{"id":1576,"date":"2015-11-28T03:43:14","date_gmt":"2015-11-28T03:43:14","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1576"},"modified":"2024-08-23T14:33:24","modified_gmt":"2024-08-23T14:33:24","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-principio-de-pascal-e-prensa-hidraulica","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/hidrostatica\/principio-de-pascal\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-principio-de-pascal-e-prensa-hidraulica\/","title":{"rendered":"Princ\u00edpio de Pascal e Prensa hidr\u00e1ulica – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre<\/b><\/span><\/span><\/span><\/p>\n

Princ\u00edpio de Pascal e Prensa hidr\u00e1ulica<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

\u00a0<\/span>01-<\/b><\/span><\/span><\/span>\u00a0 <\/b><\/span><\/span><\/span>Princ\u00edpio enunciado por Pascal, f\u00edsico e matem\u00e1tico franc\u00eas (1623 \u2013 1663), conhecido como<\/b><\/span><\/span><\/span>\u00a0princ\u00edpio de Pascal<\/b><\/span><\/span><\/span>:<\/b><\/span><\/span><\/span><\/p>\n

\u201c O acr\u00e9scimo de press\u00e3o exercido num ponto de um l\u00edquido ideal em equil\u00edbrio \u00e9 transmitida integralmente a todos os pontos desse l\u00edquido e tamb\u00e9m \u00e0s paredes do recipiente onde est\u00e1 contido\u201d\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Veja o esquema de um freio hidr\u00e1ulico na figura abaixo;<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A press\u00e3o \u00e9 a mesma em todos os pontos, inclusive em cada cilindro (Princ\u00edpio de Pascal)\u00a0 —<\/b><\/span><\/span><\/span>\u00a0 R- C<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>For\u00e7a. A for\u00e7a que age na roda \u00e9 muito maior do que aquela aplicada no pedal (Princ\u00edpio de Pascal), devido \u00e0 diferen\u00e7a de \u00e1rea entre os \u00eambolos e tamb\u00e9m devido ao sistema de alavancas do pedal do freio.<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Sendo a press\u00e3o em cada \u00eambolo a mesma\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/A<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/A<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 R- E<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/4S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=4F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 como o sistema est\u00e1 em equil\u00edbrio\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=4F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=P\u00a0 —\u00a0 veja nas figuras<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

abaixo que o trabalho realizado por F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0para o \u00eambolo menor descer \u00e9 o mesmo realizado por F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0para o \u00eambolo maior subir\u00a0 —\u00a0 W=F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.h<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.h<\/b><\/span><\/span><\/span>2\u00ad\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0 se o trabalho realizado \u00e9 o mesmo, a energia mec\u00e2nica do sistema se conserva\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Estando o \u00eambolo maior em equil\u00edbrio, a for\u00e7a aplicada pelo l\u00edquido sobre ele, equilibra o peso do autom\u00f3vel\u00a0 —\u00a0 F=p=mg=3.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.10\u00a0 — p=3.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>N\u00a0 —\u00a0 a press\u00e3o exercida pelo cilindro \u00e9 a mesma que a exercida pelo \u00eambolo P=F\/S=p\/S=3.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\/6.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 P=0,5.10<\/b><\/span><\/span><\/span>7<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u00a0P=5.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>N\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

b) O volume de l\u00edquido expulso pelo cilindro ao descer h<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 o mesmo que faz o \u00eambolo subir h<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.h<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.h<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 3.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>.h<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span><\/p>\n

6.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>.0,2\u00a0 —\u00a0 h<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=12.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>\/3.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0 h<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=4.10\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0h<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=40m=4.000cm<\/b><\/span><\/span><\/span>\u00a0 — (a v\u00e1lvula impede o retorno do l\u00edquido, fazendo com que esse deslocamento corresponda a muitos \u201cvai e vem\u201d do pist\u00e3o)\u00a0<\/b><\/span><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Peso do caminhoneiro\u00a0 —\u00a0 p=mg=80.10\u00a0 —\u00a0 p=800N=F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 — F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 800\/\u03c0R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/\u03c0R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

