{"id":3294,"date":"2017-10-01T00:18:49","date_gmt":"2017-10-01T00:18:49","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3294"},"modified":"2024-09-04T14:38:52","modified_gmt":"2024-09-04T14:38:52","slug":"resolucoes-estatica-2013-e-2014","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/vestibulares-recentes\/mecanica\/estatica-2014-2013\/resolucoes-estatica-2013-e-2014\/","title":{"rendered":"Resolu\u00e7\u00f5es – Est\u00e1tica – 2013 e 2014"},"content":{"rendered":"

Est\u00e1tica \u2013 2013 \u2013 2014<\/b><\/span><\/span><\/span><\/p>\n

RESOLU\u00c7\u00d5ES<\/b><\/span><\/span><\/span><\/p>\n

01-\u00a0<\/b><\/span><\/span><\/span>Equil\u00edbrio na vertical<\/b><\/span><\/span><\/span>\u00a0—\u00a0<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0+ N<\/b><\/span><\/span><\/span>B\u00a0<\/b><\/span><\/span><\/sub><\/span>= P.<\/b><\/span><\/span><\/span><\/p>\n

Equil\u00edbrio de rota\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span>— a\u00a0<\/b><\/span><\/span><\/span>soma dos momentos de cada for\u00e7a em rela\u00e7\u00e3o ao polo (colocado em A)<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

deve ser nula.<\/b><\/span><\/span><\/span><\/p>\n

M<\/b><\/span><\/span><\/span>NA<\/b><\/span><\/span><\/sub><\/span>=N<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.0=0 — M<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>= + P.1= p — M<\/b><\/span><\/span><\/span>NB<\/b><\/span><\/span><\/sub><\/span>= – N<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.3= – 3N<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0— 0 + P \u2013 3N<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0= 0<\/b><\/span><\/span><\/span>\u00a0—\u00a0<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0= P\/3 (I).<\/b><\/span><\/span><\/span><\/p>\n

N<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0+ N<\/b><\/span><\/span><\/span>B\u00a0<\/b><\/span><\/span><\/sub><\/span>= 3 — N<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0+ P\/3 =\u00a0<\/b><\/span><\/span><\/span>P — N<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=P \u2013 P\/3= (3P \u2013 P)\/3\u00a0<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=2.(P\/3) (II).<\/b><\/span><\/span><\/span><\/p>\n

Comparando (I) com (II) —\u00a0<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0= 2 N<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A componente do peso\u00a0<\/b><\/span><\/span><\/span>(P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>)\u00a0<\/b><\/span><\/span><\/span>paralela ao plano inclinado, respons\u00e1vel pela descida do bloco, \u00e9 fornecida por P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=Psen\u04e8=m.g.sen\u04e8=2,8,10,0,8 —\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=22,4N\u00a0<\/b><\/span><\/span><\/span>— intensidade da\u00a0<\/b><\/span><\/span><\/span>for\u00e7a de atrito est\u00e1tico m\u00e1xima\u00a0<\/b><\/span><\/span><\/span>(imin\u00eancia de movimento) \u00e9 dada por F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=\u03bcN=\u03bcPcos\u04e8=\u03bcmgcos\u04e8 — F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=0,5×2,8x10x0,6\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=8,4N\u00a0<\/b><\/span><\/span><\/span>—<\/b><\/span><\/span><\/span><\/p>\n

a intensidade da for\u00e7a el\u00e1stica (F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>) \u00e9 fornecida por F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=K.x, onde x \u00e9 de quanto a mola est\u00e1 distendida —<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=350.x\u00a0<\/b><\/span><\/span><\/span>— essas for\u00e7as est\u00e3o colocadas na figura abaixo — como, pelo enunciado, o sistema se<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

encontra em\u00a0<\/b><\/span><\/span><\/span>equil\u00edbrio<\/b><\/span><\/span><\/span>, a for\u00e7a resultante sobre o bloco \u00e9 nula\u00a0<\/b><\/span><\/span><\/span>(F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=0)\u00a0<\/b><\/span><\/span><\/span>e, assim P<\/b><\/span><\/span><\/span>p\u00a0<\/b><\/span><\/span><\/sub><\/span>= F<\/b><\/span><\/span><\/span>el<\/b><\/span><\/span><\/sub><\/span>\u00a0+ F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>\u00a0— 22,4 = 350 + 8,4 — 350.x = 14 — x=14\/350 —\u00a0<\/b><\/span><\/span><\/span>x=<\/b><\/span><\/span><\/span>0,04m<\/b><\/span><\/span><\/span>=4,0cm\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

03-\u00a0<\/b><\/span><\/span><\/span>Teoria:\u00a0<\/b><\/span><\/span><\/span>Plano inclinado com atrito\u00a0<\/b><\/span><\/span><\/span><\/p>\n

