Question

A box is on a ramp that is at angle$\mathit{\theta }$to the horizontal. As$\mathit{\theta }$ is increased from zero, and before the box slips, do the following increase, decrease, or remain the same: (a) the component of the gravitational force on the box, along the ramp, (b) the magnitude of the static frictional force on the box from the ramp, (c) the component of the gravitational force on the box, perpendicular to the ramp, (d) the magnitude of the normal force on the box from the ramp, and (e) the maximum value${\mathit{f}}_{\mathbf{s}\mathbf{,}\mathbf{m}\mathbf{a}\mathbf{x}}$ of the static frictional force?

Short answer

a) The component of the gravitational force on the box, along with the ramp increases.

b) The magnitude of the static frictional force on the box from the ramp decreases.

c) The component of the gravitational force on the box, perpendicular to the ramp decreases.

d) The magnitude of the normal force on the box from the ramp

e) The maximum value of the static frictional force remains the same.

Step-by-step solution

The given data

(a) A box is on a ramp at an angle to the horizontal.

(b) The angle is increased from zero.

Understanding the concept of the force and friction

To observe which force will increase, decrease or remain the same while increasing the angle from 0, we have to draw the free body diagram of the box on the ramp and then apply Newton's 2nd law of motion. This determines the behavior upon the application of net forces acting on the body.

Formulae:

The force according to Newton’s second law,

F=ma (1)

The gravitational force acting on a body,

${\mathit{F}}_{\mathbf{g}}\mathbf{=}\mathit{m}\mathit{g}$ (2)

The static frictional force acting on a body,

${\mathbf{f}}_{\mathbf{s}}\mathbf{=}\mathbf{\mu N}$ (3)

a) Calculation of the component of the gravitational force along the ramp

Free body diagram of the box on-ramp:

The component of the gravitational force along the ramp using equation (2) is given by, the Mgsin$\left(\theta \right)$ function increases with an increase in the angle $\theta$.

Hence, the gravitational force component along the ramp increases with an increase in the angle.

b) Calculation of the magnitude of the static frictional force

To find the static frictional force on the box we have to use Newton’s 2nd law of motion along the vertical direction, we use equation (i) for the net forces as follows:

(Here, there is no motion along the vertical direction so $\mathrm{a}=0\mathrm{m}/{\mathrm{s}}^2$)

$$\begin{gathered} \underset{}{\sum \mathrm{F}=\mathrm{ma}} \\ \mathrm{N}-\mathrm{Mg}\cos \left(\mathrm{\theta }\right)=0 \\ \mathrm{N}=\mathrm{Mg}\cos \left(\mathrm{\theta }\right) \end{gathered}$$

The relation between the frictional force and Normal force using the above data in equation (3) is given as:

${\mathrm{f}}_{\mathrm{s}}=\mathrm{\mu Mg}\cos \left(\mathrm{\theta }\right)................\left(4\right)$

A cosine function decreases with an increase in angle $\theta$.

Hence, the frictional force ${\mathrm{f}}_{\mathrm{s}}$also decreases with the increase $\mathrm{\theta }$.

c) Calculation of the component of the gravitational force perpendicular to the ramp

The component of the gravitational force perpendicular to the ramp using equation (2) is given by $\mathrm{Mg}\cos \left(\mathrm{\theta }\right)$.

As the cosine function decreases with increase $\mathrm{\theta }$.

Hence, this perpendicular component of gravitational force also decreases.

d) Calculation of the normal force on the box along the ramp

The normal force on the box from the ramp is defined from the part (b) calculations as,

$\mathrm{N}=\mathrm{Mg}\cos \left(\mathrm{\theta }\right)$

The normal force also decreases with increasing angles$\mathrm{\theta }$ considering the cosine function.

e) Calculation of the maximum value of static frictional force

The maximum value of static frictional force when$\mathrm{\theta }=0°$ is given using equation (4) as follows:

${\mathrm{f}}_{\mathrm{s},\max }=\mathrm{\mu Mg}$

From this, we can say that it does not depend on the angle of the inclination so it remains the same.