物題計算2題

Q:
一輛公車以等加速度率前進,車內以質量可忽略的細繩懸吊一質量為2kg的磚塊,繩與垂直線成30度角,則公車的加速度率a=??細繩上的張力T=?(g取10m/s^2)

A:
Tcos30 = mg
T = mg/cos30
T = 2kg x 10m/s^2 / 0.866
T = 23.1N

Tsin30 = ma
a = Tsin30/m
a = 23.1N x 0.5 / 2kg
a = 5.77m/s^2

Therefore,
公車的加速度率a=5.77m/s^2
細繩上的張力T = 23.1N

2006-10-18 23:39:15 補充：
A2:Let R be normal reaction acting on m,Let u be the coeff. of frictionAcceleration of mass m = F/(M+m)Consider direction perpendicular to plane,mF/(M+m)xsin(theta) + R = mgcos(theta)Consider direction parallel to plane.mgsin(theta) + mF/(M+m)xcos(theta) = uR