A 40 N crate rests on a rough horizontal floor. A 12 N horizontal force is then applied to it. If the coecients of friction are μs=0.5 and μk=0.4, the magnitude of the frictional force on the crate is:

Question

Giovanny Suarez

Physics

26 June, 07:05

A 40 N crate rests on a rough horizontal floor. A 12 N horizontal force is then applied to it. If the coecients of friction are μs=0.5 and μk=0.4, the magnitude of the frictional force on the crate is:

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Layla · Answer 1

Fr = 20 (N)

Explanation: See atached file (free body diagram)

As for Newton's low

∑ Fy = 0

-mg + N = 0 ⇒ - 40 + N = 0 ⇒ N = 40 [Newtons]

by definition : Fr = μs * N ⇒ Fr = 0,5 * 40 ⇒ Fr = 20 (N)

∑ Fx = 0 body is at rest

Fe - Fr = 0

Fr > Fe

Fr > 12 (N) the body is at rest