultante das for\u00e7as que agem sobre ela \u00e9 nula\u00a0 —\u00a0 equil\u00edbrio na horizontal\u00a0 —\u00a0 F=N1\u221a3\/2\u00a0 —\u00a0 equil\u00edbrio na vertical\u00a0 —\u00a0 N<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/2=P\u00a0 —\u00a0 N<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=2P\u00a0 —\u00a0\u00a0 F=N<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u221a3\/2=2P.\u221a3\/2\u00a0 —\u00a0 F=mg\u221a3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R-C.<\/b><\/span><\/span>\"\"<\/p>\n

 <\/p>\n

\u00a033- <\/b><\/span><\/span><\/span>Para calcular a intensidade da for\u00e7a resultante que age sobre a part\u00edcula c\u00f3smica voc\u00ea pode decompor as for\u00e7as nas dire\u00e7\u00f5es norte e leste\u00a0 —\u00a0 observe na sequ\u00eancia abaixo que a intensidade da<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

for\u00e7a resultante \u00e9 de 1N no sentido leste.<\/b><\/span><\/span><\/p>\n

Como a velocidade inicial da part\u00edcula tem intensidade V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=1200m\/s do norte para o sul e a for\u00e7a resultante sobre ela tem intensidade 1N do oeste para leste, o movimento da part\u00edcula tem as caracter\u00edsticas de composi\u00e7\u00e3o de dois movimentos, um no sentido leste e outro no sentido sul (veja figura)\u00a0 —\u00a0 no sentido leste, a proje\u00e7\u00e3o da velocidade inicial \u00e9 nula V<\/b><\/span><\/span>oL<\/b><\/span><\/span><\/sub>=0 e ela se desloca sob a\u00e7\u00e3o de uma for\u00e7a resultante de valor F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=1N e com acelera\u00e7\u00e3o\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=m.a\u00a0 —\u00a0 1=2.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.a\u00a0 —\u00a0 a= 500m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 sua velocidade nessa dire\u00e7\u00e3o ap\u00f3s t=1s ter\u00e1 intensidade\u00a0 —\u00a0 V<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>oL<\/b><\/span><\/span><\/sub>\u00a0+ a<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>.t=0 + 500.1\u00a0 —\u00a0 V<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>=500m\/s\u00a0 —\u00a0 no sentido sul ela ser\u00e1 lan\u00e7ada para baixo com V<\/b><\/span><\/span>oS<\/b><\/span><\/span><\/sub>=1200m\/s, acelerando com acelera\u00e7\u00e3o da gravidade g=10m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 ap\u00f3s t=1s, sua velocidade nessa dire\u00e7\u00e3o ser\u00e1\u00a0 —\u00a0 V<\/b><\/span><\/span>S<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>oS<\/b><\/span><\/span><\/sub>\u00a0+ g.t=1200 + 10.1=1210m\/s\u00a0 —<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0observe na figura que essas duas velocidades s\u00e3o perpendiculares e, aplicando Pit\u00e1goras voc\u00ea obter\u00e1 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= V<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ V<\/b><\/span><\/span>S<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>=(500)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ (1210)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V=\u221a(1714100)\u00a0 —\u00a0 V=1309m\/s=1,3km\/s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

34- <\/b><\/span><\/span><\/span>Pelo princ\u00edpio da in\u00e9rcia, se as for\u00e7as deixarem de atura, a for\u00e7a resultante sobre ela ser\u00e1 nula e, ap\u00f3s esse innstante, por in\u00e9rcia, ela seguir\u00e1 em MRU com velocidade constante de 1,3km\/s\u00a0 —\u00a0 observe na resolu\u00e7\u00e3o do exerc\u00edcio anterior (08) que, antes de 1s a trajet\u00f3ria era parab\u00f3lica\u00a0 —\u00a0\u00a0R- D.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

