Question

A box of mass 12 kg is at rest on a flat floor. The coefficient of static friction between the box and floor is 0.42. What is the minimum force needed to set the box in motion across the floor?

Answer 1

Answer: 50.4N

Explanation:

Frictional force (Ff) is the force that opposes the moving force acting on a body.

This frictional force (Ff) is the product of the coefficient of friction(n) and normal reaction (R)

Mathematically, Fm = Ff = nR (since both forces are equal and opposite forces along the horizontal axis.)

NB: Fm = Ff (true for static bodies only)

Also note that the weight (W) of the body will be equal to the normal reaction (R) (forces acting along the y axis on the body) i.e W = R

R = Weight = mg = 12×10 = 120N

n = 0.42

Substituting the datas given into the formula of friction force

Ff = nR = Fm

Fm = 0.42×120

Fm = 50.4N

Therefore the minimum force needed to set the box in motion across the floor is 50.4N

Answer 2

The maximum static force that can be applied is equal to the normal force*the frictional force. the normal force on the box is equal to mg since the floor is flat using 9.81m/s^2 for gravity 12kg*9.81m/s^2 = 118N multiplying the normal force by the frictional force you get a 118*.42= 49.6N so overcome the force of static friction on the box a minimum of 49.6N would need to be applied.