kinetic friction

用戶98670

2009-10-04 8:01 am

Suppose the coefficient of kinetic friction between m(A) and the plane in the figure is u_k = 0.10, and that m(A) = m(B) = 2.7kg
As m(B) moves down, determine the magnitude of the acceleration of m(A)and m(B), given theta = 32 degree.
What smallest value of u_k will keep the system from accelerating?
u_k is kinetic friction

http://tuhsphysics.ttsd.k12.or.us/Tutorial/NewIBPS/PS4_8/PS4_8.htm
the picture is problem 62

回答 (1)

天同

2009-10-06 4:31 am

✔ 最佳答案

The acceleration of the masses can be found from the method given in the link,
acceleration a = g[1-sin(theta) - u.cos(theta)]/2 -------- (1)
where g is the acceleration due to gravity(=9.8 m/s2)
u is the coefficient of kinetic friction(=0.1)
"theta" is the slope angle (=32 degrees)
hence, a = (9.8)[1-sin(32)- 0.1.cos(32)]/2 m/s2 = 1.89 m/s2
If the acceleration a = 0 m/s2, then equation (1) becomes,
0 = g[1-sin(theta) - u.cos(theta)]/2
i.e. 1-sin(theta) - u.cos(theta) = 0
u = [1-sin(theta)]/cos(theta) = (1-sin(32))/cos(32) = 0.55

👍 0 👎 0