{"id":1343,"date":"2015-09-15T15:26:39","date_gmt":"2015-09-15T15:26:39","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1343"},"modified":"2024-08-23T13:11:51","modified_gmt":"2024-08-23T13:11:51","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-forcas-no-movimento-circular","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/dinamica\/forcas-no-movimento-circular-em-trajetorias-curvas\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-forcas-no-movimento-circular\/","title":{"rendered":"For\u00e7as no Movimento Circular"},"content":{"rendered":"

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

For\u00e7as no Movimento Circular<\/b><\/span><\/span><\/span><\/p>\n

\u00a001<\/b><\/span><\/span>– F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=m.W<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.R\u00a0 —\u00a0 F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=1.000.(10)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.20\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=2.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>N\u00a0 R- A<\/b><\/span><\/span><\/p>\n

02<\/b><\/span><\/span>– A for\u00e7a de atrito est\u00e1tico ser\u00e1 m\u00e1xima quando o carro estiver na imin\u00eancia\u00a0 de escorregar para fora da pista e nesse caso F<\/b><\/span><\/span>atem\u00e1ximo<\/b><\/span><\/span><\/sub>=\u03bc<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>N=m<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>P=\u03bc<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>mg\u00a0 —\u00a0 F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>atm\u00e1x<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R=\u03bc<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>mg\u00a0 —\u00a0\u00a0\"\", que \u00e9 a m\u00e1xima velocidade com que ele consegue fazer a curva sem derrapar\u00a0 —\u00a0 V=\u221am.R.g\u00a0 —\u00a0 (15)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=(\u221am.50.10)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 225=500m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>m=0,45\u00a0 R-C<\/b><\/span><\/span><\/p>\n

03<\/b><\/span><\/span>– R- E (vide teoria)<\/b><\/span><\/span><\/p>\n

04<\/b><\/span><\/span>– A velocidade m\u00ednima para que o carv\u00e3o n\u00e3o caia ocorrer\u00e1 quando a for\u00e7a de compress\u00e3o no ponto superior for nula, ou seja, quando N=0\u00a0 —\u00a0 F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=P + N\u00a0 —\u00a0 F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=P + 0\u00a0 —\u00a0 F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=P\u00a0 —\u00a0 m.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R=m.g\u00a0\u00a0 —\u00a0 V=\u221a(R.g)\u00a0 —\u00a0 V=\u221a0,8.10\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V=\u221a8=2\u221a2m\/s\u00a0 R- C<\/b><\/span><\/span><\/p>\n

05<\/b><\/span><\/span>– R-A (vide teoria)<\/b><\/span><\/span><\/p>\n

06<\/b><\/span><\/span>– a<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 a<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=9\/2\u00a0 —\u00a0 a<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=4,5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 R- D<\/b><\/span><\/span><\/p>\n

07<\/b><\/span><\/span>– Como a curva \u00e9 horizontal, a for\u00e7a resultante centr\u00edpeta que \u00e9 radial e dirigida para o centro da circunfer\u00eancia dever\u00e1 ser tamb\u00e9m horizontal. R- A<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

a\u00a0\u00e9 o \u00e2ngulo de inclina\u00e7\u00e3o das asas em rela\u00e7\u00e3o ao plano horizontal\u00a0 —\u00a0\u00a0\"\"\u00a0\u00e9 a for\u00e7a de sustenta\u00e7\u00e3o aplicada pelo ar e que \u00e9 perpendicular \u00e0s asas\u00a0 —\u00a0\u00a0\"\"\u00a0\u00e9 o peso.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

tga=cateto oposto\/cateto adjacente\u00a0 —\u00a0 tga=F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\/P\u00a0 —\u00a0 tga\u03b1=(m.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R)\/mg\u00a0<\/b><\/span><\/span><\/p>\n

\"\"\u00a0\u00a0 —\u00a0\u00a0\u00a0<\/b><\/span><\/span>substituindo os valores\u00a0 —\u00a0 100=\u221a(R.9,5.\u221a3)\u00a0 —\u00a0 10.000=R.9,5.1,7\u00a0 —\u00a0 R=10.000\/16,15\u00a0 —\u00a0 R=619,2m\u00a0 R- A<\/b><\/span><\/span><\/p>\n

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

 <\/p>\n

\"\"<\/p>\n

a) P=m.g\u00a0 —\u00a0 P=0,5.10\u00a0 —\u00a0 P=5N\u00a0 —\u00a0 Da figura da direita\u00a0 —\u00a0\u00a0 cos\u03c6=P\/T\u00a0 —\u00a0 0,5=5\/T\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=10N<\/b><\/span><\/span><\/p>\n

b) Da figura da direita – tgj=F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\/P\u00a0 —\u00a0 tg\u03c6=(mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R)\/mg\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>tg\u03c6=V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/Rg\u00a0\u00a0 —\u00a0 R=Lsen\u03c6\u00a0\u00a0<\/b><\/span><\/span>—\u00a0\u00a0<\/b><\/span><\/span>tg\u03c6=V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/L.sen\u03c6.g<\/b><\/span><\/span>\u00a0 — em fun\u00e7\u00e3o da velocidade angular W=V\/R\u00a0 —\u00a0 V=W.R\u00a0 —\u00a0 sen\u03c6\/cos\u03c6=W<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R.g\u00a0 —\u00a0 sen\u03c6\/cos\u03c6=W<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.R\/.g\u00a0 —\u00a0 sen\u03c6\/cos\u03c6=W<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.Lsen\u03c6\/.g\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"\u00a0\u00a0\u00a0\u00a0 —\u00a0\u00a0 substituindo os valores\u00a0 —\u00a0 W=\u221a(10\/1,25.0,5)\u00a0 —\u00a0 W=\u221a16\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 W=4rad\/s<\/b><\/span><\/span><\/p>\n