As\u00a0<\/b><\/span><\/span><\/span>for\u00e7as que agem\u00a0<\/b><\/span><\/span><\/span>sobre um corpo apoiado sobre um plano inclinado s\u00e3o\u00a0<\/b><\/span><\/span><\/span>seu peso\"\", vertical e para baixo e a for\u00e7a normal\u00a0\"\", perpendicular \u00e0 superf\u00edcie de contato entre o bloco e o plano.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u03b1\u00a0\u00e9 o \u00e2ngulo de inclina\u00e7\u00e3o do plano.\u00a0<\/b><\/span><\/span><\/span>Como\u00a0\"\"\u00a0e\u00a0\"\"\u00a0n\u00e3o tem a mesma dire\u00e7\u00e3o, vamos\u00a0<\/b><\/span><\/span><\/span>decompor o peso\u00a0\"\"\u00a0em duas parcelas:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"\u00a0\u00a0<\/span>parcela do peso paralela \u00e0 superf\u00edcie do plano inclinado\u00a0<\/b><\/span><\/span><\/span>(respons\u00e1vel pela tentativa de descida do bloco).<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>parcela do peso perpendicular \u00e0 superf\u00edcie do plano inclinado\u00a0<\/b><\/span><\/span><\/span>(for\u00e7a que comprime o bloco contra o plano)<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

sen<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u03b1 =cateto oposto\/hipotenusa\u00a0 —\u00a0 sen<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u03b1 =P<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>\/P\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=P.sen\u03b1<\/b><\/span><\/span><\/span><\/p>\n

cos<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u03b1 =cateto adjacente\/hipotenusa\u00a0 —\u00a0 cos<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u03b1 =P<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>\/P\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>=P.cos\u03b1<\/b><\/span><\/span><\/span><\/p>\n

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

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

As\u00a0<\/b><\/span><\/span><\/span>duas for\u00e7as acima\u00a0<\/b><\/span><\/span><\/span>substituem o peso e\u00a0<\/b><\/span><\/span><\/span>podemos tir\u00e1-lo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A componente\u00a0<\/b><\/span><\/span><\/span>normal do peso\u00a0\"\"\u00a0anula a rea\u00e7\u00e3o normal do apoio\u00a0\"\",\u00a0<\/b><\/span><\/span><\/span>assim temos:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A for\u00e7a de atrito \u00e9 sempre contr\u00e1ria ao movimento ou \u00e0 sua tend\u00eancia e \u00e9 fornecida por F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=\u03bcN\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>=N\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=\u03bcP<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=\u03bcPcosa\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>=\u03bcmgcos\u03b1<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Se\u00a0<\/b><\/span><\/span><\/span>o bloco estiver em equil\u00edbrio (est\u00e1tico ou din\u00e2mico), a for\u00e7a resultante sobre ele seria nula e, nesse caso,\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>at\u00a0<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Para se\u00a0<\/b><\/span><\/span><\/span>determinar o coeficiente de atrito est\u00e1tico (F<\/b><\/span><\/span><\/span>ate<\/b><\/span><\/span><\/sub><\/span>) de um corpo, que \u00e9 o valor m\u00e1ximo do coeficiente de atrito<\/b><\/span><\/span><\/span>, devemos colocar esse corpo sobre um plano inclinado e ir aumentando o \u00e2ngulo at\u00e9 que ele esteja\u00a0<\/b><\/span><\/span><\/span>na imin\u00eancia de movimento e medir esse \u00e2ngulo.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Na\u00a0<\/b><\/span><\/span><\/span>imin\u00eancia de movimento P<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>ate<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 Psen\u03b1=\u03bcePcos\u03b1\u00a0 —\u00a0\u00a0\u03bc<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=sena\/cos\u03b1\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=tg\u03b1<\/b><\/span><\/span><\/span><\/p>\n

N0 caso do exerc\u00edcio —\u00a0<\/b><\/span><\/span><\/span>\u03bc<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=tg\u03b1<\/b><\/span><\/span><\/span>=tg30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0<\/b><\/span><\/span><\/span>= \u221a3\/3<\/b><\/span><\/span><\/span><\/p>\n

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

04-\u00a0<\/b><\/span><\/span><\/span>Veja a figura abaixo onde foi calculada a\u00a0<\/b><\/span><\/span><\/span>soma vetorial pela regra do paralelogramo\u00a0<\/b><\/span><\/span><\/span>— aplicando<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

a lei dos cossenos — F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.F.F.cos60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=2F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.(1\/2) — F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=\u221a(3F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>) —\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=\u221a(3) .F<\/b><\/span><\/span><\/span><\/p>\n

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

05-\u00a0<\/b><\/span><\/span><\/span>Na primeira figura est\u00e3o colocadas as for\u00e7as e, considerando a prateleira de densidade homog\u00eanea o peso e todas as outras for\u00e7as est\u00e3o aplicadas no centro (meio) da prateleira e essas for\u00e7as s\u00e3o:<\/b><\/span><\/span><\/span><\/p>\n

Peso\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0— vertical, para baixo e de intensidade P=mg=2.10 —\u00a0<\/b><\/span><\/span><\/span>P=20N<\/b><\/span><\/span><\/span><\/p>\n

Tens\u00e3o (tra\u00e7\u00e3o)\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0no cabo aplicada na prateleira pela parede<\/b><\/span><\/span><\/span><\/p>\n

