質量2KG之木塊以一細繩懸起,以一水平力推之,使繩與鉛質方向成37度,求:(a)該推力之大小?(b)繩中之張力?
質量分別為4M,2M,及8M之三木塊連接如圖5.26所示,兩繩中之張力分別為T1及T2,設無摩擦,試以M,g,θ及表示答案:(a)加速度;(b)T1-T2.另,(c)若M=1Kg,θ=45度則加速度及T1-T2各若干?
兩題物理不會請交教我吧?
回答 (1)
2015-11-29 9:28 pm
1.W = 2 kg木塊以細繩懸起,水平力推之,繩與鉛質方向成37度
(a) F = ?
Lami's law:
W/sin(90+37) = F/sin(90+53) = T/sin90
F = 2*cos53/cos37 = 2*0.6/0.8 = 1.5 kg
(b) T = ?
T = 2/cos37 = 2/0.8 = 2.5 kg
2.質量4M,2M,8M三木塊連接,兩繩張力T1及T2,無摩擦,以M,g,Q表示答案
(a) a = ?
8mg - T1 = 8ma
T1 - T2 - 2mg*sinQ = 2ma
T2 - 4mg*sinQ = 4ma
Sum up: a = (4 - 3sinQ)g/7
(b) T1 - T2 = ?
= 2ma + 2mg*sinQ ;;; by Eq.(2)
= 2m(a + g*sinQ)
= 2m[(4 - 3sinQ)/7 + g*sinQ]
= 8m(1 + g*sinQ)/7
(c) 若M=1 kg, Q=45度, 則加速度及T1-T2各若干?
a = (4 - 3*sin45)*9.8/7 = 2.63 m/s^2
T1 - T2 = 8*1(1 + 9.8*sin45)/7 = 9.06 N
(a) F = ?
Lami's law:
W/sin(90+37) = F/sin(90+53) = T/sin90
F = 2*cos53/cos37 = 2*0.6/0.8 = 1.5 kg
(b) T = ?
T = 2/cos37 = 2/0.8 = 2.5 kg
2.質量4M,2M,8M三木塊連接,兩繩張力T1及T2,無摩擦,以M,g,Q表示答案
(a) a = ?
8mg - T1 = 8ma
T1 - T2 - 2mg*sinQ = 2ma
T2 - 4mg*sinQ = 4ma
Sum up: a = (4 - 3sinQ)g/7
(b) T1 - T2 = ?
= 2ma + 2mg*sinQ ;;; by Eq.(2)
= 2m(a + g*sinQ)
= 2m[(4 - 3sinQ)/7 + g*sinQ]
= 8m(1 + g*sinQ)/7
(c) 若M=1 kg, Q=45度, 則加速度及T1-T2各若干?
a = (4 - 3*sin45)*9.8/7 = 2.63 m/s^2
T1 - T2 = 8*1(1 + 9.8*sin45)/7 = 9.06 N
收錄日期: 2021-05-01 15:27:06
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20151128232909AAMadof