A bedroom bureau with a mass of$\mathbf{45}\mathbf{}\mathit{k}\mathit{g}$, including drawers and clothing, rests on the floor. (a) If the coefficient of static friction between the bureau and the floor is$\mathbf{0}\mathbf{.}\mathbf{45}$, what is the magnitude of the minimum horizontal force that a person must apply to start the bureau moving? (b) If the drawers and clothing, with$\mathbf{17}\mathbf{}\mathit{k}\mathit{g}$mass, are removed before the bureau is pushed, what is the new minimum magnitude?

Mass,$M=45 \text{kg}$

Coefficient of static friction,${\mu }_s=0.45$

Mass of the drawers and clothes,$m=17 \text{kg}$

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. According to Newton's 2nd law of motion, a force applied to an object at rest causes it to accelerate in the direction of the force.

Formula:

$F_{net}=\sum Ma$

Here, F is the net force, is mass anda is an acceleration.

In this case Normal force consider the equation for the force as:

$N=Mg$

Substitute the values and solve as:

$\begin{array}{rcl}N& =& \left(45\right)\left(9.81\right)\\ & =& 441.45\text{N}\\ & & \end{array}$

So, the frictional force is,

$\begin{array}{rcl}{F}_s& =& {\mu }_{s}N\\ & =& \left(0.45\right)\left(441.45\right)\\ & =& 198.65\text{N}\\ & & \end{array}$

Thus, the magnitude of the minimum horizontal force that a person must apply to start bureau moving is$198.65\text{N}$

Ifthe drawers and clothes are removed, then Normal force is as follows:

$\begin{array}{rcl}N\text{'}& =& M\text{'}g\\ & =& \left(28\right)\left(9.81\right)\\ & =& 274.68\text{N}\\ & & \end{array}$

And frictional force is as follows:

$\begin{array}{rcl}{F}_s\text{'}& =& {\mu }_{s}N\text{'}\\ & =& \left(0.45\right)\left(274.68\right)\\ & =& 123.61\text{N}\\ & & \end{array}$

Thus, the magnitude of the minimum horizontal force that a person must apply to startbureau moving is$123.61\text{N}$ .