Normal\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>— aplicada pela parede na prateira (horizontal e para a direita)<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Na segunda figura a\u00a0<\/b><\/span><\/span><\/span>tra\u00e7\u00e3o\u00a0\"\"\u00a0foi decomposta em duas parcelas\u00a0<\/b><\/span><\/span><\/span>de intensidades\u00a0<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span>Tcos60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=<\/b><\/span><\/span><\/span>0,5T\u00a0<\/b><\/span><\/span><\/span>e\u00a0<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span>Tsen60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=(\u221a3\/2)T<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Na terceira figura, como o sistema est\u00e1 em\u00a0<\/b><\/span><\/span><\/span>equil\u00edbrio<\/b><\/span><\/span><\/span>, na vertical,\u00a0<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=P\u00a0<\/b><\/span><\/span><\/span>— 0,5T=20 —\u00a0<\/b><\/span><\/span><\/span>T=40N<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

Obs: Se fosse pedido a rea\u00e7\u00e3o normal\u00a0\"\"\u00a0aplicada pela parede na prateira voc\u00ea teria que na horizontal tamb\u00e9m existe equil\u00edbrio e N=F<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=(\u221a3\/2)T=(\u221a3\/2).40 —\u00a0<\/b><\/span><\/span><\/span>N=20\u221a3N<\/b><\/span><\/span><\/span><\/p>\n

06-\u00a0<\/b><\/span><\/span><\/span>Peso de cada esfera<\/b><\/span><\/span><\/span>\u00a0— P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>g=0,3.10 —\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=3N\u00a0<\/b><\/span><\/span><\/span>— P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>g=0,2.10 —\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=2N\u00a0<\/b><\/span><\/span><\/span>— P<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>g=0,1.10 —\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=1N<\/b><\/span><\/span><\/span><\/p>\n

Observe na\u00a0<\/b><\/span><\/span><\/span>figura todas as for\u00e7as que agem sobre cada esfera\u00a0<\/b><\/span><\/span><\/span>e, como cada uma est\u00e1\u00a0<\/b><\/span><\/span><\/span>em equil\u00edbrio\u00a0<\/b><\/span><\/span><\/span>a\u00a0<\/b><\/span><\/span><\/span>for\u00e7a resultante sobre cada uma ser\u00e1 nula<\/b><\/span><\/span><\/span>, ent\u00e3o voc\u00ea ter\u00e1:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Esfera 3<\/b><\/span><\/span><\/span>\u00a0—\u00a0<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=1N\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>esfera 2\u00a0<\/b><\/span><\/span><\/span>— T<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 2 = T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0— 1 + 2=T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=3N\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>esfera 1\u00a0<\/b><\/span><\/span><\/span>— T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 3 —<\/b><\/span><\/span><\/span><\/p>\n

T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=3 + 3 —\u00a0<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=6N<\/b><\/span><\/span><\/span><\/p>\n

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

07-\u00a0<\/b><\/span><\/span><\/span>Se\u00a0<\/b><\/span><\/span><\/span>o peso da crian\u00e7a (II) \u00e9 duas vezes maior que o da crian\u00e7a (I),\u00a0<\/b><\/span><\/span><\/span>a crian\u00e7a (II), para que haja equil\u00edbrio, deve estar a uma dist\u00e2ncia duas vezes menor que a crian\u00e7a (I)\u00a0<\/b><\/span><\/span><\/span>do apoio.<\/b><\/span><\/span><\/span><\/p>\n

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

08-\u00a0<\/b><\/span><\/span><\/span>Veja teoria a seguir:<\/b><\/span><\/span><\/span><\/p>\n

Se voc\u00ea\u00a0<\/b><\/span><\/span><\/span>pendurar um quadro na parede por meio de dois fios<\/b><\/span><\/span><\/span>,\u00a0<\/b><\/span><\/span><\/span>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.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

09-<\/b><\/span><\/span><\/span>Lei de Hooke — F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=kx\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>k=60N\/m\u00a0<\/b><\/span><\/span><\/span>e no dia de menor rendimento\u00a0<\/b><\/span><\/span><\/span>x=0,8m\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=60.0,8<\/b><\/span><\/span><\/span>=48N.<\/b><\/span><\/span><\/span><\/p>\n

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

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

O1. Correta\u00a0<\/b><\/span><\/span><\/span>— possui\u00a0<\/b><\/span><\/span><\/span>maior \u00e1rea de contato\u00a0<\/b><\/span><\/span><\/span>com o solo oferecendo mais estabilidade para caminhar.<\/b><\/span><\/span><\/span><\/p>\n

02. Correta\u00a0<\/b><\/span><\/span><\/span>— o salto alto\u00a0<\/b><\/span><\/span><\/span>desloca o peso\u00a0<\/b><\/span><\/span><\/span>e consequentemente o\u00a0<\/b><\/span><\/span><\/span>centro de gravidade<\/b><\/span><\/span><\/span>. Veja figura<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

