Chapter 5.2, Problem 19PP

To determine

The horizontal force to be applied on a given object kept on a given surfacesuch that it is just to start its motion.

Answer to Problem 19PP

The horizontal force to be applied is $F_A=73.7\text{ N}$

Explanation of Solution

Given:

A box along with books is kept on a pavement and it weighs 134 N . The coefficient of static friction between the two surfaces such that box is about to move is 0.55.

Formula Used:

Here, a force is to be applied horizontally on the box full of books just to make it start moving along the carpet. This represents the limiting case of static friction.

There will be a force opposing the applied force which is due to static friction in this case, since the body is just about to start its motion. According to Newton’s third laws, the frictional force $F_{f,static}$ and the applied force $F_A$ are equal and they oppose each other.

This can be expressed as,

  $F_A=-F_{f,static}$ …… (1)

Now, the static friction force will be proportional to the normal force acting in the body. In this case the normal force is equal to the weight of the body as there are no other forces along the normal.

  $F_{\text{N}}=mg$ …… (2)

Thus,static friction force can be expressed as,

  $F_{f,static}={\mu }_s{F}_{\text{N}}$ …… (3)

From this, the magnitude of applied horizontal force can be calculated as,

  $F_A={\mu }_s{F}_{\text{N}}$ …… (4)

Here ${\mu }_s$ represents the coefficient of kinetic friction.

Calculation:

Given the weight of the box along with the books as 135 N. Thus based on (2), the normal force is

  $F_{\text{N}}=134\text{ N}$

Now, substituting the coefficient of static friction and the normal force, the horizontal force to be applied on the box to start its motion is obtained from (4) as,

  $\begin{array}{c}{F}_A=0.55\times 134\\ =73.7\text{ N}\end{array}$

Conclusion :

Thus, the horizontal force to be applied on the box on the limiting case of static friction is 73.7 N.