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A 45.0 kg crate sits in the bed of the truck accelerating at 3.24m/s^2 . the...

A 45.0 kg crate sits in the bed of the truck accelerating at 3.24m/s^2 . the coefficient of the static friction between the crate and the truck bed is 0.422, and the coefficient of the kinetic friction 0.383. if the crate is not sliding on the truck bed, what is the frictional force acting on the crate?

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Answer #1

As crate is at rest with respect to truck therefore acceleration of crate is also 3.24 m/s2 in horizontal direction. This acceleration will be provided by friction force acting on the crate.

For crate in horizontal direction,

Net force= mass × acceleration

Friction force = mass × acceleration

f = ma

= (45 kg)×(3.24 m/s2)

= 146 N ---Answer

( Since this is less than limiting values of static friction force , hence block does not slide on the truck surface.

Limiting friction force =

= 0.422×45×9.8=186 N)

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