04. Falsa\u00a0<\/b><\/span><\/span><\/span>— veja 02.<\/b><\/span><\/span><\/span><\/p>\n

08. Correta\u00a0<\/b><\/span><\/span><\/span>— com\u00a0<\/b><\/span><\/span><\/span>salto o peso \u00e9 deslocado para a frente\u00a0<\/b><\/span><\/span><\/span>e\u00a0<\/b><\/span><\/span><\/span>sem salto, num plano inclinado o peso<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

tamb\u00e9m \u00e9 deslocado para a frente.<\/b><\/span><\/span><\/span><\/p>\n

16. Falsa\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>press\u00e3o=for\u00e7a (no caso, o peso)\/\u00e1rea de contato com o solo) —\u00a0<\/b><\/span><\/span><\/span>P=P\/S\u00a0<\/b><\/span><\/span><\/span>— observe que a\u00a0<\/b><\/span><\/span><\/span>press\u00e3o P \u00e9 inversamente proporcional \u00e0 \u00e1rea de contato\u00a0<\/b><\/span><\/span><\/span>— com o sapato B,\u00a0<\/b><\/span><\/span><\/span>S \u00e9 menor, e a press\u00e3o sobre o solo \u00e9 maior.<\/b><\/span><\/span><\/span><\/p>\n

Soma (01 + 02 + 08)= 11<\/b><\/span><\/span><\/span><\/p>\n

11-\u00a0<\/b><\/span><\/span><\/span>Como os blocos se movem com\u00a0<\/b><\/span><\/span><\/span>velocidade constante<\/b><\/span><\/span><\/span>, est\u00e3o em\u00a0<\/b><\/span><\/span><\/span>MRU<\/b><\/span><\/span><\/span>\u00a0e a intensidade da\u00a0<\/b><\/span><\/span><\/span>for\u00e7a resultante sobre cada um deles \u00e9 nula.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Bloco de\u00a0<\/b><\/span><\/span><\/span>massa M\u00a0<\/b><\/span><\/span><\/span>— T=P —\u00a0<\/b><\/span><\/span><\/span>T=Mg (I)<\/b><\/span><\/span><\/span><\/p>\n

Bloco de\u00a0<\/b><\/span><\/span><\/span>massa m\u00a0<\/b><\/span><\/span><\/span>— P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>\u00a0= T + F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>\u00a0— mgsen\u03b8 = T + \u03bc<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>mgcos\u03b8 —\u00a0<\/b><\/span><\/span><\/span>T=mgsen\u03b8 – \u03bc<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>mgcos\u03b8 (II)<\/b><\/span><\/span><\/span><\/p>\n

(I) = (II)\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>Mg = mgsen\u03b8 – \u03bc<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>mgcos\u03b8 —\u00a0<\/b><\/span><\/span><\/span>M= msen\u03b8 – \u03bc<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>mcos\u03b8<\/b><\/span><\/span><\/span><\/p>\n

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

12-\u00a0<\/b><\/span><\/span><\/span>Observe as\u00a0<\/b><\/span><\/span><\/span>figuras abaixo\u00a0<\/b><\/span><\/span><\/span>onde as for\u00e7as s\u00e3o:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"\u00a0<\/sub><\/span>\u2013\u00a0<\/span>for\u00e7a que a\u00a0<\/b><\/span><\/span><\/span>m\u00e3o aplica aplica na parede<\/b><\/span><\/span><\/span>\u00a0e\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>\u2013<\/b><\/span><\/span><\/span>\u00a0for\u00e7a com que a\u00a0<\/b><\/span><\/span><\/span>parede reage na m\u00e3o e consequentemente na pessoa.<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>\u2013\u00a0<\/span>peso da pessoa\u00a0<\/b><\/span><\/span><\/span>cuja rea\u00e7\u00e3o est\u00e1 no centro da Terra.<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>\u2013\u00a0<\/span>for\u00e7a que a\u00a0<\/b><\/span><\/span><\/span>pessoa aplica no solo\u00a0<\/b><\/span><\/span><\/span>e\u00a0\"\"\u00a0– for\u00e7a com que o\u00a0<\/b><\/span><\/span><\/span>solo reage sobre a pessoa\u00a0<\/b><\/span><\/span><\/span>(<\/b><\/span><\/span><\/span>essa for\u00e7a \u00e9 denominada de for\u00e7a normal\u00a0\"\"<\/b><\/span><\/span><\/span>)<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>\u2013\u00a0<\/span>for\u00e7a com que o\u00a0<\/b><\/span><\/span><\/span>p\u00e9 empurra o solo\u00a0<\/b><\/span><\/span><\/span>e\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>for\u00e7a com que o\u00a0<\/b><\/span><\/span><\/span>solo reage sobre o p\u00e9\u00a0<\/b><\/span><\/span><\/span>(essas for\u00e7as s\u00e3o as\u00a0<\/b><\/span><\/span><\/span>for\u00e7a de atrito\u00a0<\/b><\/span><\/span><\/span>trocadas entre o p\u00e9 e o solo).<\/b><\/span><\/span><\/span><\/p>\n

Na\u00a0<\/b><\/span><\/span><\/span>figura da direita\u00a0<\/b><\/span><\/span><\/span>est\u00e3o colocadas apenas as\u00a0<\/b><\/span><\/span><\/span>for\u00e7as que agem sobre o indiv\u00edduo e que se\u00a0<\/b><\/span><\/span><\/span>anulam na vertical e na horizontal, pois o indiv\u00edduo encontra-se em repouso (<\/b><\/span><\/span><\/span>for\u00e7a resultante nula<\/b><\/span><\/span><\/span>).<\/b><\/span><\/span><\/span><\/p>\n

