{"id":3691,"date":"2018-02-13T16:38:36","date_gmt":"2018-02-13T16:38:36","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3691"},"modified":"2024-08-20T18:38:43","modified_gmt":"2024-08-20T18:38:43","slug":"resolucao-comentada-estatico-de-um-corpo-extenso","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/enem\/estatico-de-um-corpo-extenso\/resolucao-comentada-estatico-de-um-corpo-extenso\/","title":{"rendered":"Est\u00e1tico de um corpo extenso – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00e3o Comentada<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Est\u00e1tico de um corpo extenso<\/b><\/span><\/span><\/span><\/p>\n

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

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe na figura abaixo que a for\u00e7a\u00a0 peso\u00a0\"\"\u00a0do port\u00e3o colocada em seu centro de gravidade (vertical e para baixo) e a for\u00e7a\u00a0\"\"\u00a0, aplicada pelo garoto na extremidade livre da direita (tamb\u00e9m <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

vertical e para baixo) tendem a fazer o port\u00e3o girar no sentido hor\u00e1rio, fazendo-o puxar a dobradi\u00e7a em A (tentando arranc\u00e1-la) e comprimi-la em B\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>bandeja \u2013 b\u00a0 m<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>=0,4kg\u00a0 —\u00a0 prato -A – \u2013 m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=0,5kg\u00a0 —\u00a0 copo \u2013 B \u2013 m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=0,1kg\u00a0 —\u00a0 garrafa \u2013 C \u2013 m<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=1,0kg\u00a0 — c\u00e1lculo da abscissa do centro de massa\u00a0 —\u00a0 X<\/b><\/span><\/span><\/span>CG<\/b><\/span><\/span><\/sub><\/span>=(m<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>.X<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.XA<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>X<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>X<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>)\/(m<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>)= (0,4..0\u00a0+ 0,5.(-2). + 0,1.(-10) + 1.(+4+)\/(0,4 + 0,5 + 0,1 + 1,0)=2\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>X<\/b><\/span><\/span><\/span>CG<\/b><\/span><\/span><\/sub><\/span>=1cm\u00a0<\/b><\/span><\/span><\/span>\u00a0—\u00a0 c\u00e1lculo do centro de massa da ordenada\u00a0 —\u00a0 Y<\/b><\/span><\/span><\/span>CG<\/b><\/span><\/span><\/sub><\/span>=(m<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>.Y<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.Y<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0+<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>Y<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>Y<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>)\/(m<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>)=(0,4.0 + 0,5.(-5) + 0,1.5 + 1.4)\/2\u00a0 —\u00a0 Y<\/b><\/span><\/span><\/span>CM<\/b><\/span><\/span><\/sub><\/span>=2\/2\u00a0 —\u00a0 Y<\/b><\/span><\/span><\/span>CM<\/b><\/span><\/span><\/sub><\/span>=1cm\u00a0 —\u00a0 CG (1,1)\u00a0 —\u00a0 \u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A figura mostra as abscissas x<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>; x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>; x<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0e x<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>\u00a0e as ordenadas y<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>; y<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>; y<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0e y<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>\u00a0dos quatro corpos; a abscissa X e a ordenada Y do centro de massa:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\u00a0R – C<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Sem as moedas o centro de gravidade do sapinho est\u00e1 em seu pr\u00f3prio corpo\u00a0 —\u00a0 quando voc\u00ea <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

coloca as moedas em suas patas dianteiras e o apoia no l\u00e1pis, essas duas moedas deslocam seu centro de gravidade para baixo de seu corpo e ele fica em equil\u00edbrio est\u00e1vel\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

Observa\u00e7\u00e3o: No boneco \u00a0\u201cJo\u00e3o Bobo\u201d, existe uma grande massa (contrapeso, normalmente chumbo), na parte inferior \u00a0do brinquedo, fazendo com que seu centro de gravidade esteja bem <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

embaixo e ele fique em equil\u00edbrio est\u00e1vel\u00a0 —\u00a0 quando voc\u00ea o retira da posi\u00e7\u00e3o de equil\u00edbrio est\u00e1vel, a tend\u00eancia do contrapeso \u00e9 retornar novamente ao equil\u00edbrio, oscilando at\u00e9 que o contrapeso fique novamente em sua parte inferior.<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como o centro de gravidade CG est\u00e1 sempre no ponto m\u00e9dio e \u00e0 medida que o cigarro vai queimando ele se desloca para a esquerda, quando o cigarro estiver na imin\u00eancia de cair, voc\u00ea ter\u00e1 a situa\u00e7\u00e3o da figura abaixo, onde voc\u00ea deve<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0observar que o cigarro queimou d=80 \u2013 50=30mm com velocidade de 5mm\/min —\u00a0 V=d\/t\u00a0 —\u00a0 5=30\/t\u00a0 —\u00a0t=6 min<\/b><\/span><\/span><\/span><\/p>\n

