Chapter 4, Problem 34P

Part (a)

To determine

The horizontal force required to move the crate at a steady speed across the floor.

Answer to Problem 34P

Solution:

64.7 N

Explanation of Solution

Given:

Mass of the crate, $m=22\text{kg}$

Coefficient of kinetic friction, ${\mu }_k=0.30$

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 of the crate.

  $N=mg$

  $F={\mu }_{k}mg$

Where F is the force acting, m is the mass, g is the acceleration due to gravity, ${\mu }_k$ is the coefficient of kinetic friction between the crate and the floor.

Calculation:

The crate is required to be moved at steady speed which means acceleration is zero. The required horizontal force would be just equal to the kinetic frictional force.

  $F={\mu }_{k}mg$

  $F=\left(0.30\right)\left(22\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)$

  $F=64.\text{7 N}$

Conclusion:

The horizontal force is 64.7 N.

Part (b)

To determine

The horizontal force if ${\mu }_k$ is zero.

Answer to Problem 34P

Solution:

0 N

Explanation of Solution

Given:

Mass of the crate, $m=22\text{kg}$

Coefficient of kinetic friction, ${\mu }_k=0$

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 of the crate.

  $N=mg$

  $F={\mu }_{k}mg$

Where F is the force acting, m is the mass, g is the acceleration due to gravity, ${\mu }_k$ is the coefficient of kinetic friction between the crate and the floor.

Calculation:

The crate is required to be moved at steady speed which means acceleration is zero. The required horizontal force would be just equal to the kinetic frictional force.

  $F={\mu }_{k}mg$

  $=\left(0\right)mg$

  $=0\text{ N}$

Conclusion:

Therefore, the force required to move the crate on a frictionless surface is zero.