10<\/b><\/span><\/span>– Sendo o movimento circular a for\u00e7a resultante sobre a mosca \u00e9 a centr\u00edpeta de dire\u00e7\u00e3o radial, sentido para o centro da circunfer\u00eancia e intensidade Fc<\/b><\/span><\/span>=<\/b><\/span><\/span><\/sub>m<\/b><\/span><\/span>.<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R. Como m e V s\u00e3o constantes,F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0e R s\u00e3o inversamente proporcionais, ou seja, quanto menor \u00e9 o raio R, maior ser\u00e1 a intensidade da for\u00e7a resultante centr\u00edpeta F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>. R- D\u00a0<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

 <\/p>\n

No tri\u00e2ngulo hachurado\u00a0 —\u00a0 tg\u03b8=cateto oposto\/cateto adjacente\u00a0 —\u00a0 tg\u03b8=F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\/P\u00a0 —\u00a0 tg\u03b8=(mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R)\/mg\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=R.g.tg\u03b8 V=\u221aR.g.tg\u03b8\u00a0 —\u00a0 m\u00e1xima velocidade que o carro deve ter para efetuar a curva sem atrito\u00a0 —\u00a0 V=\u221a200.10.0,2\u00a0 —\u00a0V=20m\/s que dividido por 3,6 nos fornece\u00a0<\/b><\/span><\/span>V=72km\/h\u00a0 R- B<\/b><\/span><\/span><\/p>\n

12<\/b><\/span><\/span>– A m\u00e1xima velocidade com que o carro consegue realizar a curva sem perder contato com o solo ocorre quando a for\u00e7a de<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

compress\u00e3o\u00a0\"\"\u00a0entre os pneus do carro e o solo \u00e9 nula, ou seja, N=0\u00a0 —\u00a0 F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=P-N\u00a0 —\u00a0 mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R=mg \u2013 0\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R=g\u00a0 —\u00a0 <\/b><\/span><\/span>V=\u221aR.g \u00a0\u00a0—\u00a0\u00a0<\/b><\/span><\/span>10=\u221aR.10\u00a0 —\u00a0 (10)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=(\u221aR.10)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 100=10R\u00a0\u00a0<\/b><\/span><\/span>\u00a0\u00a0—\u00a0\u00a0<\/b><\/span><\/span>R=10m\u00a0 R- C<\/b><\/span><\/span><\/p>\n

13<\/b><\/span><\/span>– a for\u00e7a resultante centr\u00edpeta\u00a0\u00a0\"\", que \u00e9 dirigida sempre para o centro da circunfer\u00eancia, no caso, para<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

cima, \u00e9 a diferen\u00e7a vetorial entre\u00a0\"\"\u00a0e\u00a0\u00a0\"\", de intensidade\u00a0 —\u00a0 F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=N \u2013 P\u00a0 —\u00a0\u00a0 m.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R= N \u2013 m.g\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>N=m.g + mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span><\/p>\n

N=50.10 + 50.100\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>N=3.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>N<\/b><\/span><\/span><\/p>\n

14<\/b><\/span><\/span>– Ponto A – lombada<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

A for\u00e7a resultante centr\u00edpeta que \u00e9 sempre dirigida para o centro da circunfer\u00eancia que \u00e9 para baixo, ou seja, FC=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0\u00a0\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

P \u2013 N=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0\u00a0 —\u00a0\u00a0 mg \u2013 N = mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0\u00a0 N=2.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10 \u2013 2.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.100\/100\u00a0 —\u00a0 N= 20.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0\u2013 2.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>N=18.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>N<\/b><\/span><\/span>\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

Ponto B \u2013 depress\u00e3o<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

A for\u00e7a resultante centr\u00edpeta que \u00e9 sempre dirigida para o centro da circunfer\u00eancia<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>que \u00e9 para cima, ou seja, F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0\u00a0\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

N \u2013 P=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 N=20.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>N=22.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>N<\/b><\/span><\/span><\/p>\n

Ponto C \u2013 horizontal\u00a0 —\u00a0 N=P\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>N=20.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>N<\/b><\/span><\/span><\/p>\n

Observe que N<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>>N<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>>N<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/p>\n

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

01- Falsa \u2013 em qualquer movimento circular, uniforme ou n\u00e3o, existe sempre uma for\u00e7a resultante centr\u00edpeta.<\/b><\/span><\/span><\/p>\n

02- Correta \u2013 vide 01<\/b><\/span><\/span><\/p>\n

04- Correta, a for\u00e7a resultante centr\u00edpeta tem sempre dire\u00e7\u00e3o radial e sentido para o centro da trajet\u00f3ria.<\/b><\/span><\/span><\/p>\n

08-\u00a0 Correta. Observe nas figuras abaixo que\u00a0\"\"=\"\".<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Tg45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\/P\u00a0 —\u00a0 1=F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\/10.000X10\u00a0\u00a0 —\u00a0\u00a0 F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=100.000N<\/b><\/span><\/span><\/p>\n

16- F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=m.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 100.000=10.000.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/1.000\u00a0 —\u00a0 V=\u00d610<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V=100X3,6=<\/b><\/span><\/span>360km\/h.<\/b><\/span><\/span><\/p>\n

R: 02 + 04 + 08 + 16=30<\/b><\/span><\/span><\/p>\n

16<\/b><\/span><\/span>– Para que o astronauta permane\u00e7a na mesma posi\u00e7\u00e3o \u201ccolado\u201d \u00e0 nave, ele deve ter a mesma velocidade angular W que ela e tamb\u00e9m o mesmo per\u00edodo T (tempo que ela demora para efetuar uma volta completa), que \u00e9 de 20s.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