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

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Trata-se de corpos apoiados\u00a0 —\u00a0 situa\u00e7\u00e3o I\u00a0 —\u00a0 como os tr\u00eas \u00e2ngulos s\u00e3o de 120<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0o centro de gravidade CG est\u00e1 na jun\u00e7\u00e3o \u2013 indiferente\u00a0 —\u00a0 situa\u00e7\u00e3o II\u00a0 —\u00a0 o centro de gravidade CG est\u00e1 acima da <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0jun\u00e7\u00e3o \u2013 inst\u00e1vel\u00a0 —\u00a0 situa\u00e7\u00e3o III\u00a0 — o CG centro de gravidade \u00a0est\u00e1 abaixo da jun\u00e7\u00e3o \u2013 est\u00e1vel\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A porta gira no sentido daquele que aplicar o maior momento na mesma\u00a0 —\u00a0 momento da for\u00e7a aplicada pelo homem\u00a0 — M<\/b><\/span><\/span><\/span>h<\/b><\/span><\/span><\/sub><\/span>=+F<\/b><\/span><\/span><\/span>h<\/b><\/span><\/span><\/sub><\/span>.d\u00a0 =80.10\u00a0 —\u00a0 M<\/b><\/span><\/span><\/span>h<\/b><\/span><\/span><\/sub><\/span>=800N.cm\u00a0 —\u00a0 momento da for\u00e7a aplicada pelo menino\u00a0 —\u00a0 M<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=-F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>.d= -5.200\u00a0 —\u00a0 M<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>= – 1.000N.cm\u00a0 —\u00a0 ganha o menino\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O jovem consegue soltar o parafuso, fazendo a chave girar quando o momento da\u00a0 for\u00e7a por ele <\/b><\/span><\/span><\/span><\/p>\n

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

aplicada (F<\/b><\/span><\/span><\/span>j<\/b><\/span><\/span><\/sub><\/span>=75.10=750N)valer —\u00a0 M<\/b><\/span><\/span><\/span>Fj<\/b><\/span><\/span><\/sub><\/span>=F.d=750.20\u00a0 —\u00a0 M<\/b><\/span><\/span><\/span>Fj<\/b><\/span><\/span><\/sub><\/span>=15.000N.cm\u00a0 —\u00a0 momento da for\u00e7a (F=51.10=510N) aplicada pela namorada\u00a0 —\u00a0 M<\/b><\/span><\/span><\/span>Fn<\/b><\/span><\/span><\/sub><\/span>=F.d=510.30\u00a0 —\u00a0 M<\/b><\/span><\/span><\/span>Fn<\/b><\/span><\/span><\/sub><\/span>=15.300N\u00a0 —\u00a0 observe que o momento (torque) da for\u00e7a aplicada pela namorada \u00e9 maior que o momento da for\u00e7a aplicada pelo jovem\u00a0 —\u00a0 <\/b><\/span><\/span><\/span>sim, ela consegue soltar o parafuso.<\/b><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe a figura abaixo\u00a0 —\u00a0 equil\u00edbrio de transla\u00e7\u00e3o do andaime (corpo extenso)\u00a0 —\u00a0 a resultante <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

das for\u00e7as na vertical deve ser nula,e, para que isso ocorra\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/sub><\/span>\u00a0+ T<\/b><\/span><\/span><\/span>D<\/b><\/span><\/span><\/sub><\/span>=P\u00a0 —\u00a0 equil\u00edbrio de rota\u00e7\u00e3o ( o andaime n\u00e3o deve girar)\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><\/span>E<\/b><\/span><\/span><\/sub><\/span>.(x) \u2013 T<\/b><\/span><\/span><\/span>D<\/b><\/span><\/span><\/sub><\/span>.(y)=0\u00a0 —\u00a0 \u00a0T<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/sub><\/span>.(x)=T<\/b><\/span><\/span><\/span>D<\/b><\/span><\/span><\/sub><\/span>.(y)\u00a0 —\u00a0 observe que, como x>y, T<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/sub><\/span>\u00a0deve ser menor que T\u00ad<\/b><\/span><\/span><\/span>D<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/sub><\/span><T<\/b><\/span><\/span><\/span>D<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/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><\/span><\/p>\n

a) Falsa\u00a0 —\u00a0 a resultante das for\u00e7as na dire\u00e7\u00e3o horizontal \u00e9 n\u00e3o nula, n\u00e3o havendo, portanto, equil\u00edbrio.<\/b><\/span><\/span><\/span><\/p>\n

b) Falsa\u00a0 —\u00a0 a resultante das for\u00e7as na dire\u00e7\u00e3o vertical \u00e9 n\u00e3o nula.<\/b><\/span><\/span><\/span><\/p>\n

c) Correta\u00a0 —\u00a0 a for\u00e7a resultante pode ser nula bem como o momento resultante\u00a0 —\u00a0 trata-se de uma situa\u00e7\u00e3o de poss\u00edvel equil\u00edbrio<\/b><\/span><\/span><\/span><\/p>\n

