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A 181-kg bookcase is at rest in the front office. A custodian pushes horizontally on the bookcase to try to move it. If it takes 584 N of force before the bookcase begins to move, find the coefficient of static friction between the bookcase and the floor.

1 Answer

4 votes

Answer:

The coefficient of static friction between the bookcase and the floor, μₓ = 0.3292

Step-by-step explanation:

Given data,

The mass of the bookcase, m = 181 kg

The static frictional force of the bookcase, Fₓ = 584 N

The gravitational normal force acting on the bookcase is, F = m g

= 181 x 9.8

= 1773.8 N

The formula for static frictional force is

Fₓ = μₓ F

Where μₓ - the coefficient of the static friction

μₓ = Fₓ / F

= 584 N / 1773.8 N

= 0.3292

Hence, the coefficient of static friction between the bookcase and the floor, μₓ = 0.3292

User Filip Szczepanski
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