T=20s\u00a0 —\u00a0 W=2<\/b><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span>\/T\u00a0 —\u00a0 W=2<\/b><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span>\/20=p\/10\u00a0 —\u00a0 W=\u03c0\/10rad\/s\u00a0 —\u00a0 F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=m.W<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.R\u00a0 —\u00a0 F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=70.(\u03c0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/100).10\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=7\u03c0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>N<\/b><\/span><\/span><\/p>\n

17<\/b><\/span><\/span>– a) O \u201cpeso\u201d do astronauta \u00e9 percebido pela rea\u00e7\u00e3o normal\u00a0\"\"\u00a0das paredes da nave sobre ele, que \u00e9 a pr\u00f3pria for\u00e7a resultante centr\u00edpeta de intensidade N=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R ou N=m.W<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.R, que nesse caso \u00e9 seu pr\u00f3prio peso, ou seja, N=P=m.g.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Primeiro andar \u2013 raio R\u00a0 —\u00a0 m.g=m.W<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.R\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>W=\u221a(g\/R)<\/b><\/span><\/span><\/p>\n

b) ) O \u201cpeso\u201d do astronauta \u00e9 percebido pela rea\u00e7\u00e3o normal\u00a0\"\"\u00a0das paredes da nave sobre ele\u00a0 —\u00a0 N=F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=m.W<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.(R-h)\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

 <\/p>\n

\"\"<\/p>\n

N=m.(\u221ag\/R)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.(R-h)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>N=m.g.(R-h)\/R\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>Observe na express\u00e3o N=m.W<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.R que se a nave girar com W constante (que \u00e9 a mesma para todos os andares) e como a massa do astronauta \u00e9 a mesma, a \u201cgravidade\u201d N \u00e9 diretamente proporcional ao raio R. Assim, a medida que o astronauta se aproxima do centro C de rota\u00e7\u00e3o do sistema a \u201cgravidade\u201d vai diminuindo at\u00e9 se anular no centro C, onde o astronauta tem sensa\u00e7\u00e3o de aus\u00eancia de peso (imponderabilidade).<\/b><\/span><\/span><\/p>\n

18<\/b><\/span><\/span>– a) A velocidade vetorial\u00a0\"\"tem sempre o sentido do movimento e \u00e9 sempre tangente \u00e0 trajet\u00f3ria em cada ponto.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

\"\"<\/p>\n

b) Do enunciado F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=1,5GMm\/R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>, mas sua intensidade tamb\u00e9m \u00e9 F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 m.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R=1,5.G.M.m\/R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V=\u221a(1,5.G.M)\/R<\/b><\/span><\/span><\/p>\n

19<\/b><\/span><\/span>– a) Como o ponto de entrada do fregu\u00eas coincide com o ponto de sa\u00edda, o tempo m\u00ednimo de perman\u00eancia ocorre quando o restaurante efetua um volta completa, o que ocorre no per\u00edodo T\u00a0 —\u00a0 W=2\u03c0\/T\u00a0\u00a0 —\u00a0\u00a0\u00a0p\/1800=2\u03c0\/T\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=3.600s ou T=1h<\/b><\/span><\/span><\/p>\n

b) F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=m.W<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.R=50.(\u00a0p\/1800)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.20\u00a0 —\u00a0 F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=50.(3,14)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.20\/11(1.800)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=3.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>N<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

b) Somando vetorialmente\u00a0\"\"com\u00a0\"\"\u00a0obtemos a for\u00e7a resultante centr\u00edpeta\u00a0\"\", conforme a figura abaixo.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Do tri\u00e2ngulo hachurado\u00a0 —\u00a0 tg45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\/P\u00a0 —\u00a0 1=(m.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R)\/m.g\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=R.g<\/b><\/span><\/span>\u00a0 —\u00a0 sen45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\/N\u00a0 —\u00a0\u00a0\u221a2\/2=(m.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R)\/N\u00a0 —\u00a0 \u221a2\/2=m.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R.N\u00a0\u00a0 —\u00a0\u00a0\u221a2\/2=m.R.g.\/R.N\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>N=\u221a2.m.g<\/b><\/span><\/span><\/p>\n

21<\/b><\/span><\/span>– a) Tra\u00e7\u00e3o e peso. A tra\u00e7\u00e3o sobre o fio no sentido do ponto fixo, enquanto que o peso \u00e9 vertical para baixo, conforme a figura abaixo.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

A soma vetorial da for\u00e7a de tra\u00e7\u00e3o no fio e a for\u00e7a peso nos fornece a for\u00e7a resultante centr\u00edpeta.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

No tri\u00e2ngulo da direita\u00a0 —\u00a0 tga=F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\/P=m.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/m.R.g\u00a0 —\u00a0 1,2\/1,6=V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/1,2.10\u00a0 —\u00a0 V=\u221a9\u00a0 —\u00a0<\/b><\/span><\/span>V=3m\/s<\/b><\/span><\/span><\/p>\n

sen\u03b1=F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>\/P\u00a0 —\u00a0 0,6=(m.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R)\/T\u00a0 —\u00a0 0,6= m.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R.T\u00a0 —\u00a0 0,6=0,4.9\/1,2.T\u00a0 —\u00a0 0,72T=3,6\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=5N<\/b><\/span><\/span><\/p>\n

22<\/b><\/span><\/span>–<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>O per\u00edodo corresponde ao tempo que ele demora para efetuar uma volta completa\u00a0 —\u00a0 regra de tr\u00eas\u00a0 —<\/b><\/span><\/span><\/p>\n