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

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

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Colocando as for\u00e7as que agem sobre a trave\u00a0 —\u00a0 equil\u00edbrio de rota\u00e7\u00e3o\u00a0 —\u00a0 a somat\u00f3ria dos <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

momentos de todas as for\u00e7as deve ser nula\u00a0 —\u00a0 p\u00f3lo em C\u00a0 —\u00a0 – F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 + F.d \u2013 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0+ F<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>.d<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0= 0\u00a0 —\u00a0 – 400.(0,2 + 0,1 + 0,3)\u00a0 + F.(0,1 + 0,3) \u2013 400.(0,3) + 400.(0,3) = 0\u00a0 —- 400.0,6 + F.0,4 \u2013 400.0,3 + 400.0,3 = 0\u00a0 —\u00a0 F=240\/0,4\u00a0 —\u00a0 F= 600N\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Para que a escada n\u00e3o escorregue e caia, no apoio inferior tem que ter atrito, o que n\u00e3o precisa ocorrer no apoio superior\u00a0 —\u00a0 como a parede \u00e9 lisa, ou seja, no ponto M n\u00e3o existe atrito\u00a0 —\u00a0 a figura abaixo mostra as tr\u00eas for\u00e7as que agem sobre a escada\u00a0 que<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

est\u00e1 em equil\u00edbrio\u00a0 —\u00a0\u00a0\"\"– for\u00e7a de contato com a parede\u00a0 —\u00a0\u00a0\"\"– for\u00e7a que a escada troca com o ch\u00e3o\u00a0 —\u00a0\u00a0\"\"– for\u00e7a peso (ponto m\u00e9dio da escada)\u00a0 —\u00a0 essas tr\u00eas for\u00e7as obedecem ao teorema das tr\u00eas for\u00e7as de enunciado: \u201c Se tr\u00eas for\u00e7as agem sobre um corpo de modo a mant\u00ea-lo em equil\u00edbrio, elas devem ser coplanares (mesmo plano) e concorrentes (seus prolongamentos devem se encontrar) num mesmo ponto, no caso Q\u201d\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe a figura abaixo\u00a0 —\u00a0 colocando o polo em N<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 a soma dos momentos em rela\u00e7\u00e3o ao <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

polo deve ser nula\u00a0 —\u00a0 M<\/b><\/span><\/span><\/span>N2<\/b><\/span><\/span><\/sub><\/span>=—\u00a0M<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=+P.4\u00a0 —\u00a0 M<\/b><\/span><\/span><\/span>N1<\/b><\/span><\/span><\/sub><\/span>=-N<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.6\u00a0 — M<\/b><\/span><\/span><\/span>N1<\/b><\/span><\/span><\/sub><\/span>=-6.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>N.m\u00a0 —\u00a0 +4P \u2013 6.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>=0\u00a0 —\u00a0 P=6.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\/4\u00a0 —\u00a0 P=1,5.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>N\u00a0 —\u00a0 a resultante das for\u00e7as verticais deve ser nula\u00a0 —\u00a0 N<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ N<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=1,5.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 1,0.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0+ N<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=1,5.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 N<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=0,5.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0N<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=5,0.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>N\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe 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><\/span><\/p>\n

\"\"<\/p>\n

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

15-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Colocando o polo em N e calculando, calculando o momento de cada for\u00e7a, somando-os e <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

16-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe a rela\u00e7\u00e3o F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 se d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0> d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0> F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

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

Resolu\u00e7\u00e3o Comentada\u00a0 Est\u00e1tico de um corpo extenso \u00a0 01-\u00a0Observe na figura abaixo que a for\u00e7a\u00a0 peso\u00a0\u00a0do port\u00e3o colocada em seu centro de gravidade (vertical e para baixo) e a for\u00e7a\u00a0\u00a0, aplicada pelo garoto na extremidade livre da direita (tamb\u00e9m vertical e para baixo) tendem a fazer o port\u00e3o girar no sentido hor\u00e1rio, fazendo-o puxar a dobradi\u00e7a em A (tentando arranc\u00e1-la)<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3689,"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-3691","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3691","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=3691"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3691\/revisions"}],"predecessor-version":[{"id":10650,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3691\/revisions\/10650"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3689"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3691"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}