At the end of a factory production line, boxes start from rest and slide down a 30 degree ramp 5.2 m long.?
If the slide is to take no more than 4.3 s , what's the maximum allowed frictional coefficient?
- oubaasLv 74 months agoFavorite Answer
acceleration = 2d/t^2 = 10.4/4.3^2 = 0.562
accel. a = (motive force-friction force)/mass
0.562 = m*g*(sin30-cos30*μ)/m
mass m cross
0.562 = 4.90-8.49*μ
μ = (4.90-0.562)/8.49 = 0.511
- kuiperbelt2003Lv 74 months ago
First, we find the maximum acceleration. Since the box starts from rest and accelerates down the plane, we can use
s = 1/2 a t^2
to solve for acceleration, since you know the distance, s, is 5.2m and t is 4.3 s Solve for a, we will use it below.
the forces acting along the ramp are the component of gravity down the ramp, given by m g sin (theta), and the frictional force up the ramp which is given by u m g cos(theta) where u is the coeff of friction.
Your book should derive these equations fully.
Newton's second law tells you that ma equals the sum of forces acting on the box, so we have
m a = m g sin(theta) - u m g cos(theta)
where we have chosen down the ramp as the positive direction
this tells us that
a = g( sin (theta) - u cos(theta))
now, use the value of a you computed above with theta = 30 degrees and g = 9.8 m/s/s and solve for u