A box of mass 22 kg is at rest on a flat floor. The coefficient of static fricti

Physics
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A box of mass 22 kg is at rest on a flat floor. The coefficient of static friction between the box and floor is 0.37. What is the minimum force needed to set the box in motion across the floor?

a.  112 N

b.  80 N

c.  56 N

d.  216 N

Apr 24th, 2015

In order to solve this question, you must know how the coefficient of static friction is defined. The coefficient of static friction often denoted with the greek letter mu, is defined as the ratio of the maximum static frictional force to the normal force. Mathematically, this means, mu = max frictional force/normal force, or mu * normal force = max frictional force. So in order to set the box in motion we must apply a force greater than the maximum frictional force, because this will create an unbalanced force on the object and accelerate it. Now, let's find the maximum static friction force. 

Max static friction force = mu * normal

normal = mg = (22*9.8)

(This is because the forces in the vertical direction must sum to zero because the object isn't accelerating vertically)

max static friction force = .37*(22*9.8) = 79.772

So B is the answer.

Apr 24th, 2015

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