60 voltas \u2013 30s<\/b><\/span><\/span><\/p>\n

1 volta\u00a0\u00a0 –\u00a0\u00a0 T<\/b><\/span><\/span><\/p>\n

60T=30\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 T=1\/2s<\/b><\/span><\/span>\u00a0 —\u00a0 W=2\u03c0\/T\u00a0 —\u00a0 W=(2\u03c0)\/1\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>W=4\u03c0rad\/s\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 Como n\u00e3o h\u00e1 movimento vertical, peso e normal se anulam e, na horizontal a tra\u00e7\u00e3o no fio \u00e9 a pr\u00f3pria forca resultante centr\u00edpeta\u00a0 —\u00a0 F<\/b><\/span><\/span>C\u00ad<\/b><\/span><\/span><\/sub>=T=m.W<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.R=0,5.(4\u03c0)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.1\u00a0 —\u00a0 T=0,5.16.10.1\u00a0 —\u00a0<\/b><\/span><\/span>T=80N\u00a0 R- B<\/b><\/span><\/span><\/p>\n

23<\/b><\/span><\/span>–\u00a0<\/b><\/span><\/span>a)\u00a0<\/b><\/span><\/span>2,0<\/b><\/span><\/span>.<\/b><\/span><\/span><\/sup>10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>b)\u00a0<\/b><\/span><\/span>2,6<\/b><\/span><\/span>.<\/b><\/span><\/span><\/sup>10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>c)<\/b><\/span><\/span>\u00a03,0<\/b><\/span><\/span>.<\/b><\/span><\/span><\/sup>10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>d)<\/b><\/span><\/span>2,0<\/b><\/span><\/span>.<\/b><\/span><\/span><\/sup>10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>e)\u00a0<\/b><\/span><\/span>3,0<\/b><\/span><\/span>.<\/b><\/span><\/span><\/sup>10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.<\/b><\/span><\/span><\/p>\n

Como ele est\u00e1 em movimento circular existe uma for\u00e7a\u00a0<\/b><\/span><\/span>resultante centr\u00edpeta que \u00e9 sempre dirigida para o centro da circunfer\u00eancia\u00a0<\/b><\/span><\/span>que \u00e9 para cima, ou seja, F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 ou\u00a0 F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=mW<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.R\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>N \u2013 P=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 ou N \u2013 P=mW<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>R.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

N \u2013 2.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.10=2.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.(30)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/3.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>N=2,6.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>N<\/b><\/span><\/span><\/p>\n

24<\/b><\/span><\/span>– R- B (vide exerc\u00edcio 13)<\/b><\/span><\/span><\/p>\n

25<\/b><\/span><\/span>– Como o movimento \u00e9 circular e uniforme, a for\u00e7a centr\u00edpeta\u00a0\"\"=\"\"\u00a0e a acelera\u00e7\u00e3o centr\u00edpeta\u00a0\"\"=\"\"\u00a0 tem a mesma dire\u00e7\u00e3o<\/b><\/span><\/span><\/p>\n

(radial) e o mesmo sentido (para o centro da circunfer\u00eancia) e ambas s\u00e3o sempre perpendiculares ao vetor velocidade\u00a0\"\", que \u00e9 sempre tangente \u00e0 trajet\u00f3ria em cada ponto e tem o sentido do movimento. R- D<\/b><\/span><\/span><\/p>\n

26<\/b><\/span><\/span>– Nesse caso, a for\u00e7a centr\u00edpeta \u00e9 o F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0que evita com que ele escorregue e saia pela tangente. A for\u00e7a de atrito est\u00e1tico ser\u00e1 m\u00e1xima quando o carro estiver na imin\u00eancia\u00a0 de escorregar para fora da pista e nesse caso F<\/b><\/span><\/span>atem\u00e1ximo<\/b><\/span><\/span><\/sub>=\u03bc<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>N=m<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>P=\u03bc<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>mg\u00a0 —\u00a0 F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>atm\u00e1x<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R=\u03bc<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>mg\u00a0 —\u00a0\u00a0\"\", que \u00e9 a m\u00e1xima velocidade com que ele consegue fazer a curva sem derrapar e,<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

nesse caso, o coeficiente de atrito deve ser m\u00ednimo\u00a0 —\u00a0 20=\u221a\u03bc<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>.100.10\u00a0 —\u00a0 400=1.000.\u03bc<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span>\u03bc<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=0,4\u00a0 R-\u00a0<\/b><\/span><\/span>E\u00a0<\/b><\/span><\/span><\/p>\n

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

 <\/p>\n

\"\"<\/p>\n

figura 1 \u2013 repouso \u2013 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=P\u00a0\u00a0\u00a0 —\u00a0\u00a0\u00a0 figura 2 \u2013 F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u2013 P\u00a0 —\u00a0 T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u2013 P = m.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=P + m.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>>T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/p>\n

28<\/b><\/span><\/span>– A velocidade m\u00ednima para que o sistema n\u00e3o caia ocorrer\u00e1 quando a compress\u00e3o no ponto A for nula, ou seja, quando N=0\u00a0 —\u00a0 F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=P + N\u00a0 —\u00a0 F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=P + 0 \u00a0—\u00a0 F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=P\u00a0 —\u00a0 m.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R=m.g\u00a0 —\u00a0 V=\u221aR.g\u00a0 —\u00a0 velocidade m\u00ednima do sistema (moto + motociclista) no ponto mais alto para fazer a curva e n\u00e3o perder contato com a superf\u00edcie interna do globo, ou seja, n\u00e3o cair.<\/b><\/span><\/span><\/p>\n

