Question

A box of weight 120 N sits on a level surface. It is initially at rest. The coefficient of kinetic friction is 0.30 and the coefficient of static friction is 0.50. A person goes up to the box and pushes horizontally with a force of 45 N. What is the force of friction acting on the box during the push?

Answer 1

When horizontal Force pushes the block, then Using Force Balance in vertical direction:

Since initially block is at rest, So

Normal force(N) = W(Weight of block) = 120 N

Now maximum friction force on block will be:

Fr_max = friction force on block = s*N

N = Applied force on block = 120 N

s = static friction coefficient = 0.50

So,

Fr = 0.5*120 = 60 N

Now in vertical direction, Since maximum static friction force on block is greater than Horizontal force(40 N), So block will remain at

rest. [Fr > F]

Which means acceleration of block = 0 m/sec^2

If in some other case W > Fr, then there will be kinetic force applied and box will start moving, in that case there will be acceleration.

Now, by force balance in horizontal firection,

Fr - F = m*a

from above part, a = 0

then, Fr = friction acting on the box during the push = F(Horizontal force)

Fr = 45 N

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