Est\u00e1tica de um ponto material<\/b><\/span><\/span><\/span><\/p>\n

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

38<\/b><\/span><\/span><\/span>–\u00a0Determinando\u00a0 o valor da constante el\u00e1stica k da mola pela figura 1\u00a0 —\u00a0 quando x=4,0cm=4.10<\/span><\/span><\/span>-2<\/span><\/span><\/sup><\/span>m, F=20N\u00a0 —\u00a0 F<\/span><\/span><\/span>e<\/span><\/span><\/sub><\/span>=kx\u00a0 —\u00a0 20=k.4.10<\/span><\/span><\/span>-2\u00a0\u00a0<\/span><\/span><\/sup><\/span>—\u00a0 k=5.10<\/span><\/span><\/span>2\u00a0<\/span><\/span><\/sup><\/span>N\/m\u00a0 —\u00a0 for\u00e7a aplicada pela mola quando o sistema est\u00e1 em equil\u00edbrio e a mola deformada de x=6,4cm=6,4.10<\/span><\/span><\/span>-2<\/span><\/span><\/sup><\/span>m\u00a0 —\u00a0 F=kx=5.10<\/span><\/span><\/span>2<\/span><\/span><\/sup><\/span>.6,4.10<\/span><\/span><\/span>-2<\/span><\/span><\/sup><\/span>=32N\u00a0 —\u00a0 observe a seq\u00fc\u00eancia das figuras abaixo onde<\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

foram colocadas todas as for\u00e7as no ponto de interse\u00e7\u00e3o dos fios (I)\u00a0 —\u00a0 em (II) foram decompostas as for\u00e7as inclinadas\u00a0 —\u00a0 T<\/span><\/span><\/span>x<\/span><\/span><\/sub><\/span>=Tcos\u03b1=0,8T\u00a0 —\u00a0 T<\/span><\/span><\/span>y<\/span><\/span><\/sub><\/span>=Tsen\u03b1=0,6T\u00a0 —\u00a0 F<\/span><\/span><\/span>x<\/span><\/span><\/sub><\/span>=Fcos\u03b2=32.0,6=19,2N\u00a0 —\u00a0 F<\/span><\/span><\/span>y<\/span><\/span><\/sub><\/span>=Fsen\u03b2=32.0,8=25,6N\u00a0 —\u00a0 em (III) est\u00e3o colocadas apenas as for\u00e7as que agem nas dire\u00e7\u00f5es horizontal e vertical\u00a0 —\u00a0 estando o sistema em equil\u00edbrio essas for\u00e7as se anulam em cada dire\u00e7\u00e3o\u00a0 —\u00a0 na horizontal\u00a0 —\u00a0 0,8T=19,2\u00a0 —\u00a0 T=24N\u00a0 —\u00a0 na vertical\u00a0 —\u00a0 0,6T + 25,6 = 10m\u00a0 —\u00a0<\/span><\/span><\/span><\/p>\n

0,6.24 + 25,6 = 10m\u00a0 —\u00a0 14,4 + 25,6 = 10m\u00a0 —\u00a0 m=40\/10=4kg\u00a0 —\u00a0\u00a0<\/span><\/span><\/span>R- C.<\/b><\/span><\/span><\/span><\/p>\n

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

Corpo extenso \u2013 centro de massa e tipos de equil\u00edbrio<\/b><\/span><\/span><\/span><\/p>\n

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

\u00a021-<\/b><\/span><\/span><\/span>\u00a0Com duas pessoas no carro a deforma\u00e7\u00e3o do sistema de molas \u00e9 x=2cm=2.10<\/span><\/span><\/span>-2<\/span><\/span><\/sup><\/span>m e\u00a0 a for\u00e7a que deformadora \u00e9 o peso das duas pessoas\u00a0 —\u00a0 P=2mg=2.60.10=12.10<\/span><\/span><\/span>2<\/span><\/span><\/sup><\/span>N\u00a0 —\u00a0 esse peso corresponde \u00e0 for\u00e7a el\u00e1stica\u00a0 —\u00a0 F<\/span><\/span><\/span>e<\/span><\/span><\/sub><\/span>P=kx\u00a0 —\u00a0<\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

12.10<\/span><\/span><\/span>2<\/span><\/span><\/sup><\/span>=k.2.10<\/span><\/span><\/span>-2<\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 k=6.10<\/span><\/span><\/span>4<\/span><\/span><\/sup><\/span>N\/m (constante el\u00e1stica do sistema de molas que \u00e9 constante independente da for\u00e7a deformadora)\u00a0 —\u00a0 com\u00a0 4 pessoas a deforma\u00e7\u00e3o \u00e9 x\u2019 e o peso \u00e9 P\u2019=4mg=4.70.10=28.10<\/span><\/span><\/span>2<\/span><\/span><\/sup><\/span>N\u00a0 —\u00a0\u00a0 P\u2019=kx\u2019\u00a0 —\u00a0 28.10<\/span><\/span><\/span>2<\/span><\/span><\/sup><\/span>=6.10<\/span><\/span><\/span>4<\/span><\/span><\/sup><\/span>x\u2019\u00a0 —\u00a0 x\u2019=4,66.10<\/span><\/span><\/span>-2<\/span><\/span><\/sup><\/span>m=4,66cm\u00a0 —\u00a0\u00a0R- D.<\/span><\/span><\/span><\/p>\n

