{"id":4891,"date":"2020-01-10T03:15:28","date_gmt":"2020-01-10T03:15:28","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=4891"},"modified":"2024-09-04T14:40:05","modified_gmt":"2024-09-04T14:40:05","slug":"estatica-resolucao-comentada-2018-2019","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/vestibulares-recentes\/mecanica\/estatica-2018-2019\/estatica-resolucao-comentada-2018-2019\/","title":{"rendered":"Est\u00e1tica \u2013 Resolu\u00e7\u00e3o Comentada \u2013 2018\/2019"},"content":{"rendered":"

Est\u00e1tica \u2013 Resolu\u00e7\u00e3o Comentada \u2013 2018\/2019<\/span><\/span><\/span><\/p>\n

01-<\/span><\/span><\/span><\/p>\n

Observe a imagem abaixo:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Em todos os exerc\u00edcios de equil\u00edbrio precisamos\u00a0<\/b><\/span><\/span>verificar os dois tipos \u2013 translacional e rotacional \u2013<\/b><\/span><\/span>\u00a0come\u00e7ando pelo translacional, no caso s\u00f3 em X, pois em Y ela se equilibra com a normal do contato com a quina:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Observe que pelo desenho temos:<\/b><\/span><\/span><\/span><\/span><\/p>\n

PX\u00a0= P.sen\u0398<\/b><\/span><\/span><\/span><\/span><\/p>\n

PY\u00a0= P.cos\u0398<\/b><\/span><\/span><\/span><\/span><\/p>\n

TX\u00a0= T.cos\u0398<\/b><\/span><\/span><\/span><\/span><\/p>\n

TY\u00a0= T.sen\u0398<\/b><\/span><\/span><\/span><\/span><\/p>\n

Agora o equil\u00edbrio rotacional<\/b><\/span><\/span>, lembre-se que, diferente das for\u00e7as, a dist\u00e2ncia ao centro de massa importa:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Ap\u00f3s bastante esfor\u00e7o, a alternativa correta \u00e9 a E.<\/b><\/span><\/span><\/span><\/span><\/p>\n

02-<\/span><\/span><\/span><\/p>\n

Colocando as for\u00e7as que agem sobre cada fio (figura 1) e decompondo-as (figura 2).<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

Observe\u00a0<\/b><\/span><\/span>na figura 3 que as proje\u00e7\u00f5es de T na horizontal de anulam\u00a0<\/b><\/span><\/span>(equil\u00edbrio horizontal).<\/b><\/span><\/span><\/span><\/span><\/p>\n

Proje\u00e7\u00e3o de T na vertical\u00a0\u00a0Ty\u00a0= T.sen30o<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 2k.0,5\u00a0\u00a0<\/b><\/span><\/span>Ty\u00a0= 1 kN.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Equil\u00edbrio na vertical\u00a0\u00a02Ty\u00a0= P\u00a0<\/b><\/span><\/span>\u00a02.1k = P\u00a0\u00a0<\/b><\/span><\/span>P\u00a0<\/b><\/span><\/span>= 2 kN\u00a0<\/b><\/span><\/span>= 2000 N.<\/b><\/span><\/span><\/span><\/span><\/p>\n

P = m.g\u00a0<\/b><\/span><\/span>\u00a02000 = m.10\u00a0\u00a0m = 2000\/10\u00a0\u00a0m = 200 kg.<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

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

Inicialmente as balan\u00e7as marcam o mesmo valor 40 kg\u00a0<\/b><\/span><\/span>que \u00e9\u00a0<\/b><\/span><\/span>igual \u00e0 metade da massa do homem.<\/b><\/span><\/span>\u00a0Isso significa que\u00a0<\/b><\/span><\/span>ele est\u00e1 na metade da plataforma, equidistante 4 m de cada extremidade.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Situa\u00e7\u00e3o depois\u00a0<\/b><\/span><\/span>quando ele\u00a0<\/b><\/span><\/span>se aproxima de uma das extremidades<\/b><\/span><\/span>, por exemplo, a\u00a0<\/b><\/span><\/span>da direita, as balan\u00e7as passam a marcar 20 kg (Pe\u00a0= 20.10 = 200 N) a da esquerda e 60 kg (Pd\u00a0= 60.10 = 600 N) a da direita.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Colocando o\u00a0<\/b><\/span><\/span>polo O na balan\u00e7a da esquerda,\u00a0<\/b><\/span><\/span>calculando<\/b><\/span><\/span>\u00a0m\u00f3dulo do momento de cada for\u00e7a em rela\u00e7\u00e3o ao mesmo e estabelecendo o sentido hor\u00e1rio de rota\u00e7\u00e3o como positivo:<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

Ent\u00e3o,\u00a0<\/b><\/span><\/span>ao se deslocar ele passou da posi\u00e7\u00e3o 4 m\u00a0<\/b><\/span><\/span>(antes)\u00a0<\/b><\/span><\/span>para a posi\u00e7\u00e3o 6 m<\/b><\/span><\/span>\u00a0(depois), se\u00a0<\/b><\/span><\/span>deslocando<\/b><\/span><\/span><\/span><\/span><\/p>\n

