Question

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?

Short answer

a) The magnitude of the frictional force${\mathrm{f}}_{\mathrm{s}}$ on the box remains the same.

b) The magnitude of the normal force on the box from the floor increases.

c) The maximum value ${\mathrm{f}}_{\mathrm{s}}^{\max }$of the magnitude of the static frictional force on the box increases.

d) The box eventually slides.

Step-by-step solution

The given data

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.

Understanding the concept of the free body diagram and force

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)

a) Calculation of the magnitude of the frictional force

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.

b) Calculation of the magnitude of the normal force

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.

c) Calculation of the magnitude of the maximum static frictional force

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.

d) Calculation of the sliding condition of the box

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