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

22-<\/b><\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>Leia a teoria abaixo:<\/span><\/span><\/span><\/p>\n

Associa\u00e7\u00e3o de polias ou roldanas<\/span><\/span><\/span><\/p>\n

a) Uma polia fixa e outra m\u00f3vel<\/span><\/span><\/span><\/p>\n

A polia de cima, fixa, n\u00e3o interfere no valor da for\u00e7a aplicada pela pessoa, serve apenas para sua comodidade, levantando o bloco ao puxar o fio para baixo.<\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A polia de baixo, m\u00f3vel, reduz \u00e0 metade a for\u00e7a aplicada pela pessoa (metade do peso do bloco). Lembre-se de que, se a pessoa puxar o fio de uma distancia d, o bloco subir\u00e1 d\/2.<\/span><\/span><\/span><\/p>\n

\u00a0b) Uma polia fixa e v\u00e1rias polias m\u00f3veis (talha exponencial)<\/span><\/span><\/span><\/p>\n

Na figura abaixo, onde temos 3 polias m\u00f3veis e uma fixa, o bloco de peso P \u00e9 mantido em equil\u00edbrio pela pessoa. Observe que<\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

a for\u00e7a que a pessoas aplica tem intensidade 8 vezes menor que o peso do bloco e que cada polia m\u00f3vel reduz pela metade a for\u00e7a aplicada nela. Esse tipo de associa\u00e7\u00e3o \u00e9 chamado de talha exponencial e a for\u00e7a exercida pela pessoa, se tivermos\u00a0n polias m\u00f3veis, corresponde a 2n\u00a0do peso do bloco com n=1,2,3… .<\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Assim, se o bloco da figura acima tiver peso de 80N, a pessoa deve fazer uma for\u00e7a de apenas 10N para mant\u00ea-lo em equil\u00edbrio, mas, se ele puxar a corda de 1m, o bloco subir\u00e1 apenas 0,125m (8 vezes menor).<\/span><\/span><\/span><\/p>\n

Observe na figura do exerc\u00edcio que existem duas polias m\u00f3veis\u00a0 —\u00a0 F=P\/2<\/span><\/span><\/span>2<\/span><\/span><\/sup><\/span>=P\/4\u00a0 —\u00a0\u00a0R- D.<\/span><\/span><\/span><\/p>\n

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

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

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

66-<\/b><\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>Trata-se de uma alavanca interfixa\u00a0\u2013 o apoio est\u00e1 entre a for\u00e7a potente e a for\u00e7a resistente\u00a0 —\u00a0 observe a alavanca<\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0interfixa acima\u00a0 —\u00a0 havendo equil\u00edbrio, a soma dos momentos das for\u00e7as em rela\u00e7\u00e3o ao apoio deve ser nula\u00a0 —\u00a0 F<\/span><\/span><\/span>o<\/span><\/span><\/sub><\/span>.2x=Fr.x\u00ad\u00a0 \u00ad—\u00a0 F<\/span><\/span><\/span>r<\/span><\/span><\/sub><\/span>\/F<\/span><\/span><\/span>o<\/span><\/span><\/sub><\/span>=2x\/x\u00ad\u00a0\u00a0—\u00a0 F<\/span><\/span><\/span>r<\/span><\/span><\/sub><\/span>=2F<\/span><\/span><\/span>p<\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 observe que quanto maior for o bra\u00e7o potente, maior ser\u00e1 a vantagem mec\u00e2nica, ou seja, maior ser\u00e1 a for\u00e7a transmitida\u00a0 —\u00a0\u00a0R- A.<\/span><\/span><\/span><\/p>\n

67-<\/b><\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>O exerc\u00edcio quer o diagrama das for\u00e7as que as dobradi\u00e7as aplicam na porta:<\/span><\/span><\/span><\/p>\n

Na horizontal: ao girar, a porta traciona\u00a0 (puxa) para a direita a dobradi\u00e7a superior e esta, pelo princ\u00edpio da a\u00e7\u00e3o e rea\u00e7\u00e3o reage na porta com uma for\u00e7a de mesma intensidade, mesma dire\u00e7\u00e3o, mas sentido contr\u00e1rio,\u00a0tracionando-a (puxando-a) para a esquerda com uma for\u00e7a\u00a0\"\"\u00a0\u00a0—\u00a0 analogamente ao girar, a porta comprime\u00a0 (empurra) para a esquerda a dobradi\u00e7a superior e esta, pelo princ\u00edpio da a\u00e7\u00e3o e rea\u00e7\u00e3o reage na porta com uma for\u00e7a de mesma intensidade, mesma dire\u00e7\u00e3o mas sentido contr\u00e1rio,\u00a0comprimindo-a (empurrando-a) para a direita com uma for\u00e7a\u00a0\"\"<\/span><\/span><\/span>.<\/span><\/span><\/sub><\/span>\u00a0<\/span><\/span><\/span><\/p>\n

