{"id":3285,"date":"2017-10-01T00:14:01","date_gmt":"2017-10-01T00:14:01","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3285"},"modified":"2024-09-04T14:39:30","modified_gmt":"2024-09-04T14:39:30","slug":"resolucoes-estatica-2015-2016-e-2017","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/vestibulares-recentes\/mecanica\/estatica-2017-2016-2015\/resolucoes-estatica-2015-2016-e-2017\/","title":{"rendered":"Resolu\u00e7\u00f5es – Est\u00e1tica – 2015 – 2016 e 2017"},"content":{"rendered":"

Est\u00e1tica de um ponto material 2017 \u2013 2016 – 2015 <\/span><\/span><\/span><\/p>\n

Resolu\u00e7\u00f5es<\/span><\/span><\/span><\/p>\n

 <\/p>\n

01<\/b><\/span><\/span>-Denominando\u00a0<\/b><\/span><\/span>de ml\u00a0a massa de cada cubo laranja e de ma\u00a0a massa de cada cubo azul\u00a0<\/b><\/span><\/span>e, estando as\u00a0<\/b><\/span><\/span>balan\u00e7as em equil\u00edbrio,\u00a0<\/b><\/span><\/span>voc\u00ea ter\u00e1:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Balan\u00e7a\u00a0<\/b><\/span><\/span>da esquerda\"\"2ma\u00a0+ ml\u00a0= 2 (I)<\/b><\/span><\/span><\/span><\/span><\/p>\n

Balan\u00e7a\u00a0<\/b><\/span><\/span>da direita\u00a0\"\"ma\u00a0+ 3 = 2ml\u00a0(II)<\/b><\/span><\/span><\/span><\/span><\/p>\n

Isolando ma\u00a0em (II) e substituindo em (I<\/b><\/span><\/span>)\u00a0\"\"<\/b><\/span><\/span>ma\u00a0= 2ml\u00a0\u2013 3\u00a0\"\"2(2ml\u00a0\u2013 3) + ml\u00a0= 2\u00a0\"\"<\/b><\/span><\/span>\u00a04ml\u00a0\u2013 6 + ml\u00a0= 2\u00a0\"\"5ml\u00a0= 8\u00a0\"\"<\/b><\/span><\/span>ml<\/b><\/span><\/span>= 8\/5\u00a0<\/b><\/span><\/span>= 1,6 kg.<\/b><\/span><\/span><\/span><\/span><\/p>\n

2ma\u00a0+ 1,6 =2\u00a0\"\"<\/b><\/span><\/span>\u00a0ma\u00a0= 0,4\/2\u00a0\"\"<\/b><\/span><\/span>ma\u00a0= 0,2 kg.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Assim, a\u00a0<\/b><\/span><\/span>massa de um cubo laranja supera a de um cubo azul\u00a0<\/b><\/span><\/span>em\u00a0<\/b><\/span><\/span>ml\u00a0\u2013 ma<\/b><\/span><\/span>= 1,6 \u2013 0,2\u00a0<\/b><\/span><\/span>= 1,4 kg.<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

\"\"<\/p>\n

No\u00a0<\/b><\/span><\/span>equil\u00edbrio de transla\u00e7\u00e3o,\u00a0<\/b><\/span><\/span>para que ocorra equil\u00edbrio na\u00a0<\/b><\/span><\/span>vertical a resultante das for\u00e7as\u00a0<\/b><\/span><\/span>deve ser\u00a0<\/b><\/span><\/span>nula,<\/b><\/span><\/span>\u00a0ou seja,\u00a0<\/b><\/span><\/span>NA\u00a0+ NB\u00a0= NN\u00a0\"\"<\/b><\/span><\/span>\u00a075 + 150 = NN\u00a0\"\"<\/b><\/span><\/span>\u00a0NN\u00a0= 225 N<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