S =<\/b><\/span><\/span>\u00a06 \u2013 4\u00a0<\/b><\/span><\/span>= 2m<\/b><\/span><\/span><\/span><\/span><\/p>\n

R \u2013 D<\/b><\/span><\/span><\/span><\/span><\/p>\n

04-<\/span><\/span><\/span><\/p>\n

Colocando as for\u00e7as que agem sobre cada ponto\u00a0<\/b><\/span><\/span><\/span>da arara:<\/b><\/span><\/span><\/span><\/p>\n

Peso da arara colocado no centro de massa da mesma localizado a da\u00a0= 0,5 m de A e 0,5 m de B\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Pa\u00a0= m.g\u00a0<\/b><\/span><\/span><\/span>= 1.10\u00a0<\/b><\/span><\/span><\/span>= 10 N<\/b><\/span><\/span><\/span><\/p>\n

Peso do conjunto 1\u00a0\u00a0P1\u00a0= m1.g\u00a0<\/b><\/span><\/span><\/span>= 1.10\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>P1\u00a0= 10 N com d1\u00a0= 0,1 m de A<\/b><\/span><\/span><\/span><\/p>\n

Peso do conjunto 2\u00a0\u00a0P2\u00a0= m2.g\u00a0<\/b><\/span><\/span><\/span>= 0,5.10\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>P2\u00a0= 5 N com d2\u00a0= 0,3 m de A<\/b><\/span><\/span><\/span><\/p>\n

Peso do conjunto 3\u00a0\u00a0P3\u00a0= m3.g\u00a0<\/b><\/span><\/span><\/span>= 1,5.10\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>P3\u00a0= 15 N com d3\u00a0= 0,8 m de A<\/b><\/span><\/span><\/span><\/p>\n

NA<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0rea\u00e7\u00e3o normal em A com dA\u00a0= 0<\/b><\/span><\/span><\/span><\/p>\n

NB<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0rea\u00e7\u00e3o normal em B com dB\u00a0= 1m de A.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Colocando o polo<\/b><\/span><\/span><\/span>\u00a0(eixo de rota\u00e7\u00e3o)\u00a0<\/b><\/span><\/span><\/span>em A, calculando o momento de cada for\u00e7a em rela\u00e7\u00e3o ao polo\u00a0<\/b><\/span><\/span><\/span>e estabelecendo o\u00a0<\/b><\/span><\/span><\/span>sentido hor\u00e1rio de rota\u00e7\u00e3o em torno de A como positivo:<\/b><\/span><\/span><\/span><\/p>\n

MNA<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= NA.0\u00a0<\/b><\/span><\/span><\/span>= 0<\/b><\/span><\/span><\/span><\/p>\n

MP1\u00a0= P1.d1<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= 10.0,1\u00a0<\/b><\/span><\/span><\/span>= 1 N.m<\/b><\/span><\/span><\/span><\/p>\n

MP2\u00a0= P2.d2<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= 5.0,3\u00a0<\/b><\/span><\/span><\/span>= 1,5 N.m<\/b><\/span><\/span><\/span><\/p>\n

Ma\u00a0= Pa.da<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= 10.0,5\u00a0<\/b><\/span><\/span><\/span>= 5 N.m<\/b><\/span><\/span><\/span><\/p>\n

MP3\u00a0= P3.d3<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= 15.0,8\u00a0<\/b><\/span><\/span><\/span>= 12 N.m<\/b><\/span><\/span><\/span><\/p>\n

MNB\u00a0= \u2013 NB.1 = \u2013 NB<\/b><\/span><\/span><\/span><\/p>\n

No\u00a0<\/b><\/span><\/span><\/span>equil\u00edbrio de rota\u00e7\u00e3o a soma dos momentos de cada for\u00e7a em rela\u00e7\u00e3o ao polo deve ser nula\u00a0<\/b><\/span><\/span><\/span><\/p>\n

0 + 1,0 + 1,5 + 5,0 + 12 \u2013 NB\u00a0= 0\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>NB\u00a0= 19,5 N<\/b><\/span><\/span><\/span><\/p>\n

No\u00a0<\/b><\/span><\/span><\/span>equil\u00edbrio de transla\u00e7\u00e3o<\/b><\/span><\/span><\/span>\u00a0a soma das for\u00e7as para cima de anular a soma das for\u00e7as para baixo\u00a0<\/b><\/span><\/span><\/span><\/p>\n

10 + 5 + 10 + 15 = NA\u00a0+ NB\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>40 = NA\u00a0+ 19,5\u00a0\u00a0NA\u00a0= 40 \u2013 19,5\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>NA\u00a0= 20,5 N<\/b><\/span><\/span><\/span><\/p>\n

