Question

A mug rests on an inclined surface, as shown in (Figure 1) , θ=10∘.

A.) What is the magnitude of the frictional force exerted on the mug?

B.) What is the minimum coefficient of static friction required to keep the mug from sliding?

Answer 1

Concepts and reason

The concepts needed to solve this problem are the Newton’s second law of motion and static friction.

Draw a useful force diagram to solve for reaction force and the component of the weight of the mug along the inclined plane. Use the concept that the static friction force is self-adjusting friction force and solve for the static friction force on the mug.

Use the equation for static friction force and use the solved static friction force and solve for the required coefficient of static friction.

Fundamentals

From the Newton’s second law of motion, the net force $\sum F$ acting on a body is equal to its mass m multiplied by the acceleration a of the body.

$\sum F = ma$

The expression for the static friction ${f_s}$ is,

${f_s} = {\mu _s}N$

Here, ${\mu _s}$ is the coefficient of static friction and N is the normal force on the body exerted by the surface.

The weight w of a mass of m can be calculated as follows:

$w = mg$

Here, g is the acceleration due to gravity.

(a)

The force diagram for the given situation is as follows:

In the figure, N is the reaction force on the mug exerted by the inclined surface, m is the mass of the mug, g is the acceleration due to gravity, and $\theta$ is the angle of inclination.

The net force equation for the mug along the inclined plane can be written as follows:

$\begin{array}{c}\\\sum F = ma\\\\mg\sin \theta - f = ma\\\end{array}$

As there is no motion along the inclined surface, the acceleration of the mug should be equal to zero.

$a = 0$

Substitute 0 for a in the equation $mg\sin \theta - f = ma$ , and solve for friction force.

$\begin{array}{c}\\mg\sin \theta - f = m\left( 0 \right)\\\\f = mg\sin \theta \\\end{array}$

Substitute 0.27 kg for m, $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g, and $10^\circ$ for $\theta$ .

$\begin{array}{c}\\f = \left( {0.27{\rm{ kg}}} \right)\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\sin 10^\circ \\\\ = {\rm{0}}{\rm{.459 N}}\\\\ = {\rm{0}}{\rm{.46 N}}\\\end{array}$

(b)

The expression for the static friction ${f_s}$ is,

${f_s} = {\mu _s}N$

There are two forces N and $mg\cos \theta$ are acting on the mug in the normal direction to the inclined plane. The net force on the mug in the normal direction to the incline is,

$\sum F = N - mg\cos \theta$

As the mug is not moving in this direction, the net force on the mug should be equal to zero.

$\sum F = 0$

Substitute 0 for $\sum\limits_{} F$ in the equation $\sum F = N - mg\cos \theta$ .

$\begin{array}{c}\\0 = N - mg\cos \theta \\\\N = mg\cos \theta \\\end{array}$

Substitute $mg\cos \theta$ for N in the equation ${f_s} = {\mu _s}N$ .

${f_s} = {\mu _s}\left( {mg\cos \theta } \right)$

Rearrange the equation for ${\mu _s}$ .

${\mu _s} = \frac{{{f_s}}}{{mg\cos \theta }}$

Substitute 0.459 N for f, 0.27 kg for m, $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g, and $10^\circ$ for $\theta$ .

$\begin{array}{c}\\{\mu _s} = \frac{{0.459{\rm{ N}}}}{{\left( {0.27{\rm{ kg}}} \right)\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\cos 10^\circ }}\\\\ = {\rm{0}}{\rm{.176}}\\\\ = {\rm{0}}{\rm{.18}}\\\end{array}$

Ans: Part a

The friction force on the mug is ${\rm{0}}{\rm{.46 N}}$ .