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

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

04- <\/b><\/span><\/span><\/span>A)\u00a0<\/b><\/span><\/span>Colocando o\u00a0<\/b><\/span><\/span>polo\u00a0<\/b><\/span><\/span>(eixo de rota\u00e7\u00e3o) no\u00a0<\/b><\/span><\/span>ponto O<\/b><\/span><\/span>, e estabelecendo o <\/b><\/span><\/span>sentido de tend\u00eancia<\/b><\/span><\/span>\u00a0de\u00a0<\/b><\/span><\/span>rota\u00e7\u00e3o em torno de O como positivo para o hor\u00e1rio e negativo para o anti<\/b><\/span><\/span><\/span><\/span><\/p>\n

hor\u00e1rio,\u00a0<\/b><\/span><\/span>vamos\u00a0<\/b><\/span><\/span>calcular<\/b><\/span><\/span>\u00a0o\u00a0<\/b><\/span><\/span>momento (torque) de cada for\u00e7a em rela\u00e7\u00e3o a O,<\/b><\/span><\/span>\u00a0sendo\u00a0<\/b><\/span><\/span>dP\u00a0e dm\u00a0as dist\u00e2ncias entre o ponto de aplica\u00e7\u00e3o das for\u00e7as O at\u00e9 as for\u00e7as FP\u00a0e Fm:<\/b><\/span><\/span><\/span><\/span><\/p>\n

MP\u00a0=<\/b><\/span><\/span><\/span>\u00a0P.d = P.0\u00a0<\/b><\/span><\/span><\/span>= 0<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

MFP<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span>= + FP.dP<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

MFm\u00a0= \u2013 Fm.dm<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span><\/span>condi\u00e7\u00e3o de equil\u00edbrio de rota\u00e7\u00e3o \u00e9 de que a soma dos momentos (torques) de cada for\u00e7a seja nul<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>0 + FP.dP\u00a0\u2013 Fm.dm\u00a0=\u00a0<\/b><\/span><\/span><\/span>0\u00a0\"\"\u00a0FP.dP\u00a0= Fm.dm\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>Fm\u00a0= FP.dP\/dm.<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

B) O momento (torque) exercido pelo prego sobre o martelo (e vice versa) \u00e9 fornecido e tem intensidade MP\u00a0= FP.dP\u00a0= 30 Nm<\/b><\/span><\/span><\/span>\u00a0com\u00a0<\/b><\/span><\/span><\/span>dm\u00a0= 0,2 m,\u00a0<\/b><\/span><\/span><\/span>que substituidos em\u00a0<\/b><\/span><\/span><\/span>Fm\u00a0= FP.dP\/dm<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span>fornrcem<\/b><\/span><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span>Fm\u00a0=\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>Fm\u00a0= 150 N<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

C)\u00a0<\/b><\/span><\/span><\/span>Aqu\u00ed o\u00a0<\/b><\/span><\/span><\/span>peso<\/b><\/span><\/span><\/span>\u00a0de intensidade\u00a0<\/b><\/span><\/span><\/span>P\u00a0<\/b><\/span><\/span><\/span>participa e sua\u00a0<\/b><\/span><\/span><\/span>dist\u00e2ncia ao polo \u00e9 d<\/b><\/span><\/span><\/span>, e as<\/b><\/span><\/span><\/span>tr\u00eas for\u00e7as\u00a0<\/b><\/span><\/span><\/span>que influem na\u00a0<\/b><\/span><\/span><\/span>rota\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span>tem\u00a0<\/b><\/span><\/span><\/span>intensidades P, Fp\u00a0e Fm,\u00a0<\/b><\/span><\/span><\/span>cujos<\/b><\/span><\/span><\/span>momentos (torques)\u00a0<\/b><\/span><\/span><\/span>em rela\u00e7\u00e3o \u00e0 O,\u00a0<\/b><\/span><\/span><\/span>estabelecendo o\u00a0<\/b><\/span><\/span>sentido de tend\u00eancia<\/b><\/span><\/span>\u00a0de\u00a0<\/b><\/span><\/span>rota\u00e7\u00e3o em torno de O como positivo para o hor\u00e1rio e negativo para o anti hor\u00e1rio,\u00a0<\/b><\/span><\/span>ser\u00e3o:<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

