Chapter 5.2, Problem 21PP

To determine

The coefficient of static friction between sofa and carpet.

Answer to Problem 21PP

The coefficient of static friction is ${\mu }_s=0.392$

Explanation of Solution

Given:

A force of 403N is applied on a sofa of mass 105-kg to make it just start moving on the carpet.

Formula Used:

Here, a force is applied horizontally on the sofa 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 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)

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

Calculation:

The coefficient of static friction can be deduced from (3) as,

  ${\mu }_s=\frac{F_{f,static}}{F_{\text{N}}}$

Given the mass of the sofa is 105-k. Thus based on (2), the normal force is

  $F_{\text{N}}=105g$

Here, g represents the acceleration due to gravity that is $g=9.8{\text{ m/s}}^2$ . Thus, the normal force is

  $\begin{array}{c}{F}_{\text{N}}=105\times 9.8\\ =1029\text{ N}\end{array}$

Now, the applied force is $F_A=403\text{-N}$ . Thus, based on (1), the magnitude of static friction force is

  $F_{f,static}=403\text{-N}$ .

Substituting for the static force of friction and the normal force, the coefficient of static friction is obtained as,

  $\begin{array}{c}{\mu }_s=\frac{403}{1029}\\ =0.392\end{array}$

Conclusion :

Thus, the coefficient of static friction between the given surfaces is 0.392.