In three experiments, three different horizontal forces are applied to the same block lying on the same countertop. The force magnitudes are${\mathit{F}}_{\mathbf{1}}\mathbf{=}\mathbf{12}\mathbf{N}$,${\mathit{F}}_{\mathbf{2}}\mathbf{=}\mathbf{8}\mathbf{N}$, ${\mathit{F}}_{\mathbf{3}}\mathbf{=}\mathbf{4}\mathbf{N}$. In each experiment, the block remains stationary in spite of the applied force. Rank the forces according to (a) the magnitude of the static frictional force on the block from the countertop and (b) the maximum value role="math" localid="1660904123305" ${\mathit{f}}_{\mathbf{s}\mathbf{,}\mathbf{max}}$of that force, greatest first.

The magnitudes of the force are: ${\mathrm{F}}_1=12\mathrm{N}$, ${\mathrm{F}}_2=8\mathrm{N}$, and ${\mathrm{F}}_3=4\mathrm{N}$.

We have to use Newton’s 2nd law of motion along with x and y directions on each block. Now, using the given data of the magnitudes of the force, we can get the behavior of the frictional force and the maximum frictional force.

Formulae:

The force according to Newton’s second law,

F=ma (1)

Free body diagram of blocks:

Blocks are stationary therefore the acceleration of the blocks along x is 0.

By using Newton’s 2nd law along the x-direction, the net of the horizontal force for the first block can be given using equation (1) as:

(Case-1)

$$\begin{gathered} \sum _{}{\mathrm{F}}_{\mathrm{x}}={\mathrm{ma}}_{\mathrm{x}} \\ {\mathrm{F}}_1-{\mathrm{f}}_{{\mathrm{s}}_1}=0 \\ {\mathrm{f}}_{{\mathrm{s}}_1}={\mathrm{F}}_1 \\ {\mathrm{f}}_{{\mathrm{s}}_1}=12\mathrm{N} \end{gathered}$$

By using Newton’s 2nd law along the x-direction, the net of the horizontal force for the first block can be given using equation (1) as:

(Case-2)

$$\begin{gathered} \sum _{}{\mathrm{F}}_{\mathrm{x}}={\mathrm{ma}}_{\mathrm{x}} \\ {\mathrm{F}}_2={\mathrm{f}}_{{\mathrm{s}}_2}=0 \\ {\mathrm{f}}_{{\mathrm{s}}_2}={\mathrm{F}}_2 \\ {\mathrm{f}}_{{\mathrm{s}}_2}=8\mathrm{N} \end{gathered}$$

By using Newton’s 2nd law along the x-direction, the net of the horizontal force for the third block can be given using equation (1) as:

(Case-3)

$$\begin{gathered} \sum _{}{\mathrm{F}}_{\mathrm{x}}={\mathrm{ma}}_{\mathrm{x}} \\ {\mathrm{F}}_3={\mathrm{f}}_{{\mathrm{s}}_3}=0 \\ {\mathrm{f}}_{{\mathrm{s}}_3}={\mathrm{F}}_3 \\ {\mathrm{f}}_{{\mathrm{s}}_3}=4\mathrm{N} \end{gathered}$$

So, the block with ${\mathrm{F}}_1=12\mathrm{N}$force has the greatest frictional force ${\mathrm{f}}_{{\mathrm{s}}_1}$.

Hence, the magnitude of the forces is ordered as ${\mathrm{f}}_{{\mathrm{s}}_1}>{\mathrm{f}}_{{\mathrm{s}}_2}>{\mathrm{f}}_{{\mathrm{s}}_3}$.

The value of the magnitude ${\mathrm{f}}_{\mathrm{s}}^{\max }$of the static frictional force is defined as,

${\mathrm{f}}_{\mathrm{s}}^{\max }=\mathrm{\mu N}$

So, $f_s^{max}$is directly proportional to the normal force N.

By using Newton’s 2nd law along the vertical direction, the net vertical forces acting on the body can be given using equation (1) as (since blocks are stationary $a_y=0$.)

$$\begin{gathered} \sum _{}{F}_y=m{a}_y \\ \mathrm{N}-\mathrm{Mg}=0 \\ \mathrm{N}=\mathrm{Mg} \end{gathered}$$

Here, the mass of the blocks (M) is the same, and g is constant so normal force also remains the same in all blocks.

Therefore, $f_s^{max}$is the same for all blocks that imply the magnitude values as ${\mathrm{f}}_{{\mathrm{s}}_1}^{\max }={\mathrm{f}}_{{\mathrm{s}}_2}^{\max }={\mathrm{f}}_{{\mathrm{s}}_3}^{\max }$.