MP\u00a0= \u2013 P.d<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

MFm\u00a0= \u2013 Fm.dm\u00a0=\u00a0<\/b><\/span><\/span><\/span>– Fm.20d\u00a0<\/b><\/span><\/span><\/span>= \u2013 20dFm<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

MFP<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span>= + FP.dP\u00a0=\u00a0<\/b><\/span><\/span><\/span>+ FP.4d\u00a0<\/b><\/span><\/span><\/span>= + 4dFP<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

No\u00a0<\/b><\/span><\/span><\/span>equil\u00edbrio de rota\u00e7\u00e3o a soma desses momentos\u00a0<\/b><\/span><\/span><\/span>deve ser\u00a0<\/b><\/span><\/span><\/span>nula<\/b><\/span><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>– P.d \u2013 20dFm\u00a0+ 4dFP\u00a0= 0\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>20Fm\u00a0= P \u2013 4<\/b><\/span><\/span><\/span>FP\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Fm\u00a0=<\/b><\/span><\/span><\/span>\u00a0(P \u2013 4FP\/20 =\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0\u2013\u00a0\"\"<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

 <\/p>\n

05- <\/b><\/span><\/span><\/span>Para esse exerc\u00edcio p<\/b><\/span><\/span>recisamos analisar a f\u00f3rmula do momento (ou torque):<\/b><\/span><\/span><\/span><\/span><\/p>\n

M = F.d.sen\"\"<\/b><\/span><\/span><\/span><\/span><\/p>\n

Sendo:<\/b><\/span><\/span><\/span><\/span><\/p>\n

M \u00e9 o momento ou torque<\/b><\/span><\/span><\/span><\/span><\/p>\n

F \u00e9 a for\u00e7a exercida<\/b><\/span><\/span><\/span><\/span><\/p>\n

d \u00e9 a dist\u00e2ncia entre a for\u00e7a e aonde ela \u00e9 executada<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>\u00e9 o \u00e2ngulo entre a for\u00e7a e aonde ela est\u00e1 sendo executada<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Ao mantermos o momento e o \u00e2ngulo constantes,\u00a0<\/b><\/span><\/span>como o jardineiro aumentou a dist\u00e2ncia ao prolongar o cabo, a for\u00e7a, consequentemente, ser\u00e1 menor. Alternativa a correta.<\/b><\/span><\/span><\/span><\/span><\/p>\n

 <\/p>\n

06- <\/b><\/span><\/span><\/span>Calculando o momento de cada for\u00e7a\u00a0<\/b><\/span><\/span>com o\u00a0<\/b><\/span><\/span>polo (eixo de rota\u00e7\u00e3o) na posi\u00e7\u00e3o ind<\/b><\/span><\/span>icada e estabelecendo o\u00a0<\/b><\/span><\/span>sentido hor\u00e1rio de rota\u00e7\u00e3o como positivo e anti- hor\u00e1rio como negativo:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

MPA<\/b><\/span><\/span>\u00a0= \u2013 850.<\/b><\/span><\/span>4 = \u2013 3400 N.m\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>MPG\u00a0= MN\u00a0= 0\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>MPC\u00a0= + Pc.1,5\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>MPB<\/b><\/span><\/span>\u00a0= + 400.4 =\u00a0<\/b><\/span><\/span>+ 1600 N.m.<\/b><\/span><\/span><\/span><\/span><\/p>\n

No\u00a0<\/b><\/span><\/span>equil\u00edbrio de rota\u00e7\u00e3o\u00a0<\/b><\/span><\/span>a\u00a0<\/b><\/span><\/span>soma dos momentos de cada for\u00e7a deve ser nulo\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>– 3400 + 0 + 1,5PC\u00a0+<\/b><\/span><\/span><\/span><\/span><\/p>\n