Na vertical,\u00a0devido ao seu peso, a porta aplica em cada dobradi\u00e7a uma for\u00e7a puxando-as para baixo e, elas reagem\u00a0sobre a porta com for\u00e7as\u00a0\"\"\u00a0e\u00a0\"\", verticais e para cima.<\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A soma vetorial dessas for\u00e7as, fornece as for\u00e7as resultantes\u00a0\"\"\u00a0e\u00a0\"\"que cada dobradi\u00e7a exerce sobre a porta (figura)\u00a0\u00a0<\/span><\/span><\/span>R- D.<\/b><\/span><\/span><\/span><\/p>\n

68-<\/b><\/span><\/span><\/span>\u00a0Peso do chumbo quadrado\u00a0 —\u00a0 P=m.g=0,030×10=0,3N\u00a0 —\u00a0 peso dos tri\u00e2ngulos de chumbo\u00a0 —\u00a0 P=m.9=0,010×10=0,1N\u00a0 —\u00a0 colocando as for\u00e7as na haste maior e considerando o ponto O como p\u00f3lo (figura)\u00a0 —\u00a0 equil\u00edbrio de rota\u00e7\u00e3o\u00a0 —\u00a0 a soma dos momentos de cada for\u00e7a em rela\u00e7\u00e3o ao p\u00f3lo O deve ser nula\u00a0 —\u00a0 M<\/span><\/span><\/span>0,3N<\/span><\/span><\/sub><\/span>= – F,d = –<\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

0,3.x\u00a0 —\u00a0 M<\/span><\/span><\/span>0,2N<\/span><\/span><\/sub><\/span>\u00a0= + F.d=+0,2.30 =6N.cm\u00a0 —\u00a0 M<\/span><\/span><\/span>0,3N\u00a0<\/span><\/span><\/sub><\/span>+ M<\/span><\/span><\/span>0,2N<\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 – 0,3x + 6=0\u00a0 —\u00a0 x=6\/0,3\u00a0 —\u00a0 x=20cm\u00a0 —\u00a0\u00a0\u00a0<\/span><\/span><\/span>\u00a0R- C.<\/b><\/span><\/span><\/span><\/p>\n

69<\/span><\/span><\/span>–<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>Se voc\u00ea chamar as rea\u00e7\u00f5es em A e B de N<\/span><\/span><\/span>A<\/span><\/span><\/sub><\/span>\u00a0e N<\/span><\/span><\/span>B<\/span><\/span><\/sub><\/span>, a rea\u00e7\u00e3o de contato da esfera Q na barra de P<\/span><\/span><\/span>Q<\/span><\/span><\/sub><\/span>, e considerar o\u00a0<\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

peso da barra (P<\/span><\/span><\/span>b<\/span><\/span><\/sub><\/span>) atuando no centro geom\u00e9trico da mesma, aplicando a condi\u00e7\u00f5es de equil\u00edbrio de rota\u00e7\u00e3o: somat\u00f3rio de momentos em rela\u00e7\u00e3o ao ponto A (p\u00f3lo) igual a zero, voc\u00ea ter\u00e1\u00a0 —\u00a0\u00a0 M<\/span><\/span><\/span>NA<\/span><\/span><\/sub><\/span>=N<\/span><\/span><\/span>A<\/span><\/span><\/sub><\/span>.d=0\u00a0 —\u00a0 M<\/span><\/span><\/span>pQ<\/span><\/span><\/sub><\/span>=+P<\/span><\/span><\/span>Q<\/span><\/span><\/sub><\/span>.d=+80.0,4=32N.m\u00a0 —\u00a0 M<\/span><\/span><\/span>Pb<\/span><\/span><\/sub><\/span>=+P<\/span><\/span><\/span>b<\/span><\/span><\/sub><\/span>.d\u00a0 — M<\/span><\/span><\/span>Pb<\/span><\/span><\/sub><\/span>=50.1=50N.m\u00a0 —\u00a0 M<\/span><\/span><\/span>NB<\/span><\/span><\/sub><\/span>=- N<\/span><\/span><\/span>B<\/span><\/span><\/sub><\/span>.d=-N<\/span><\/span><\/span>B<\/span><\/span><\/sub><\/span>.2\u00a0 — 0 + 32 + 50 – 2N<\/span><\/span><\/span>B<\/span><\/span><\/sub><\/span>\u00a0= 0\u00a0 —\u00a0 N<\/span><\/span><\/span>B<\/span><\/span><\/sub><\/span>=82\/2\u00a0 —\u00a0 N<\/span><\/span><\/span>B<\/span><\/span><\/sub><\/span>=41N\u00a0 —\u00a0\u00a0<\/span><\/span><\/span>R- B.\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

