In Fig. 6-13, horizontal force${\overrightarrow{\mathbf{F}}}_{\mathbf{1}}$of magnitude 10N is applied to a box on a floor, but the box does not slide. Then, as the magnitude of vertical force${\overrightarrow{\mathbf{F}}}_{\mathbf{2}}$is increased from zero, do the following quantities increase, decrease, or stay the same: (a) the magnitude of the frictional force${\overrightarrow{\mathbf{f}}}_{\mathbf{s}}$on the box; (b) the magnitude of the normal force on the box from the floor; (c) the maximum value${\overrightarrow{\mathbf{f}}}_{\mathbf{s}\mathbf{.}\mathbf{m}\mathbf{a}\mathbf{x}}$of the magnitude of the static frictional force on the box? (d) Does the box eventually slide?

a) The magnitude of the horizontal force,${\mathrm{F}}_1=10\mathrm{N}$

b) The magnitude of the vertical force${\mathrm{F}}_2$ is increased from zero.

To observe which quantity will increase or decrease we have to draw the free body diagram of the given system and then apply Newton’s 2nd law of motion. Using this concept of force, we can get the values of the frictional forces and the normal force. This determines the nature of the sliding of the box.

Formulae:

The force according to Newton’s second law, $\overrightarrow{\mathbf{F}}\mathbf{=}\mathbf{m}\overrightarrow{\mathbf{a}}$ (1)

The static frictional force acting on a body, ${\mathbf{f}}_{\mathbf{s}}\mathbf{=}{\mathbf{\mu }}_{\mathbf{s}}{\mathbf{F}}_{\mathbf{N}}$ (2)

Free body diagram of the box

Initially, the box is at rest therefore the acceleration of the box along x is 0.

By using Newton’s 2nd law along the x-direction, the net x-component forces are given using equation (1) as,

$$\begin{gathered} \underset{}{\sum {\mathrm{F}}_{\mathrm{x}}={\mathrm{ma}}_{\mathrm{x}}} \\ {\mathrm{F}}_1-{\mathrm{f}}_{\mathrm{s}}=0 \\ {\mathrm{f}}_{\mathrm{s}}={\mathrm{f}}_1 \\ {\mathrm{f}}_{\mathrm{s}}=10\mathrm{N} \end{gathered}$$

As ${\mathrm{F}}_1$is constant, the magnitude of frictional force${\mathrm{f}}_{\mathrm{s}}$ also remains the same.

Initially, the box is at rest therefore the acceleration of the box along y is 0.

By using Newton’s 2nd law along the y-direction, the net vertical forces can be given using equation (1) as:

$$\begin{gathered} \underset{}{\sum {\mathrm{F}}_{\mathrm{y}}={\mathrm{ma}}_y} \\ {\mathrm{F}}_N-F_2=0 \\ {F}_{\mathrm{N}}=F_2 \end{gathered}$$

From the above relation, we get that if we increase${\mathrm{F}}_2$ ,role="math" localid="1657167546271" ${\mathrm{F}}_{\mathrm{N}}$ it also increases.

The Normal force${\mathrm{F}}_{\mathrm{N}}$ increases with an increase in${\mathrm{F}}_2$ . The relation between${\mathrm{f}}_{\mathrm{s}}$and${\mathrm{F}}_{\mathrm{N}}$ is given using equation (2) as:${\mathrm{f}}_{\mathrm{s}}^{\max }={\mathrm{\mu F}}_{\mathrm{N}}$

So,${\mathrm{f}}_{\mathrm{s}}^{\max }$is directly proportional to the normal force${\mathrm{F}}_{\mathrm{N}}$ .

Hence, an increase in${\mathrm{F}}_2$ will increase the value of maximum static frictional force.

The box does not slide because of the increase in the value of the maximum static frictional force ${\mathrm{f}}_{\mathrm{s}}^{\max }$.