contestada

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:

Respuesta :

Answer:

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

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