35- <\/b><\/span><\/span><\/span>Colocando as for\u00e7as\u00a0 —\u00a0 bloco 1\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=\u03bcN=\u03bcP<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=\u03bcm<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>g\u00a0 —\u00a0 bloco2\u00a0 —\u00a0 P<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>g\u00a0 —\u00a0 P<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>g\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

decompondo a tra\u00e7\u00e3o T\u00a0 —T<\/b><\/span><\/span>horizontal<\/b><\/span><\/span><\/sub>=mgcos\u03b8\u00a0 —\u00a0 T<\/b><\/span><\/span>vertical<\/b><\/span><\/span><\/sub>=mgsen\u03b8\u00a0 —\u00a0 equil\u00edbrio na horizontal\u00a0 —\u00a0 \u03bcm<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>g=m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>gcos60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0 \u03bcm<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>g=m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>g.(1\/2)\u00a0 —\u00a0 \u03bc=m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\/2m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —<\/b><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

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

b) Se o conjunto n\u00e3o deve cair a for\u00e7a resultante sobre ele tanto na vertical como na horizontal deve ser nula\u00a0 —\u00a0 equil\u00edbrio na vertical\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=P + T\u00a0 —\u00a0 P=m<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>g\u00a0 —\u00a0 T=Mg\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>=<\/b><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub>N\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub>N = m<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>g + Mg (I)\u00a0 —\u00a0 equil\u00edbrio na horizontal\u00a0 —\u00a0 F<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>=N (II)\u00a0 —\u00a0 (II0 em (I)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub>F<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>\u00a0= m<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>g + Mg\u00a0 —\u00a0 Mg =\u00a0<\/b><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub>F<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>\u00a0\u2013 m<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>g\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0M= (<\/b><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub>F<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>\u00a0\u2013 m<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>g)\/g ou M=<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub>F<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>\/g – m<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0—\u00a0 para esse valor de M o conjunto est\u00e1 na imin\u00eancia de cair\u00a0 —\u00a0 para qualquer valor de M menor que esse, o conjunto n\u00e3o cai.<\/b><\/span><\/span><\/p>\n

37- <\/b><\/span><\/span><\/span>C\u00e1lculo de d na figura 1 —\u00a0 tri\u00e2ngulo ret\u00e2ngulo \u2013 Pit\u00e1goras\u00a0 —\u00a0 d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= 30<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 40<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 d=50cm\u00a0 —\u00a0 c\u00e1lculo de h<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 tri\u00e2ngulo amarelo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

da figura 1\u00a0 —\u00a0 sen\u03b1=h<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/30\u00a0 —\u00a0 tri\u00e2ngulo inteiro\u00a0 —\u00a0 sen\u03b1=40\/d\u00a0 —\u00a0 h<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/30=40\/50\u00a0 —\u00a0 h<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=24cm\u00a0 —\u00a0 aplicando Pit\u00e1goras no tri\u00e2ngulo amarelo da figura 2\u00a0 —\u00a0 35<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= (d\/2)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ h<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 1225 = 25<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ h<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 h=\u221a(600)\u00a0 —\u00a0 h\u224824,5cm\u00a0 —\u00a0 o trabalho da for\u00e7a peso \u00e9 fornecido pela express\u00e3o\u00a0 —\u00a0 W<\/b><\/span><\/span>peso<\/b><\/span><\/span><\/sub>=m.g.\u2206h=10.10.(0,245 \u2013 0,24)\u00a0 —\u00a0 W<\/b><\/span><\/span>peso<\/b><\/span><\/span><\/sub>=0,5J\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/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 Ponto Material \u00a0 01-\u00a0Sobre o homem atuam tr\u00eas for\u00e7as, seu peso (vertical e para baixo) e as duas for\u00e7as de tra\u00e7\u00e3o aplicadas pelo teto atrav\u00e9s dos fios, agindo sobre suas m\u00e3os \u00a0—\u00a0 R- A 02-\u00a0A parede da direita aplica em B, sobre a esfera 2 uma for\u00e7a horizontal e para<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1480,"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-1484","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1484","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=1484"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1484\/revisions"}],"predecessor-version":[{"id":10875,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1484\/revisions\/10875"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1480"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1484"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}