1600 = 0\u00a0\"\"<\/b><\/span><\/span>\u00a01,5PC\u00a0= 1800\u00a0\"\"\u00a0<\/b><\/span><\/span>PC<\/b><\/span><\/span>\u00a0=\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>= 1200 N.<\/b><\/span><\/span><\/span><\/span><\/p>\n

PC\u00a0= mC.g\u00a0\"\"<\/b><\/span><\/span>\u00a01200 = mC.10\u00a0\"\"\u00a0<\/b><\/span><\/span>mC\u00a0= 120 kg.<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

 <\/p>\n

07-<\/b><\/span><\/span><\/span> Se voc\u00ea n\u00e3o domina a teoria, veja as informa\u00e7\u00f5es a seguir:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Momento (ou torque) de uma for\u00e7a<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

O\u00a0<\/b><\/span><\/span>sinal do momento da for\u00e7a pode ser negativo ou positi<\/b><\/span><\/span>vo\u00a0e, por\u00a0conven\u00e7\u00e3o,\u00a0vamos adotar o<\/b><\/span><\/span>sentido hor\u00e1rio de rota\u00e7\u00e3o em torno de O como positivo e anti-hor\u00e1rio como negativo.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Condi\u00e7\u00f5es de equil\u00edbrio de um corpo extenso<\/b><\/span><\/span><\/span><\/span><\/p>\n

S\u00e3o duas as condi\u00e7\u00f5es para que um corpo extenso r\u00edgido esteja em equil\u00edbrio<\/b><\/span><\/span>:<\/b><\/span><\/span><\/span><\/span><\/p>\n

1a<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>Equil\u00edbrio de transla\u00e7\u00e3o\u00a0\u00a0\"\"<\/b><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span>A resultante do sistema de for\u00e7as deve ser nula.<\/b><\/span><\/span><\/span><\/span><\/p>\n

2a<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0Equil\u00edbrio\u00a0de rota\u00e7\u00e3o\u00a0\u00a0\"\"\u00a0 A\u00a0soma alg\u00e9brica dos momentos das for\u00e7as<\/b><\/span><\/span>\u00a0que agem sobre o sistema, em rela\u00e7\u00e3o \u00e0 qualquer ponto\u00a0<\/b><\/span><\/span>(polo O),deve ser nula.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Resolvendo o sistema\u00a0<\/b><\/span><\/span>composto pelas duas equa\u00e7\u00f5es acima voc\u00ea chega\u00a0<\/b><\/span><\/span>\u00e0 resolu\u00e7\u00e3o\u00a0<\/b><\/span><\/span>do exerc\u00edcio.<\/b><\/span><\/span><\/span><\/span><\/p>\n

No\u00a0<\/b><\/span><\/span>caso do exerc\u00edcio<\/b><\/span><\/span>,\u00a0<\/b><\/span><\/span>vamos<\/b><\/span><\/span>\u00a0colocar as for\u00e7as que agem sobre a barra de intensidades:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Parroz\u00a0(peso do arroz)\u00a0\"\"Pbarrapeso da barra (no meio da barra, pois ela \u00e9 homog\u00eanea)\u00a0\"\"\u00a0N (rea\u00e7\u00e3o normal que o apoio aplica na barra)
\n\"\"<\/b><\/span><\/span><\/span><\/span><\/p>\n

Momento de cada for\u00e7a em rela\u00e7\u00e3o ao polo 0:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Mbarra\u00a0= + Pbarra.1x\u00a0\"\"\u00a0MN\u00a0= N.0 = 0\u00a0\"\"Marroz\u00a0= \u2013 Parroz.3x.<\/b><\/span><\/span><\/span><\/span><\/p>\n