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

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

Sendo que o\u00a0<\/b><\/span><\/span><\/span>enunciado afirma que o peso e consequentemente a massa (P = m.g, com g constante) dos objetos (tri\u00e2ngulos)\u00a0<\/b><\/span><\/span><\/span>\u00e9\u00a0<\/b><\/span><\/span><\/span>proporcional \u00e0 sua \u00e1rea,\u00a0<\/b><\/span><\/span><\/span>vamos\u00a0<\/b><\/span><\/span><\/span>calcular a \u00e1rea\u00a0<\/b><\/span><\/span><\/span>de\u00a0<\/b><\/span><\/span><\/span>cada tri\u00e2ngulo.<\/b><\/span><\/span><\/span><\/p>\n

Como o enunciado afirma que a\u00a0<\/b><\/span><\/span><\/span>hipotenusa do tri\u00e2ngulo ret\u00e2ngulo\u00a0<\/b><\/span><\/span><\/span>tem o\u00a0<\/b><\/span><\/span><\/span>mesmo comprimento que os lados do tri\u00e2ngulo equil\u00e1tero, vamos denomina-los de\u00a0.<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo da \u00e1rea S do tri\u00e2ngulo equil\u00e1tero<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

C\u00e1lculo da \u00e1rea S do tri\u00e2ngulo is\u00f3sceles<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

06-<\/span><\/span><\/span><\/p>\n

<\/p>\n

Causa da sustenta\u00e7\u00e3o de um avi\u00e3o<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

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

07-<\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

Equil\u00edbrio na horizontal\u00a0\u00a0N1\u00a0= T.<\/b><\/span><\/span><\/span><\/p>\n

Equil\u00edbrio na vertical\u00a0\u00a0P = N.<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo do\u00a0<\/b><\/span><\/span><\/span>momento de cada for\u00e7a\u00a0<\/b><\/span><\/span><\/span>que \u00e9 o valor da\u00a0<\/b><\/span><\/span><\/span>for\u00e7a multiplicada\u00a0<\/b><\/span><\/span><\/span>pela\u00a0<\/b><\/span><\/span><\/span>dist\u00e2ncia do prolongamento da for\u00e7a at\u00e9 o polo 0 (colocado em B).<\/b><\/span><\/span><\/span><\/p>\n

Com o\u00a0<\/b><\/span><\/span><\/span>polo em B\u00a0<\/b><\/span><\/span><\/span>vamos\u00a0<\/b><\/span><\/span><\/span>calcular o momento de cada for\u00e7a\u00a0<\/b><\/span><\/span><\/span>estabelecento o\u00a0<\/b><\/span><\/span><\/span>sentido hor\u00e1rio d rota\u00e7\u00e3o como positivo:<\/b><\/span><\/span><\/span><\/p>\n

MN1\u00a0= \u2013 N1.AM = \u2013 T.1,5<\/b><\/span><\/span><\/span><\/p>\n

MP\u00a0= + P.BN = 20..cos60o<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= 20..\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span>MP\u00a0= 15 N.m<\/b><\/span><\/span><\/span><\/p>\n

No\u00a0<\/b><\/span><\/span><\/span>equil\u00edbrio de rota\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span>a\u00a0<\/b><\/span><\/span><\/span>soma dos momentos de cada for\u00e7a\u00a0<\/b><\/span><\/span><\/span>em\u00a0<\/b><\/span><\/span><\/span>rela\u00e7\u00e3o ao polo deve ser nula\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\u2013 T.1,5\u00a0+ 15 = 0\u00a0<\/b><\/span><\/span><\/span>\u00a0T =\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>=\u00a0\u00a0=\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span>T =\u00a0\u00a0N.<\/b><\/span><\/span><\/span><\/p>\n

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

08-<\/span><\/span><\/span><\/p>\n

<\/p>\n

09-<\/span><\/span><\/span><\/p>\n

As\u00a0<\/b><\/span><\/span><\/span>for\u00e7as peso do bloco P e a Terra (centro da Terra) constituem par a\u00e7\u00e3o e rea\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span>e as\u00a0<\/b><\/span><\/span><\/span>for\u00e7as trocadas entre a m\u00e3o e o bloco constituem par a\u00e7\u00e3o e rea\u00e7\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n

R- B<\/b><\/span><\/span><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"

Est\u00e1tica \u2013 Resolu\u00e7\u00e3o Comentada \u2013 2018\/2019 01- Observe a imagem abaixo: Em todos os exerc\u00edcios de equil\u00edbrio precisamos\u00a0verificar os dois tipos \u2013 translacional e rotacional \u2013\u00a0come\u00e7ando pelo translacional, no caso s\u00f3 em X, pois em Y ela se equilibra com a normal do contato com a quina: Observe que pelo desenho temos: PX\u00a0= P.sen\u0398 PY\u00a0= P.cos\u0398 TX\u00a0= T.cos\u0398 TY\u00a0= T.sen\u0398<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":4889,"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-4891","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4891","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=4891"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4891\/revisions"}],"predecessor-version":[{"id":4892,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4891\/revisions\/4892"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4889"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=4891"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}