800\/\u03c0(5.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/\u03c0.(3.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=28.800N\u00a0 —\u00a0 massa dessa carga\u00a0 —\u00a0 p=mg\u00a0 — 28.800=m.10\u00a0 — m=2.880kg\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/A<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 2\/\u03c0(5.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/\u03c0(1.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=2.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>\/25.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=0,08.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=8N \u00a0—\u00a0 essa for\u00e7a equilibra uma massa\u00a0 —\u00a0 p=Mg\u00a0 —\u00a0 8=M.10\u00a0 —\u00a0 M=0,8kg=800g\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>L e M\u00a0 —\u00a0 P\/A=P<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\/2A\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>=2P\u00a0 —\u00a0 L e N\u00a0 —\u00a0 P\/A=P<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>\/3A\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>=3P\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=8.A<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F\/A<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=400\/8.A<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0F=50N<\/b><\/span><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como as unidades s\u00e3o as mesmas e est\u00e3o, na equa\u00e7\u00e3o, em cada lado da igualdade, n\u00e3o \u00e9 preciso transform\u00e1-las em m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.g\/a=m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.g\/b\u00a0 —\u00a0 4\/80=m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\/20\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=1kg\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/\u03c0(d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/2)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/\u03c0(R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 100\/\u03c0(0,05\/2)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=10.000\/\u03c0(R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 100\/0,0625=10.000\/(R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=\u221a6,25\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=2,5m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=5,0m<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 a press\u00e3o \u00e9 a mesma nos dois \u00eambolos e vale\u00a0 —\u00a0 P=10.000\/\u03c0(6,25)\u00a0 —\u00a0 P=10.000\/3,1416.6,25\u00a0 —\u00a0 P=\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0P=509,3Pa\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0independe da unidade\u00a0 —\u00a0 A<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=50\u00a0 —\u00a0 P=F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/A<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/50\u00a0 —\u00a0\u00a0 A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=2\u00a0 —\u00a0\u00a0 P=F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=30\/2=15\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/50=<\/b><\/span><\/span><\/span><\/p>\n

15\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=750N\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=30d<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 Velocidade constante\u00a0 —\u00a0 equil\u00edbrio din\u00e2mico\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=p<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=m.g\u00a0 —\u00a0\u00a0 F<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>d<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=30d<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>\/S<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span>\/SF\u00a0 —\u00a0 m.g\/\u03c0(d<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>\/2)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=F\/(d<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span>\/2)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 9.000×10\/(30d<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=F\/(d<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 90.000\/900=F\u00a0 —\u00a0 F=100N\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\/S<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/\u03c0D<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=F<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\/\u03c0(4D)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/1=F<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\/16\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0F<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/F<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=16<\/b><\/span><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>200\/25=F\/2.000\u00a0 — F=p=400.000\/25\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0F=p=16.000N<\/b><\/span><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 1200\/(S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/2)=F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=400N\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>For\u00e7a (F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>) ampliada pelo sistema de articula\u00e7\u00f5es quando no pedal \u00e9 aplicada a for\u00e7a F=50N\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

colocando o p\u00f3lo O na articula\u00e7\u00e3o central e lembrando que a soma dos momentos das for\u00e7as em rela\u00e7\u00e3o ao p\u00f3lo \u00e9 nula\u00a0 —\u00a0 – F.d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

– 50.0,2 + F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.0,04=0\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=250N\u00a0 —\u00a0 essa for\u00e7a \u00e9 transmitida at\u00e9 o \u00eambolo 1 de \u00e1rea S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=40mm<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=4.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0que est\u00e1 ligado, por meio de vaso comunicante at\u00e9 o \u00eambolo 2 de \u00e1rea S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=80mm<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=8.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 aplicando o teorema de Pascal<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

nos \u00eambolos 1 e 2\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 250\/4.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>=F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/8.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=500N\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/span><\/p>\n

20-\u00a0<\/b><\/span><\/span><\/span>Colocando o p\u00f3lo no ponto fixo, calculando o momento de cada for\u00e7a em rela\u00e7\u00e3o \u00e0 ele e<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

igualando a zero\u00a0 —\u00a0\u00a0 – 10.0,2+ 1N.0,1=0\u00a0 —\u00a0 N=20N\u00a0 —\u00a0 aplicando o teorema de Pascal\u00a0 —\u00a0 N\/a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=F\/5.a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 20\/1=F\/5\u00a0 —\u00a0 F=100N\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

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

21-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Qualquer acr\u00e9scimo de press\u00e3o efetuado em um ponto de um fluido em equil\u00edbrio \u00e9 transmitido integralmente aos demais pontos desse l\u00edquido\u00a0 —\u00a0 Teorema de Pascal\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= 4.000 N\u00a0 —\u00a0 A<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= 10 cm<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0= 0,05 cm<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 \u00a0L = 1 m\u00a0 —\u00a0 \u00a0m = 1 g = 10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\u00a0kg\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

quando voc\u00ea aplica a for\u00e7a \"\"\u00a0sobre o pist\u00e3o de \u00e1rea A<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>, atrav\u00e9s do ar comprimido, transmite-se sobre o chumbinho de \u00e1rea A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>, a for\u00e7a\u00a0\"\"\u00a0\u00a0—\u00a0 a rela\u00e7\u00e3o entre essas grandezas \u00e9 dada pelo princ\u00edpio de Pascal\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/A<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 4.000\/10=F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/0,05\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=20N\u00a0 —\u00a0 essa for\u00e7a\u00a0\"\", tamb\u00e9m suposta constante, realiza trabalho sobre o chumbinho ao longo do cano da espingarda, aumentando sua velocidade a partir do repouso (v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0= 0)\u00a0 —\u00a0 pelo teorema da energia cin\u00e9tica, considerando\u00a0\"\"\u00a0como a for\u00e7a resultante sobre o chumbinho\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>F2<\/b><\/span><\/span><\/sub><\/span>=\u0394E<\/b><\/span><\/span><\/span>cin<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 F.L=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 20.1=10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 V=200m\/s\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