70<\/b><\/span><\/span><\/span>–<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>Peso do homem\u00a0 —\u00a0 P<\/span><\/span><\/span>H<\/span><\/span><\/sub><\/span>=m.g=80.10=800N\u00a0 —\u00a0 c\u00e1lculo do momento de cada for\u00e7a colocando o polo (eixo <\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

de rota\u00e7\u00e3o) em B\u00a0 —\u00a0 N<\/span><\/span><\/span>NA<\/span><\/span><\/sub><\/span>=N<\/span><\/span><\/span>A<\/span><\/span><\/sub><\/span>.d<\/span><\/span><\/span>AB<\/span><\/span><\/sub><\/span>=2N<\/span><\/span><\/span>A<\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 M<\/span><\/span><\/span>PH<\/span><\/span><\/sub><\/span>= – P<\/span><\/span><\/span>H<\/span><\/span><\/sub><\/span>.d<\/span><\/span><\/span>HB<\/span><\/span><\/sub><\/span>= – 800.1,8= – 1440N.m\u00a0 —\u00a0 M<\/span><\/span><\/span>NB<\/span><\/span><\/sub><\/span>=N<\/span><\/span><\/span>B<\/span><\/span><\/sub><\/span>.d=N<\/span><\/span><\/span>B<\/span><\/span><\/sub><\/span>.0=0\u00a0 —\u00a0 equil\u00edbrio de rota\u00e7\u00e3o\u00a0 —\u00a0 a soma dos momentos de cada for\u00e7a em rela\u00e7\u00e3o ao p\u00f3lo \u00e9 nula\u00a0 —\u00a0 2N<\/span><\/span><\/span>A<\/span><\/span><\/sub><\/span>\u00a0\u2013 1440 + 0 = 0\u00a0 —\u00a0 N<\/span><\/span><\/span>A<\/span><\/span><\/sub><\/span>=1440\/2\u00a0 —\u00a0 N<\/span><\/span><\/span>A<\/span><\/span><\/sub><\/span>=720 N\u00a0 —\u00a0\u00a0<\/span><\/span><\/span>R- D.<\/b><\/span><\/span><\/span><\/p>\n

71-<\/b><\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>Projetando as for\u00e7as e as dist\u00e2ncias nas dire\u00e7\u00f5es perpendiculares \u00e0 viga\u00a0 —\u00a0 em (I), for\u00e7as\u00a0 —\u00a0 cos60<\/span><\/span><\/span>o<\/span><\/span><\/sup><\/span>=F\/80\u00a0 —\u00a0 1\/2 =<\/span><\/span><\/span><\/p>\n

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

F\/80\u00a0 —\u00a0 2F=80\u00a0 —\u00a0 F=40N\u00a0 —\u00a0 em (II), dist\u00e2ncias\u00a0 —\u00a0 cos60<\/span><\/span><\/span>o<\/span><\/span><\/sup><\/span>=2\/d\u00a0 —\u00a0 1\/2 = 2\/d\u00a0 —\u00a0 d=4m\u00a0 —\u00a0 veja na figura (III) como fica o esquema pronto para voc\u00ea calcular o momento de cada for\u00e7a com o p\u00f3lo em A\u00a0 —\u00a0 M<\/span><\/span><\/span>PE<\/span><\/span><\/sub><\/span>= – 40.16= – 640<\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0N.m\u00a0 —\u00a0 M<\/span><\/span><\/span>NB<\/span><\/span><\/sub><\/span>= + N<\/span><\/span><\/span>D<\/span><\/span><\/sub><\/span>x12\u00a0 —\u00a0 M<\/span><\/span><\/span>PC<\/span><\/span><\/sub><\/span>= – 40.8= – 320N\u00a0 —\u00a0 M<\/span><\/span><\/span>F<\/span><\/span><\/sub><\/span>= + 30.4\u00a0 —\u00a0 M<\/span><\/span><\/span>NA<\/span><\/span><\/sub><\/span>=N<\/span><\/span><\/span>A<\/span><\/span><\/sub><\/span>.0=0\u00a0 —\u00a0 a soma dos momentos de cada for\u00e7a deve ser nula\u00a0 —\u00a0 -640 +12N<\/span><\/span><\/span>D<\/span><\/span><\/sub><\/span>\u00a0\u2013 320 +120 + 0=0\u00a0 —\u00a0 12N<\/span><\/span><\/span>D<\/span><\/span><\/sub><\/span>=840\u00a0 —\u00a0 N<\/span><\/span><\/span>D<\/span><\/span><\/sub><\/span>=70N\u00a0 —\u00a0\u00a0<\/span><\/span><\/span>R- C.<\/b><\/span><\/span><\/span><\/p>\n

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

Est\u00e1tica \u2013 2013 \u2013 2014 RESOLU\u00c7\u00d5ES 01-\u00a0Equil\u00edbrio na vertical\u00a0—\u00a0NA\u00a0+ NB\u00a0= P. Equil\u00edbrio de rota\u00e7\u00e3o\u00a0— a\u00a0soma dos momentos de cada for\u00e7a em rela\u00e7\u00e3o ao polo (colocado em A) deve ser nula. MNA=NA.0=0 — MP= + P.1= p — MNB= – NB.3= – 3NB\u00a0— 0 + P \u2013 3NB\u00a0= 0\u00a0—\u00a0NB\u00a0= P\/3 (I). NA\u00a0+ NB\u00a0= 3 — NA\u00a0+ P\/3 =\u00a0P — NA=P \u2013<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3292,"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-3294","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3294","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=3294"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3294\/revisions"}],"predecessor-version":[{"id":11083,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3294\/revisions\/11083"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3292"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3294"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}