Question

A box of bananas weighing 40.0 NN rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.40 and the coefficient of kinetic friction is 0.20.
(a) If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on the box?
(b) What is the magnitude of the friction force if a monkey applies a horizontal force of 6.0N to the box and the box is initially at rest?
(c) What minimum horizontal force must a monkey apply to start the box in motion?

Answer 1

Answer:

a) 0 N

b) 6 N

c) So the applied force must be greater than 16 newton.

Explanation:

Given:

weight of the box, $w=40\ N$

coefficient of static friction between the box and horizontal surface, $\mu_s=0.4$

coefficient of kinetic friction, $\mu_k=0.2$

Here the given value of coefficient of static friction is understated, it is given here as the maximum value of the coefficient of static friction which is variable according to the applied force on the body tending to promote a relative motion between the surfaces.

a)

When no horizontal force is applied on the box and the box is at rest then according to the Newton's first law of motion the box tends to have its initial state of rest and all the forces acting on the box are in equilibrium. Hence no frictional force acts on the box.

b)

When a monkey applies a force of 6 N on the box and it does not move then the contact surface friction also applies the equal amount of force in the opposite direction which balances out the applied force, this will happens upto the maximum coefficient of friction which is stated as coefficient of static friction in the question.

c)

the minimum force required to move the box must be greater than the maximum force of friction which resists the relative motion.

We have the maximum frictional force as:

$f=\mu_s.w$

$f=0.4\times 40$

$f=16\ N$

So the applied force must be greater than 16 newton.