V=\u221a4,9.10\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V=7m\/s\u00a0 R- C<\/b><\/span><\/span><\/p>\n

29<\/b><\/span><\/span>– No ponto mais alto do morro as for\u00e7as que agem sobre os dois carros s\u00e3o o peso\u00a0\"\"\u00a0e a for\u00e7a de contato com o solo\u00a0\"\".<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Na imin\u00eancia de perder contato com o solo, N=0\u00a0 —\u00a0 F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=P \u2013 N\u00a0 —\u00a0 P \u2013 N=m.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 P \u2013 0=m.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 mg=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 — V=\u221aR.g\u00a0 —\u00a0 verifique pela f\u00f3rmula que essa velocidade n\u00e3o depende da massa\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

30<\/b><\/span><\/span>– R- C (veja exerc\u00edcio 27)<\/b><\/span><\/span><\/p>\n

31-\u00a0<\/b><\/span><\/span>a e b) Observe a figura abaixo\u00a0 —\u00a0 T \u2013 intensidade da for\u00e7a de tra\u00e7\u00e3o no fio\u00a0 —\u00a0 P \u2013 intensidade do peso do acrobata\u00a0 —\u00a0 F<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/p>\n

\"\"<\/p>\n

\u00a0 \u2013 intensidade da for\u00e7a resultante centr\u00edpeta\u00a0 —\u00a0 sen45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=P\/T\u00a0 —\u00a0 Tsen45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=P\u00a0 —\u00a0 T.\u221a2\/2=mg\u00a0 —\u00a0 cos45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\/T\u00a0 —\u00a0 F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=Tcos45<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 — mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R=T.\u221a2\/2\u00a0 —\u00a0 mg=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R=g\u00a0 —\u00a0 V=\u221aRg=\u221a4,9.10\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>V=7m\/s\u00a0 —\u00a0 uma volta completa com velocidade de 7m\/s\u00a0 —\u00a0 V=\u0394S\/T\u00a0 —\u00a0 7=2\u03c0R\/T\u00a0 —\u00a0 7=2.3,14.4,9\/T\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T\u22484,4s<\/b><\/span><\/span><\/p>\n

c) Sabemos que T.cos45\u00b0 = m.g\u00a0 —\u00a0 T.\u221a2\/2 = 60.10\u00a0 —\u00a0 T.0,71 = 600\u00a0 —\u00a0\u00a0\u00a0<\/b><\/span><\/span>T \u2248845 N<\/b><\/span><\/span><\/p>\n

Nas condi\u00e7\u00f5es do teste de seguran\u00e7a\u00a0 —\u00a0 3.T = M<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.g\u00a0 —\u00a0 M<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0= 3.845\/10\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0M<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u2248253,5 kg<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

32-\u00a0\"\"<\/b><\/span><\/span>\u00a0Observe a a figura da direita abaixo\u00a0 —\u00a0 R=4 + 10sen\u03b8=4 + 10.1\/2\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R=9m<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"\u00a0Observe a figura da esquerda acima\u00a0 —\u00a0 na vertical\u00a0 —\u00a0 Tcos\u03b8=Mg\u00a0 —\u00a0 Tcos30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=1000.10\u00a0 —\u00a0 T=10.000\/0.87\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T=11.494N<\/b><\/span><\/span><\/p>\n

\"\"A for\u00e7a resultante centr\u00edpeta F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>, horizontal, \u00e9 a componente Tsen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=Tsen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=11.494.1\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=5.747N\u00a0<\/b><\/span><\/span>
\n<\/b><\/span><\/span>33-\u00a0<\/b><\/span><\/span>O chamado estado de imponderabilidade ocorre exatamente porque todos os corpos dentro da esta\u00e7\u00e3o espacial e a pr\u00f3pria esta\u00e7\u00e3o espacial est\u00e3o sujeitos a mesma acelera\u00e7\u00e3o\u00a0 —\u00a0 o que valida a alternativa A\u00a0 —\u00a0 como existe acelera\u00e7\u00e3o (gravitacional) da Terra sobre a esta\u00e7\u00e3o e tudo mais que ela cont\u00e9m, o astronauta possui peso, o que invalida a alternativa B\u00a0 —\u00a0 alternativa c, absurda\u00a0 —\u00a0\u00a0 o estado do astronauta, bem como de tudo mais da esta\u00e7\u00e3o n\u00e3o \u00e9 inercial, pois, como j\u00e1 dito, existe acelera\u00e7\u00e3o sobre os corpos em \u00f3rbita \u00a0—\u00a0 a velocidade do astronauta pode at\u00e9 ser momentaneamente maior ou menor que a velocidade da pr\u00f3pria esta\u00e7\u00e3o, se ele se desloca dentro dela. Como ele flutua, no mesmo local da esta\u00e7\u00e3o, subentende-se que a velocidade dele \u00e9 a da esta\u00e7\u00e3o, o que invalida a alternativa E.<\/b><\/span><\/span><\/p>\n

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

34-\u00a0<\/b><\/span><\/span>O objeto solto bem como o pr\u00f3prio sat\u00e9lite est\u00e1 sujeito \u00e0 for\u00e7a gravitacional terrestre e logo ambos t\u00eam peso\u00a0 —\u00a0 admitindo que o movimento do sat\u00e9lite e do corpo s\u00e3o circulares e uniformes, a acelera\u00e7\u00e3o centr\u00edpeta ser\u00e1 a acelera\u00e7\u00e3o gravitacional\u00a0 —\u00a0 a alternativa c \u00e9 falsa, pois se uma pessoa estiver solta dentro da nave ela n\u00e3o experimenta sensa\u00e7\u00e3o de peso.<\/b><\/span><\/span><\/p>\n

