A$\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}\mathit{k}\mathit{g}$block is initially at rest on a horizontal surface. A horizontal force of magnitudeand a vertical force are then applied to the block (Fig. 6-17).The coefficients of friction for the block and surface are ${\mathit{\mu }}_{\mathbf{s}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{40}$and ${\mathit{\mu }}_{\mathbf{K}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{25}$. Determine the magnitude of the frictional force acting on the block if the magnitude of $\overrightarrow{\mathbf{\text{P}}}$is (a)$\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\mathit{N}$, (b) $\mathbf{10}\mathbf{}\mathit{N}$, and (c) $\mathbf{12}\mathbf{}\mathit{N}$.

Mass,$M=2.5\text{kg}$

Horizontal force,$F=6.0\text{N}$

Vertical force,P

Coefficient of static friction,${\mu }_s=0.40$

Coefficient of kinetic friction,${\mu }_k=0.25$

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.

Formula:

$F_{net}=\sum Ma$

Here, F is the net force, mis mass and ais an acceleration.

Free body diagram:

By using Newton’s 2nd law along y direction,

$\sum _{}^{}{F}_y=m{a}_y$

Since block is not moving along y,$a_y=0$

$P+N-mg=0$

$N=mg-P$

Maximum static frictional force is defined as,

$\begin{array}{rcl}{F}_{s,max}& =& {\mu }_s\text{N}\\ & =& {\mu }_s\left(mg-P\right)\\ & & \end{array}$

For, $P=8.0\text{N}$ substitute the values and solve as:

$\begin{array}{rcl}{F}_{s,max}& =& \left(0.40\right)\left(\left(2.5\right)\left(9.81\right)-8.0\right)\\ & =& 6.61\text{N}\\ & & \end{array}$

As$F_{s,max}$ is more than the F.So, the block cannot come into the motion.

Therefore, the frictional force on the block is equal to the $F=F_s=6.0\text{N}$.

For,$\overrightarrow{P}=10\text{N}$

Consider the formula for maximum force as:

$\begin{array}{rcl}{F}_{s,max}& =& {\mu }_s\text{N}\\ & =& {\mu }_s\left(mg-P\right)\\ & & \end{array}$

Substitute the values and solve as:

$\begin{array}{rcl}& & F_{s,max}\left(\left(2.5\right)\left(9.81\right)-10\right)\\ & =& 5.68\text{N}\\ & & \end{array}$

Here,$F_{s,max}$ is less than the F.So, the block come into the motion.

Therefore, the frictional force on the block is equal Kinetic frictional force..

$\begin{array}{rcl}{F}_s& =& {\mu }_K\text{N}\\ & =& {\mu }_k\left(mg-P\right)\\ & =& 3.63\text{N}\\ & & \end{array}$

For,$P=12\text{N}$

Consider the expression for maximum force as:

$\begin{array}{rcl}{F}_{s,max}& =& {\mu }_s\text{N}\\ & =& {\mu }_s\left(mg-P\right)\\ & & \end{array}$

Substitute the values of force as:

$\begin{array}{rcl}{F}_{s,max}& =& \left(0.40\right)\left(\left(2.5\right)\left(9.81\right)-12\right)\\ & =& 5.01 \text{N}\\ & & \end{array}$

Here,$F_{s,max}$ is less than the F.So, the block come into the motion.

Therefore, the frictional force on the block is equal Kinetic frictional force.

$\begin{array}{rcl}{F}_s& =& {\mu }_k\text{N}={\mu }_k\left(mg-P\right)\\ & =& 3.13\text{N}\\ & & \end{array}$