Question

A 20 kg crate is placed on a horizontal conveyor belt. The materials are such that μs = 0.6 and μk = 0.3.

(c) What is the maximum acceleration the belt can have without the crate slipping?
(d) If acceleration of the belt exceeds the value determined in part (c), what is the acceleration of the crate?

Answer 1

Normal force = m g
friction force static = .6 mg
for no slip = m a
.6 m g = m a
a = .6 g or about .6*9.8 = 5.88 m/s^2

F sliding friction = .3 m g = m a
a = .3 g = 2.94 g

Answer 2

thank you!

Answer 3

2.34m/s^2

Answer 4

THANKS