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

35-\u00a0<\/b><\/span><\/span>Conforme o diagrama anexo, as for\u00e7as que agem no carro s\u00e3o o peso\u00a0\"\"\u00a0e a normal\u00a0\"\". Como o movimento \u00e9 circular e uniforme, a resultante dessas for\u00e7as \u00e9 centr\u00edpeta (radial),\u00a0\"\".<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

tga = R<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\/P=ma<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\/mg\u00a0 —\u00a0 a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=gtg<\/b><\/span><\/span>\u03b1<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 como a e g s\u00e3o constantes, a acelera\u00e7\u00e3o centr\u00edpeta (radial, dirigida para o centro) tem m\u00f3dulo constante.\u00a0<\/b><\/span><\/span><\/p>\n

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

36-\u00a0<\/b><\/span><\/span>As for\u00e7as que atuam sobre o (autom\u00f3vel + motorista) est\u00e3o indicadas na figura\u00a0 —\u00a0 a for\u00e7a resultante centr\u00edpeta \u00e9 vertical e para<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

cima, pois \u00e9 dirigida para o centro da circunfer\u00eancia\u00a0 e tem intensidade F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 N \u2013 P=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 N=860.10 + 860.(20)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/20\u00a0 —\u00a0 N=25.800N\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

37-\u00a0<\/b><\/span><\/span>O enunciado sugere tratar-se de um p\u00eandulo c\u00f4nico ou p\u00eandulo composto\u00a0 —\u00a0 as figuras abaixo ilustram a situa\u00e7\u00e3o descrita\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Na Fig 1\u00a0 —\u00a0 h<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 +\u00a0 a<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 = l<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>\u00a0—\u00a0 h<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= l<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 a<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 h =\"\"\u00a0(I)\u00a0 —\u00a0 sen\u03b8=a\/l\u00a0 —\u00a0 cos\u03b8=h\/l\u00a0 —\u00a0 na Fig 2\u00a0 —\u00a0 a componente horizontal da tra\u00e7\u00e3o (F<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>) \u00e9 a resultante centr\u00edpeta (F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 F<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>\u00a0= F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 F sen q = mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/a\u00a0 —\u00a0 F.(a\/l)=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/a\u00a0 —\u00a0\u00a0 F =mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>l\/a<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>(II)\u00a0 —\u00a0 como o movimento se d\u00e1 num plano horizontal, a resultante das for\u00e7as verticais \u00e9 nula\u00a0 —\u00a0 assim, a componente vertical da tra\u00e7\u00e3o (F<\/b><\/span><\/span>y<\/b><\/span><\/span><\/sub>) equilibra o peso da esfera pendular (P)\u00a0 —\u00a0 F<\/b><\/span><\/span>y<\/b><\/span><\/span><\/sub>\u00a0= P\u00a0 —\u00a0\u00a0 F cos q = m g\u00a0 —\u00a0 F.(h\/l) = m g\u00a0 —\u00a0 F = mgl\/h(III)\u00a0 —\u00a0 igualando (II) e (III)\u00a0 —\u00a0 mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>l\/a<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=mgl\/h\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=a<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>g\/h (IV)\u00a0 —\u00a0 oper\u00edodo (T) \u00e9 o intervalo de tempo gasto em cada volta (DS = 2pR=2\u03c0a)\u00a0 —\u00a0 \u00a0<\/b><\/span><\/span><\/p>\n

v = \u0394S\/\u0394t=2\u03c0a \/T (V)\u00a0 —\u00a0 substituindo (V) em (IV)\u00a0 —\u00a0 (2\u03c0a\/T)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=a<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>g\/h\u00a0 —\u00a0 4\u03c0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>a<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=a<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>g\/h\u00a0 —\u00a0 T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=4\u03c0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>h\/g\u00a0 —\u00a0 T=2\u03c0\u221ah\/g (VI)\u00a0 –<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

38-\u00a0<\/b><\/span><\/span>Observe a figura abaixo que mostra as for\u00e7as que agem sobre a pessoa:\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

01) Correta\u00a0 —\u00a0 \u00a0for\u00e7a normal\u00a0\"\"\u00a0\u00e9 sempre perpendicular a superf\u00edcie de apoio, conforme ilustra a figura acima. Nesse caso ela \u00e9 dirigida para o centro, portanto \u00e9 uma for\u00e7a centr\u00edpeta.
\n02) Falsa\u00a0 —\u00a0 como a pessoa efetua movimento circular uniforme, na dire\u00e7\u00e3o horizontal a normal age como resultante centr\u00edpeta\u00a0\"\"\u00a0e, na dire\u00e7\u00e3o vertical, a for\u00e7a de atrito\u00a0\"\"\u00a0deve equilibrar o peso. O piso somente deve ser retirado quando a for\u00e7a de atrito est\u00e1tica m\u00e1xima for maior ou igual ao peso, caso contr\u00e1rio a pessoa escorrega pelas paredes\u00a0 —\u00a0 N=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0\u00b3 P\u00a0 —\u00a0 m N \u00b3 m g\u00a0 —\u00a0 inserindo nessa express\u00e3o a express\u00e3o anterior\u00a0 —\u00a0 \u03bcmV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\"\"mg\u00a0 —\u00a0 \u03bc\"\"Rg\/V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V\"\"\u221a(Rg)\/\u03bc\u00a0 —\u00a0 nessa express\u00e3o, vemos que a massa da pessoa n\u00e3o interfere e que a velocidade m\u00ednima com que o piso pode ser\u00a0 retirado depende apenas do raio do rotor da intensidade do campo gravitacional local e do coeficiente de atrito entre as roupas da pessoa e a parede do rotor.<\/b><\/span><\/span><\/p>\n

