Repeat Question 1 for force $\overrightarrow{\mathbf{F}}$ angled upward instead of downward as drawn.

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

(b) The force F is constant and is acting upward.

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,${\mathbf{f}}_{\mathbf{s}}\mathbf{=}{\mathbf{\mu }}_{\mathbf{s}}{\mathbf{F}}_{\mathbf{N}}$ (2)

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

Free body diagram of the box

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

Here,${\mathrm{F}}_{\mathrm{x}}=\mathrm{Fcos}\left(\mathrm{\theta }\right)$, from the free body diagram.

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

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 (i) 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$\theta$ and then $F_x$ or decrease.

So, ${\mathrm{f}}_{\mathrm{s}}$ also decreases.

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

(${\mathrm{a}}_{\mathrm{y}}=0$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. So, the net value decreases.

Hence, ${\mathrm{F}}_{\mathrm{N}}$is decreased.

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 increases $\theta$and then $F_N$ increases.

Hence,role="math" localid="1657178558684" $\mathrm{f}{}_{\mathrm{s}}^{\max }$ is also decreased.

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, ${\mathrm{F}}_{\mathrm{N}}$ decreases with increase in $\theta$, ${\mathrm{F}}_{\mathrm{K}}$is also decreased.