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

23- <\/b><\/span><\/span><\/span>a) Alavanca inter-resistente\u00a0– a for\u00e7a transmitida \u00a0localiza-se entre o p\u00f3lo (parte inferior do pedal) e a for\u00e7a aplicada pelo operador (parte superior do pedal).
\nb) Princ\u00edpio enunciado por Pascal, f\u00edsico e matem\u00e1tico franc\u00eas (1623 \u2013 1663), conhecido como\u00a0princ\u00edpio de Pascal:<\/b><\/span><\/span><\/span><\/p>\n

\u201c O acr\u00e9scimo de press\u00e3o exercido num ponto de um l\u00edquido ideal em equil\u00edbrio \u00e9 transmitida integralmente a todos os pontos desse l\u00edquido e tamb\u00e9m \u00e0s paredes do recipiente onde est\u00e1 contido\u201d\u00a0 —\u00a0 A maioria dos sistemas multiplicadores de for\u00e7as \u00e9 baseado no princ\u00edpio de Pascal e, para explic\u00e1-lo considere um l\u00edquido ideal no interior de dois cilindros verticais de se\u00e7\u00f5es diferentes e interligados. Esses cilindros, em contato com a parte superior do l\u00edquido, possuem dois \u00eambolos de \u00e1reas S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0e S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Uma for\u00e7a de intensidade F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0aplicada ao \u00eambolo de menor \u00e1rea (S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>), provocar\u00e1 um aumento de press\u00e3o dado por\u00a0\u0394P=F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0e, pelo princ\u00edpio de Pascal esse acr\u00e9scimo de press\u00e3o se transmitir\u00e1 integralmente a todos os pontos do l\u00edquido e das paredes, inclusive para o \u00eambolo de maior \u00e1rea (S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>). Ent\u00e3o, o \u00eambolo maior fica sujeito a uma for\u00e7a F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>, tal que\u00a0\u0394P=F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>. Pelo princ\u00edpio de Pascal essa varia\u00e7\u00e3o de press\u00e3o se transmite integralmente\u00a0 do \u00eambolo menor ao \u00eambolo maior e s\u00e3o iguais\u00a0 — \u00a0F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0c) No pedal\u00a0 —\u00a0 M<\/b><\/span><\/span><\/span>F1<\/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>= + 100.0,4=40N.m\u00a0 —\u00a0 M<\/b><\/span><\/span><\/span>F2<\/b><\/span><\/span><\/sub><\/span>= – F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>= – 0,2F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 a soma dos momentos de cada for\u00e7a deve ser nula\u00a0 —\u00a0 40 \u2013 0,2F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=200N\u00a0 —\u00a0 esta for\u00e7a \u00e9 transmitida at\u00e9 o pist\u00e3o 1, que atrav\u00e9s do Princ\u00edpio de<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Pascal, a aumenta para F<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0no pist\u00e3o 2\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>2=3<\/b><\/span><\/span><\/sub><\/span>\/S<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 200\/4.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>\u00a0= F<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\/16.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=800N (for\u00e7a aplicada na pastilha do freio)\u00a0<\/b><\/span><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span> <\/b>Seja F a for\u00e7a aplicada pelo motorista no pedal\u00a0 —\u00a0 pelo enunciado a alavanca tem a<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

capacidade\u00a0 de amplia\u00e7\u00e3o da for\u00e7a aplicada por um fator igual \u00e0 raz\u00e3o direta de seus bra\u00e7os, que \u00e9 de 40\/10=4 vezes\u00a0 —\u00a0 f=4F\u00a0 —\u00a0 a prensa hidr\u00e1ulica amplia a for\u00e7a f na raz\u00e3o direta de suas \u00e1reas, ou seja, de 8 vezes\u00a0 —\u00a0 F\u2019=8f\u00a0 —\u00a0 F\u2019=8.4F\u00a0 —\u00a0 F\u2019=32F\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- A.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Princ\u00edpio de Pascal e Prensa hidr\u00e1ulica   \u00a001-\u00a0 Princ\u00edpio enunciado por Pascal, f\u00edsico e matem\u00e1tico franc\u00eas (1623 \u2013 1663), conhecido como\u00a0princ\u00edpio de Pascal: \u201c O acr\u00e9scimo de press\u00e3o exercido num ponto de um l\u00edquido ideal em equil\u00edbrio \u00e9 transmitida integralmente a todos os pontos desse l\u00edquido e tamb\u00e9m \u00e0s paredes do recipiente onde<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1572,"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-1576","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1576","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=1576"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1576\/revisions"}],"predecessor-version":[{"id":10892,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1576\/revisions\/10892"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1572"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1576"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}