2<\/sub>) = F. d (1)\n<\/p>\n\n\n\nd l\u00e0 kho\u1ea3ng c\u00e1ch gi\u1eefa hai gi\u00e1 c\u1ee7a hai l\u1ef1c, kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u1ecb tr\u00ed O c\u1ee7a tr\u1ee5c quay.\n<\/p>\n\n\n\n
X\u00e9t tr\u1ee5c quay \u0111i qua O, momen c\u1ee7a ng\u1eabu l\u1ef1c l\u00fac n\u00e0y l\u00e0: \n<\/p>\n\n\n\n
M = F1<\/sub>d1<\/sub> + F2<\/sub>d2<\/sub> = F(d1<\/sub> + d2<\/sub>) = F.d (2)\n<\/p>\n\n\n\nT\u1eeb (1) v\u00e0 (2) \u2192 M = M\u2019 \u2192 momen c\u1ee7a ng\u1eabu l\u1ef1c kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u00e0o v\u1ecb tr\u00ed c\u1ee7a tr\u1ee5c quay vu\u00f4ng g\u00f3c v\u1edbi m\u1eb7t ph\u1eb3ng ch\u1ee9a ng\u1eabu l\u1ef1c (\u0111pcm). <\/p>\n\n\n\n
B\u00e0i 3 (trang 118 SGK V\u1eadt L\u00fd 10) :<\/strong> Vi\u1ebft c\u00f4ng th\u1ee9c t\u00ednh momen c\u1ee7a ng\u1eabu l\u1ef1c. Momen c\u1ee7a ng\u1eabu l\u1ef1c c\u00f3 \u0111\u1eb7c \u0111i\u1ec3m g\u00ec? <\/p>\n\n\n\nL\u1eddi gi\u1ea3i:<\/strong><\/ins><\/p>\n\n\n\nC\u00f4ng th\u1ee9c t\u00ednh momen c\u1ee7a ng\u1eabu l\u1ef1c:<\/p>\n\n\n\n
M = F.d.<\/p>\n\n\n\n
Momen c\u1ee7a ng\u1eabu l\u1ef1c ph\u1ee5 thu\u1ed9c v\u00e0o \u0111\u1ed9 l\u1edbn c\u1ee7a ng\u1eabu l\u1ef1c, v\u00e0o kho\u1ea3ng c\u00e1ch d gi\u1eefa hai gi\u00e1 c\u1ee7a hai l\u1ef1c, kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u00e0o v\u1ecb tr\u00ed tr\u1ee5c quay O.<\/p>\n\n\n\n
B\u00e0i 4 (trang 118 SGK V\u1eadt L\u00fd 10) :<\/strong> Hai l\u1ef1c c\u1ee7a m\u1ed9t ng\u1eabu l\u1ef1c c\u00f3 \u0111\u1ed9 l\u1edbn F = 5,0 N. C\u00e1nh tay \u0111\u00f2n c\u1ee7a ng\u1eabu l\u1ef1c d = 20 cm. Momen c\u1ee7a ng\u1eabu l\u1ef1c l\u00e0: <\/p>\n\n\n\nA.\t100 N.m <\/ins><\/p>\n\n\n\nB. 2,0 N.m <\/p>\n\n\n\n
C. 0,5 N.m <\/p>\n\n\n\n
D. 1,0 N.m<\/p>\n\n\n\n
L\u1eddi gi\u1ea3i:<\/strong><\/p>\n\n\n\nCh\u1ecdn D.<\/p>\n\n\n\n
\u00c1p d\u1ee5ng c\u00f4ng th\u1ee9c momen c\u1ee7a ng\u1eabu l\u1ef1c:<\/p>\n\n\n\n
M = F.d = 5.0,2 = 1 (N.m).<\/p>\n\n\n\n
<\/p>\n\n\n\n
B\u00e0i 5 (trang 118 SGK V\u1eadt L\u00fd 10) :<\/strong> M\u1ed9t ng\u1eabu l\u1ef1c g\u1ed3m hai vecto l\u1ef1c F1<\/sub> v\u00e0 F2<\/sub> c\u00f3 F1<\/sub> = F2<\/sub> = F v\u00e0 c\u00f3 c\u00e1nh tay \u0111\u00f2n d. Momen c\u1ee7a ng\u1eabu l\u1ef1c n\u00e0y l\u00e0<\/ins><\/p>\n\n\n\nA. (F1<\/sub> \u2013 F2<\/sub>).d.<\/p>\n\n\n\nB. 2Fd.<\/p>\n\n\n\n
C. Fd.<\/p>\n\n\n\n
