Chapter 4, Problem 35P

A force of 35.0 N is required to start a 6.0-kg box moving across a horizontal concrete floor. (a) What is the coefficient of static friction between the box and the floor? (b) If the 35.0-N force continues, the box accelerates at 0.60 m/s². What is the coefficient of kinetic friction?

To determine

The coefficient of static friction.

Answer to Problem 35P

Solution:

  ${\mu }_s=0.6$

Explanation of Solution

Given:

Static frictional force $=F_s=35\text{N}$

Mass of the box $=M=6\text{kg}$

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

Formula Used:

Frictional force F is obtained by:

  $F=\mu N$

Where, $\mu$ is the coefficient of friction and N is the normal reaction force.

The normal force would be equal to the weight.

  $N=Mg$

Static frictional force $F_s$ :

  ${\mu }_s=\frac{F_s}{Mg}$

  ${\mu }_s$ is the coefficient of static friction.

Calculation:

Static frictional force is the force required to make the object to start moving from rest. Since the force applied to move the box is $35\text{N}$ , the static frictional force magnitude is also the same.

Using the equation

  ${\mu }_s=\frac{F_s}{Mg}$

Substitute the values

  $\begin{array}{l}{\mu }_s=\frac{F_s}{Mg}\\ {\mu }_s=\frac{35}{6\times 9.8}\\ {\mu }_s=0.6\end{array}$

Conclusion:

The coefficient of friction is $0.6$ .

To determine

The coefficient of kinetic friction.

Answer to Problem 35P

Solution:

  ${\mu }_k=0.53$

Explanation of Solution

Given:

Applied force $=F_a=35\text{N}$

Kinetic frictional force $=F_k=?$

Mass of the box $=M=6\text{kg}$

The coefficient of kinetic friction $={\mu }_k$

Acceleration $=a=0.60{\text{m/s}}^{\text{2}}$

Formula Used:

From Newton’s second law of motion:

  $F=ma$

Where, m is the mass and a is the acceleration.

Frictional force F is obtained by:

  $F=\mu N$

Where, $\mu$ is the coefficient of friction and N is the normal reaction force.

The normal force would be equal to the weight.

  $N=Mg$

Kinetic frictional force $F_k$ :

  ${\mu }_k=\frac{F_k}{Mg}$

  ${\mu }_k$ is the coefficient of static friction.

Calculation:

Kinetic frictional force is the force that acts on the object in opposite direction of motion. Since the force applied to move the box is $35\text{N}$ , the kinetic frictional force acts in opposite direction.

From second law of motion:

  $F_a-F_k=Ma$

Substitute the values and solve.

  $\begin{array}{l}35-F_k=6(0.6)\\ F_k=31.4\text{N}\end{array}$

Substitute the value to find the coefficient of kinetic friction.

  $\begin{array}{l}{\mu }_k=\frac{F_k}{Mg}\\ {\mu }_k=\frac{31.4}{6\times 9.8}\\ {\mu }_k=0.53\end{array}$

Conclusion:

The coefficient of kinetic friction comes out to be $0.53$ .