04) Verdadeira\u00a0 — \u00a0conforme demonstra\u00e7\u00e3o no item anterior.<\/b><\/span><\/span><\/p>\n

08) Falsa\u00a0 —\u00a0 o coeficiente de atrito depende apenas das caracter\u00edsticas das superf\u00edcies em contato.<\/b><\/span><\/span><\/p>\n

16) Falsa\u00a0 — \u00a0conforme justificativa do item anterior.<\/b><\/span><\/span><\/p>\n

R-(01 + 04) = 05<\/b><\/span><\/span><\/p>\n

39-\u00a0<\/b><\/span><\/span>Dados\u00a0 —\u00a0 \u00a0v = 216 km\/h = 60 m\/s\u00a0 —\u00a0 \u00a0m = 6 kg\u00a0 —\u00a0 \u00a0r = 72 m\u00a0 —\u00a0 a for\u00e7a que o piloto deve exercer sobre o conjunto cabe\u00e7a-capacete \u00e9 a resultante centr\u00edpeta\u00a0 —\u00a0 F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R=6.(60)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/72\u00a0 —\u00a0 F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=300N\u00a0 —\u00a0 para que um corpo tenha esse mesmo peso, quando sujeito \u00e0 gravidade terrestre, sua massa deve ser\u00a0 —\u00a0 P\/g=300\/10\u00a0 —\u00a0 m=30kg\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

40-<\/b><\/span><\/span>\u00a0 Dados\u00a0 — \u00a0h = 2 m\u00a0 —\u00a0 \u00a0g = 9,8 m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u00a0D<\/b><\/span><\/span>ac<\/b><\/span><\/span><\/sub>\u00a0= 1% g = g\/100=0,098 m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 um habitante no interior da nave gira com a mesma velocidade angular (w) que ela\u00a0 —\u00a0 \u00a0\u00a0diferen\u00e7a entre as acelera\u00e7\u00f5es centr\u00edpetas nos p\u00e9s\u00a0\"\"\u00a0e na cabe\u00e7a\u00a0\"\"deve ser igual a 1% da acelera\u00e7\u00e3o da gravidade na Terra.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

P\u00e9s\u00a0 —\u00a0\u00a0\"\"= g\u00a0 —\u00a0 cabe\u00e7a\u00a0 —\u00a0\u00a0\"\"\u00a0\u00a0—\u00a0\u00a0\"\"=g\/10\u00a0 —\u00a0\u00a0\"\"=g\/10\u00a0 —\u00a0\u00a0\"\"=g\/10\u00a0 —\u00a0\u00a0\"\"\u00a0\u00a0—\u00a0 w<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= 0,049\u00a0 —\u00a0 w<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>r = g\u00a0 —\u00a0 0,049 r = 9,8\u00a0 —\u00a0 \u00a0r = 9,8\/0.049\u00a0 —<\/b><\/span><\/span>\u00a0 r=200m<\/b><\/span><\/span><\/p>\n

41-\u00a0<\/b><\/span><\/span>Dados\u00a0 —\u00a0 v<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0= 0\u00a0 —\u00a0 \u00a0m = 1.000 kg\u00a0 —\u00a0 \u00a0H = 30 m\u00a0 —\u00a0 \u00a0d = 20 m\u00a0 —\u00a0 r = 10 m\u00a0 —\u00a0 observe a figura abaixo\u00a0 —\u00a0 c\u00e1lculo da<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0velocidade no ponto B pela conserva\u00e7\u00e3o de energia mec\u00e2nica\u00a0 —\u00a0\u00a0\"\"\u00a0\u00a0—\u00a0 mgH=mgd + mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=2h(H \u2013 d)\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=20.(30 \u2013 20)\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=200\u00a0 —\u00a0 no ponto B, a resultante das for\u00e7as que agem sobre o carro s\u00e3o radiais, portanto a acelera\u00e7\u00e3o \u00e9 centr\u00edpeta\u00a0 —\u00a0 a=a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R=200\/10\u00a0 —<\/b><\/span><\/span>\u00a0a=20m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 no ponto B, a resultante \u00e9 centr\u00edpeta e a for\u00e7a vertical que o trilho exerce no carro \u00e9 a normal\u00a0\"\"\u00a0\u00a0—\u00a0 no ponto B a for\u00e7a resultante centr\u00edpeta tem dire\u00e7\u00e3o vertical e sentido para baixo (centro da circunfer\u00eancia)\u00a0 —\u00a0 F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=N + P=ma\u00a0 —\u00a0 N=ma \u2013 mg\u00a0 —\u00a0 N=1.000(20 \u2013 10)\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0N=10.000N<\/b><\/span><\/span><\/p>\n

42-\u00a0<\/b><\/span><\/span>No ponto mais alto quando N=0\u00a0 —\u00a0 F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 N + P=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 0 + mg=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 V=\u221arg\u00a0 —\u00a0 a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/r=rg\/r\u00a0 —\u00a0 a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=g\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

43-\u00a0<\/b><\/span><\/span>01. Falsa\u00a0 —\u00a0 a intensidade da for\u00e7a de tens\u00e3o em C \u00e9 maior que o peso\u00a0 —\u00a0 existe uma resultante centr\u00edpeta vertical e para cima.<\/b><\/span><\/span><\/p>\n