D.Ch\u01b0a bi\u1ebft \u0111\u01b0\u1ee3c v\u00ec c\u00f2n ph\u1ee5 thu\u1ed9c v\u00e0o v\u1ecb tr\u00ed c\u1ee7a tr\u1ee5c quay.<\/p>\n\n\n\n
L\u1eddi gi\u1ea3i:<\/strong><\/p>\n\n\n\nCh\u1ecdn C.\n<\/p>\n\n\n\n
Momen cu\u1ea3 ng\u1eabu l\u1ef1c: M = F.d<\/strong> <\/p>\n\n\n\n<\/p>\n\n\n\n
B\u00e0i 6 (trang 118 SGK V\u1eadt L\u00fd 10) :<\/strong> M\u1ed9t \nchi\u1ebfc th\u01b0\u1edbc m\u1ea3nh c\u00f3 tr\u1ee5c quay n\u1eb1m ngang \u0111i qua tr\u1ecdng t\u00e2m O c\u1ee7a th\u01b0\u1edbc. \nD\u00f9ng hai ng\u00f3n tay t\u00e1c d\u1ee5ng v\u00e0o th\u01b0\u1edbc m\u1ed9t ng\u1eabu l\u1ef1c \u0111\u1eb7t v\u00e0o hai \u0111i\u1ec3m A v\u00e0 B\n c\u00e1ch nhau 4,5 cm v\u00e0 c\u00f3 \u0111\u1ed9 l\u1edbn FA<\/sub> = FB<\/sub> = 1 N (H\u00ecnh 22.6a)<\/ins><\/p>\n\n\n\na) T\u00ednh momen c\u1ee7a ng\u1eabu l\u1ef1c.<\/p>\n\n\n\n
b) Thanh quay \u0111i m\u1ed9t g\u00f3c \u03b1 = 30o<\/sup> . Hai l\u1ef1c lu\u00f4n lu\u00f4n n\u1eb1m ngang v\u00e0 v\u1eabn \u0111\u1eb7t t\u1ea1i A v\u00e0 B (H\u00ecnh 22.6b). T\u00ednh momen c\u1ee7a ng\u1eabu l\u1ef1c.<\/p>\n\n\n\nL\u1eddi gi\u1ea3i:<\/strong><\/p>\n\n\n\na)\tMomen c\u1ee7a ng\u1eabu l\u1ef1c: M = F.d = F.AB = 1.0,045 = 0,045 (N.m).\n <\/p>\n\n\n\nb)\t \n <\/p>\n\n\n\nMomen c\u1ee7a ng\u1eabu l\u1ef1c: M\u2019 = F.BI v\u1edbi BI = AB.cos\u03b1 = 4,5.10-2<\/sup>.cos30o<\/sup> = 0,039(m)\n <\/p>\n\n\n\n\u2192 M\u2019 = 1. 0,039 = 0,039 (N.m).\n <\/ins><\/p>\n","protected":false},"excerpt":{"rendered":"B\u00e0i 22 : Ng\u1eabu l\u1ef1c C1. ( trang 117 sgk V\u1eadt L\u00fd 10): Ch\u1ee9ng minh r\u1eb1ng momen c\u1ee7a ng\u1eabu l\u1ef1c kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u00e0o v\u1ecb tr\u00ed c\u1ee7a tr\u1ee5c quay vu\u00f4ng g\u00f3c v\u1edbi m\u1eb7t ph\u1eb3ng ch\u1ee9a ng\u1eabu l\u1ef1c. Tr\u1ea3 l\u1eddi: G\u1ecdi O\u2019 l\u00e0 v\u1ecb tr\u00ed b\u1ea5t k\u1ef3 c\u1ee7a tr\u1ee5c quay vu\u00f4ng g\u00f3c v\u1edbi m\u1eb7t ph\u1eb3ng ch\u1ee9a […]<\/p>\n","protected":false},"author":12,"featured_media":44154,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1640],"tags":[],"yoast_head":"\n
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![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-bai-tap-vat-ly-10\/images\/tra-loi-cau-hoi-sgk-vat-ly-10-bai-22.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-bai-tap-vat-ly-10\/images\/bai-6-trang-118-sgk-ly-10.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-bai-tap-vat-ly-10\/images\/bai-6-trang-118-sgk-ly-10-1.PNG\")
<\/figure>\n\n\n\n