{"id":3686,"date":"2018-02-13T16:30:54","date_gmt":"2018-02-13T16:30:54","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3686"},"modified":"2024-08-20T18:36:49","modified_gmt":"2024-08-20T18:36:49","slug":"resolucao-comentada-estatica-de-um-ponto-material","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/enem\/estatica-de-um-ponto-material\/resolucao-comentada-estatica-de-um-ponto-material\/","title":{"rendered":"Est\u00e1tica de um Ponto Material – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

Est\u00e1tica de um Ponto Material<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe nas figuras abaixo\u00a0 —\u00a0 (1)\u00a0 —\u00a0 2T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=P\u00a0 —\u00a0 <\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=P\/2\u00a0 <\/b><\/span><\/span><\/span>—\u00a0 (2) 2T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>sen60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=P\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=P\/2sen60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 <\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=0,58P\u00a0 <\/b><\/span><\/span><\/span>—\u00a0 (3)<\/b><\/span><\/span><\/span>\u00a02T<\/b><\/span><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sub><\/span>sen30<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sup><\/span>=P\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sub><\/span>=P\/2sen30<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 <\/b><\/span><\/span><\/span><\/span>T<\/b><\/span><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sub><\/span>=2P\/2<\/b><\/span><\/span><\/span><\/span>=P<\/b><\/span><\/span><\/span><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/span><\/p>\n

Observa\u00e7\u00e3o: Note que quanto menor for o \u00e2ngulo com o teto, maior ser\u00e1 a for\u00e7a de tra\u00e7\u00e3o no fio.<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Colocando as for\u00e7as em cada fio e decompondo-as\u00a0 —\u00a0 equil\u00edbrio na horizontal\u00a0 —\u00a0 <\/b><\/span><\/span><\/span><\/p>\n

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

T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>cos60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>cos30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.1\/2=T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.\u221a3\/2\u00a0 —\u00a0T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=\u221a3.T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 equil\u00edbrio na vertical\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>sen60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0+ T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>sen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=P\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.\u221a3\/2 + T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.1\/2=6.10\u00a0 —\u00a0 (\u221a3.T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.\u221a3)\/2 + T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/2=60\u00a0 —\u00a0 3T<\/b><\/span><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sub><\/span>+ T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=120\u00a0 —\u00a0\u00a0T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=30N\u00a0\u00a0—\u00a0 T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=\u221a3.T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=\u221a3.30\u00a0 —\u00a0T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=30\u221a3N<\/b><\/span><\/span><\/span><\/p>\n

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

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe a figura abaixo\u00a0 —\u00a0 P=mg=60.10\u00a0 —\u00a0 P=600N\u00a0 —\u00a0 como ele est\u00e1 em equil\u00edbrio a intensidade da for\u00e7a resultante deve<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

ser nula\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 2T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=P\u00a0 —\u00a0 2T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=600\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=300N<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Observe a figura abaixo onde as duas cordas foram prolongadas at\u00e9 se encontrarem:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Para que as placas fiquem em equil\u00edbrio est\u00e1tico, resultante das for\u00e7as que agem sobre elas devem ter componentes na horizontal e na vertical para que possam se anular\u00a0 —\u00a0 observe que a \u00fanica alternativa que n\u00e3o tem componentes que possam se anular na horizontal \u00e9 a V\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/span><\/p>\n

05-\u00a0<\/b><\/span><\/span><\/span><\/span>Considere: sen 45\u00ba = 0,7; cos 45\u00ba = 0,7; sen 37\u00ba = 0,6 e \u00a0cos 37\u00ba = 0,8<\/b><\/span><\/span><\/span><\/span><\/p>\n

Observe a figura abaixo onde foram decompostas as for\u00e7a T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0e T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 Equil\u00edbrio na horizontal<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>cos45<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=T2cos37<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.0,7=T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.0,8\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=8T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/7\u00a0 —\u00a0 equil\u00edbrio na vertical\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>sen45<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0+ T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>sen37<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0=P\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.0,7 + T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.0,6=280\u00a0 —\u00a0 (8T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/7).0,7 + 0,6T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=280\u00a0 —\u00a0 0,8T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 0,6T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=280\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=280\/1,4\u00a0 —\u00a0 T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=200N\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe a figura abaixo onde as for\u00e7as de tra\u00e7\u00e3o foram decomposta na vertical\u00a0 — no equil\u00edbrio a for\u00e7a resultante \u00e9 nula\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

P=2Tcos\u03b8\u00a0 —\u00a0 T=P\/2cos\u03b8\u00a0 —\u00a0 observe nessa express\u00e3o que a tra\u00e7\u00e3o T \u00e9 inversamente proporcional a cos\u03b8, ou seja, quanto menor cos\u03b8, maior ser\u00e1 T e isso acontece quanto maior for o \u00e2ngulo \u03b1\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

Observa\u00e7\u00e3o\u00a0 —\u00a0 volte ao exerc\u00edcio 1 e veja que s\u00e3o situa\u00e7\u00f5es diferentes.<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe o diagrama abaixo que mostra as for\u00e7as atuantes no terceiro elo\u00a0 —\u00a0 Dados\u00a0 —\u00a0 \u00a0m = 200 g = 0,2 kg\u00a0 —\u00a0\u00a0 g = 10 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0P = m g = 0,2(10)\u00a0 —\u00a0\u00a0P = 2 N\u00a0 (peso de cada elo) —\u00a0F<\/b><\/span><\/span><\/span>43<\/b><\/span><\/span><\/sub><\/span>\u00a0= 2 P = 4 N\u00a0 —\u00a0\u00a0F<\/b><\/span><\/span><\/span>23<\/b><\/span><\/span><\/sub><\/span>\u00a0= 3 P = 6 N\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>Observe nas figuras abaixo onde cada for\u00e7a de tra\u00e7\u00e3o \u00e9 decomposta na vertical (T<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>=Tsen60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>) e na horizontal (T<\/b><\/span><\/span><\/span>h<\/b><\/span><\/span><\/sub><\/span>=Tcos60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>).\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como o sistema est\u00e1 em equil\u00edbrio, as componentes horizontais se anulam e na vertical voc\u00ea ter\u00e1\u00a0 —\u00a0 2T\u00ad<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>=P\u00a0 —\u00a0 2Tsen60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=P\u00a0 —\u00a0 2T.\u221a3\/2=50\u00a0 —\u00a0 T=50\/\u221a3N\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>
\n<\/span><\/h3>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o Comentada Est\u00e1tica de um Ponto Material 01-\u00a0Observe nas figuras abaixo\u00a0 —\u00a0 (1)\u00a0 —\u00a0 2T1=P\u00a0 —\u00a0 T1=P\/2\u00a0 —\u00a0 (2) 2T2sen60o=P\u00a0 —\u00a0 T2=P\/2sen60o\u00a0 —\u00a0 —\u00a0 T2=0,58P\u00a0 —\u00a0 (3)\u00a02T3sen30o=P\u00a0 —\u00a0 T3=P\/2sen30o\u00a0 —\u00a0 T3=2P\/2=P\u00a0 —\u00a0\u00a0R- B Observa\u00e7\u00e3o: Note que quanto menor for o \u00e2ngulo com o teto, maior ser\u00e1 a for\u00e7a de tra\u00e7\u00e3o no fio. 02-\u00a0Colocando as for\u00e7as em cada fio e decompondo-as\u00a0<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3684,"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-3686","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3686","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=3686"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3686\/revisions"}],"predecessor-version":[{"id":10649,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3686\/revisions\/10649"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3684"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3686"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}