02- Correta\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

04- Falsa\u00a0 —\u00a0 veja 08<\/b><\/span><\/span><\/p>\n

08- a menor velocidade e a menor energia cin\u00e9tica ocorre quando a tens\u00e3o T \u00e9 nula\u00a0 —\u00a0 T + P=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 mg=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 V=\u221aRg\u00a0 —\u00a0 V=\u221a1.10\u00a0 —\u00a0 V=\u221a10m\/s\u00a0 —\u00a0 E<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=0,5.10\/2\u00a0 —\u00a0 E<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=2,5J<\/b><\/span><\/span><\/p>\n

16- Falsa\u00a0 —\u00a0 T + mg=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 — \u00a0T=0,5.10 + 0,5.900\/2\u00a0 —\u00a0 T=225N<\/b><\/span><\/span><\/p>\n

32- Falsa\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R=0,5.900\/1\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=450N<\/b><\/span><\/span><\/p>\n

R- (2 + 8)=10<\/b><\/span><\/span><\/p>\n

44-\u00a0<\/b><\/span><\/span>A acelera\u00e7\u00e3o centr\u00edpeta \u00e9 dada pela express\u00e3o a<\/b><\/span><\/span>cp<\/b><\/span><\/span><\/sub>\u00a0= v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R. Substituindo os valores para a velocidade de m\u00f3dulo constante do carro, v = 10 km\/h, e para o raio da circunfer\u00eancia, R = 5 m = 0,005 km, obt\u00e9m-se que a<\/b><\/span><\/span>cp<\/b><\/span><\/span><\/sub>\u00a0= 20.000 km\/h<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.<\/b><\/span><\/span><\/p>\n

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

45-\u00a0<\/b><\/span><\/span>W=4.360\/2\u00a0 —\u00a0 W=720\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\/s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

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

Leia atentamente as informa\u00e7\u00f5es a seguir:<\/b><\/span><\/span><\/p>\n

Carro em pista sobrelevada de \u00e2ngulo\u00a0\u03b8 com a horizontal, sem atrito em pista circular de raio R, contida num plano horizontal.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

As duas for\u00e7as que agem sobre o carro, independente do atrito s\u00e3o seu peso\u00a0\"\"\u00a0e a rea\u00e7\u00e3o do solo\u00a0\"\". Para que o carro complete a curva a for\u00e7a resultante centr\u00edpeta\u00a0\u00a0\"\"\u00a0deve ser a soma vetorial de\u00a0\"\"\u00a0com\u00a0\"\"\u00a0e deve ser radial e dirigida para o centro C \u00a0da pista circular de raio R (veja figura abaixo).<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

No tri\u00e2ngulo hachurado\u00a0 —\u00a0 tg\u03b8=cateto oposto\/cateto adjacente\u00a0 —\u00a0 tg\u03b8=F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\/P\u00a0 —\u00a0 tg\u03b8=(mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R)\/mg\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=R.g.tg\u03b8<\/b><\/span><\/span><\/p>\n

\"\"velocidade que o carro deve ter para efetuar a curva sem atrito (ou que seus pneus percam contato (atrito) com a pista).<\/b><\/span><\/span><\/p>\n

V=\u221a(R.g.tg\u03b8)\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=R.g.tg30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 100<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=R.10.0,58\u00a0 —\u00a0 R=10000\/5,8\u00a0 —\u00a0 R=1724,14m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

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

a) Correta\u00a0 —\u00a0 a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R\u00a0 —\u00a0 observe que a acelera\u00e7\u00e3o centr\u00edpeta \u00e9 inversamente proporcional ao raio da pista\u00a0 —\u00a0 menor raio, maior acelera\u00e7\u00e3o centr\u00edpeta.<\/b><\/span><\/span><\/p>\n

b) Falsa\u00a0 —\u00a0 W=V\/R\u00a0 —\u00a0 a velocidade angular depende do raio da pista.<\/b><\/span><\/span><\/p>\n

c) Falsa\u00a0 —\u00a0 A s\u00f3 poderia acompanhar B se eles tivessem a mesma velocidade angular (W=\u2206\u03b8\/\u2206t), ou seja, eles devem \u201cvarrer\u201d o mesmo \u00e2ngulo no mesmo tempo.<\/b><\/span><\/span><\/p>\n

d) e e) Falsas\u00a0 —\u00a0 analise a express\u00e3o da for\u00e7a centr\u00edpeta\u00a0 —\u00a0 F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=m.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R.<\/b><\/span><\/span><\/p>\n

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

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

Em todo movimento circular existe sempre uma for\u00e7a resultante com dire\u00e7\u00e3o radial e sentido dirigido para o centro da circunfer\u00eancia, de intensidade Fc=mV2\/R, denominada for\u00e7a resultante centr\u00edpeta\u00a0 —\u00a0\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 For\u00e7as no Movimento Circular \u00a001– FC=m.W2.R\u00a0 —\u00a0 FC=1.000.(10)2.20\u00a0 —\u00a0\u00a0FC=2.106N\u00a0 R- A 02– A for\u00e7a de atrito est\u00e1tico ser\u00e1 m\u00e1xima quando o carro estiver na imin\u00eancia\u00a0 de escorregar para fora da pista e nesse caso Fatem\u00e1ximo=\u03bceN=meP=\u03bcemg\u00a0 —\u00a0 Fc=Fatm\u00e1x\u00a0 —\u00a0 mV2\/R=\u03bcemg\u00a0 —\u00a0\u00a0, que \u00e9 a m\u00e1xima velocidade com que ele consegue fazer a curva sem<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1339,"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-1343","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1343","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=1343"}],"version-history":[{"count":4,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1343\/revisions"}],"predecessor-version":[{"id":10832,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1343\/revisions\/10832"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1339"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1343"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}