\nCondi\u00e7\u00e3o<\/b><\/span><\/span>\u00a0de\u00a0<\/b><\/span><\/span>equil\u00edbrio de rota\u00e7\u00e3o\"\"<\/b><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span>soma dos momentos de cada for\u00e7a, em rela\u00e7\u00e3o ao ponto O<\/b><\/span><\/span>, deve ser\u00a0<\/b><\/span><\/span>nula:<\/b><\/span><\/span>
\n<\/b><\/span><\/span>Pbarra.1x + 0 \u2013 Parroz.3x = 0\u00a0\"\"<\/b><\/span><\/span>\u00a0mbarra.g= 3.marroz.g\u00a0\"\"<\/b><\/span><\/span>marroz\u00a0= 5,00 kg\u00a0<\/b><\/span><\/span>(veja desenho)\u00a0\"\"
\nmbarra.g = marroz.g.3\u00a0\"\"mbarra\u00a0= 3.5,00\u00a0\u00a0\"\"<\/b><\/span><\/span>mbarra\u00a0= 15,00 kg.<\/b><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span>
\n<\/b><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/span><\/p>\n

 <\/p>\n

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

Momento ou torque de uma for\u00e7a<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

O\u00a0<\/b><\/span><\/span>sinal do momento da for\u00e7a pode ser negativo ou positivo<\/b><\/span><\/span>\u00a0e, por conven\u00e7\u00e3o, vamos adotar o\u00a0<\/b><\/span><\/span>sentido hor\u00e1rio de rota\u00e7\u00e3o em torno de O como positivo e anti-hor\u00e1rio como negativo.<\/b><\/span><\/span><\/span><\/span><\/p>\n

S\u00e3o\u00a0<\/b><\/span><\/span>duas as condi\u00e7\u00f5es<\/b><\/span><\/span>\u00a0para que um corpo extenso r\u00edgido\u00a0<\/b><\/span><\/span>esteja em equil\u00edbrio:<\/b><\/span><\/span><\/span><\/span><\/p>\n

1a<\/b><\/span><\/span>\u00a0\"\"\u00a0Equil\u00edbrio de transla\u00e7\u00e3o\u00a0\u00a0\"\"<\/b><\/span><\/span>\u00a0 A resultante do sistema de for\u00e7as deve ser nula.<\/b><\/span><\/span><\/span><\/span><\/p>\n

2a\u00a0\"\"<\/b><\/span><\/span>\u00a0Equil\u00edbrio de rota\u00e7\u00e3o\u00a0\u00a0\"\"\u00a0\u00a0<\/b><\/span><\/span>A soma alg\u00e9brica dos momentos das for\u00e7as\u00a0<\/b><\/span><\/span>que agem sobre o sistema,<\/b><\/span><\/span>\u00a0em rela\u00e7\u00e3o ao ponto de apoio (p\u00f3lo O),deve ser nula.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

 <\/p>\n

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

Est\u00e1tica de um ponto material 2017 \u2013 2016 – 2015 Resolu\u00e7\u00f5es   01-Denominando\u00a0de ml\u00a0a massa de cada cubo laranja e de ma\u00a0a massa de cada cubo azul\u00a0e, estando as\u00a0balan\u00e7as em equil\u00edbrio,\u00a0voc\u00ea ter\u00e1: Balan\u00e7a\u00a0da esquerda2ma\u00a0+ ml\u00a0= 2 (I) Balan\u00e7a\u00a0da direita\u00a0ma\u00a0+ 3 = 2ml\u00a0(II) Isolando ma\u00a0em (II) e substituindo em (I)\u00a0ma\u00a0= 2ml\u00a0\u2013 3\u00a02(2ml\u00a0\u2013 3) + ml\u00a0= 2\u00a0\u00a04ml\u00a0\u2013 6 + ml\u00a0= 2\u00a05ml\u00a0= 8\u00a0ml=<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3283,"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-3285","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3285","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=3285"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3285\/revisions"}],"predecessor-version":[{"id":3286,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3285\/revisions\/3286"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3283"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3285"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}