Question

In Fig. 6-12, if the box is stationary and the angle $\mathit{\theta }$ between the horizontal and force $\overrightarrow{\mathbf{F}}$is increased somewhat, do the following quantities increase, decrease, or remain the same: (a) ${\mathbf{F}}_{\mathbf{x}}\mathbf{;}$(b) ${\mathbf{f}}_{\mathbf{s}}\mathbf{;}$(c) ${\mathbf{F}}_{\mathbf{N}}\mathbf{;}$(d) ${\mathit{f}}_{\mathbf{s}\mathbf{,}\mathbf{m}\mathbf{a}\mathbf{x}}$(e) If, instead, the box is sliding and $\mathit{\theta }$is increased, does the magnitude of the frictional force on the box increase, decrease, or remain the same?

Short answer

(a) The value of ${\mathrm{F}}_{\mathrm{x}}$ decreases.

(b) The value of ${\mathrm{f}}_{\mathrm{s}}$ decreases.

(c) The value of ${\mathrm{F}}_{\mathrm{N}}$ increases.

(d) The value of $f_s^{max}$ increases.

(e) The magnitude of the frictional force increases for the case of sliding and the increase in angle.

Step-by-step solution

The given data

(a) The box is stationary and $\mathrm{\theta }$is increasing.

(b) The force F is constant.

Understanding the concept of the free body diagram and force

To observe which quantity will increase or decrease we have to write the x and y components of force and draw all the forces acting on it, then use Newton’s 2nd law of motion. Using this concept of force, we can get the relation of force to the other components as given in the data. Comparing these values of the forces and their relating factor will determine their behavior.

Formulae:

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

The static frictional force acting on a body, role="math" localid="1657175579260" ${\mathbf{f}}_{\mathbf{s}}\mathbf{=}{\mathbf{\mu }}_{\mathbf{s}}{\mathbf{F}}_{\mathbf{N}}$ (2)

The kinetic frictional force acting on a body, role="math" localid="1657175587025" ${\mathbf{f}}_{\mathbf{k}}\mathbf{=}{\mathbf{\mu }}_{\mathbf{k}}{\mathbf{F}}_{\mathbf{N}}$ (3)

a) Calculation of Fx

Free body diagram of the box

We know that decreases $\cos \left(\theta \right)$ as increases$\theta$.

Here,$F_x=Fcos\left(\theta \right)$, from the free body diagram.

The value of F is constant but $\theta$is increasing gives $\cos \left(\theta \right)$ lesser and therefore ${\mathrm{F}}_{\mathrm{x}}$also decreases with increasing$\theta$ .

b) Calculation of fs

The given box is stationary so the acceleration of the box along x is 0.

Then by using Newton’s 2nd law along the x-direction, the net horizontal forces acting on the body can be given using equation (1) as,

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

From the above relation, we get that if we increase $\theta$the value of role="math" localid="1657176179248" ${\mathrm{F}}_{\mathrm{x}}$ decrease.

So, role="math" localid="1657176186713" ${\mathrm{f}}_{\mathrm{s}}$ also decreases.

c) Calculation of FN

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

($a_y=0$Since the box is stationary)

$$\begin{gathered} \sum {\mathrm{F}}_{\mathrm{y}}={\mathrm{ma}}_{\mathrm{y}} \\ {\mathrm{F}}_{\mathrm{N}}-{\mathrm{F}}_{\mathrm{g}}-\mathrm{Fsin}\left(\mathrm{\theta }\right)=0 \\ {\mathrm{F}}_{\mathrm{N}}={\mathrm{F}}_{\mathrm{g}}+\mathrm{Fsin}\left(\mathrm{\theta }\right) \end{gathered}$$

Here, m and g are constant so, ${\mathrm{F}}_{\mathrm{g}}$remain the same; but $\theta$ as increases $\sin \left(\theta \right)$function also increases. Thus, the 2nd term also $\theta$increases with increases.

Hence, $F_N$is also increased.

d) Calculation of fsmax

As the value of the maximum frictional force is given by equation (2) as:${\mathrm{f}}_{\mathrm{s}}^{\max }={\mathrm{\mu F}}_{\mathrm{N}}$

Here, $\mu$is constant but it we increase $\theta$, and then the value of ${\mathrm{F}}_{\mathrm{N}}$ increases.

Hence, role="math" localid="1657176784229" $\mathrm{f}{}_{\mathrm{s}}^{\max }$is also increased.

e) Calculation of the behavior of frictional force

If the box is sliding then there is Kinetic frictional force acting on the box is given by equation (3) as,

${\mathrm{F}}_{\mathrm{k}}={\mathrm{\mu F}}_{\mathrm{N}}$

As from part (c) calculations, ${\mathrm{F}}_{\mathrm{N}}$ it increases with an increase in$\theta$ , ${\mathrm{F}}_{\mathrm{K}}$is also increased.