Chapter 5, Problem 12SP

A steering wheel has a diameter of 40.0 cm. A force of 30.0 N is applied to its rim on the right, tangent to the wheel and in the plane of it. Determine the size of the resulting torque. [Hint: The moment arm is the radius. Watch out for units.]

To determine

The magnitude of the resulting torque on a steering wheel having a diameter of $40\text{ cm}$ if a force of $30\text{ N}$ is applied to its rim on the right, tangent to the wheel, and in the plane of it.

Answer to Problem 12SP

Solution:

$6\text{ N}\cdot \text{m}$

Explanation of Solution

Given data:

The diameter of the steering wheel is $40\text{ cm}$.

The magnitude of the applied force is $30\text{ N}$.

Formula used:

Write the expression of the resulting torque.

$\tau =rF\sin \theta$

Here, $r$ is the radial distance from the axis of the rotation, $\theta$ is the acute angle between the line of action of the force and radial distance, and $F$ is the applied force.

Explanation:

Recall the expression for $\tau$.

$\tau =rF\sin \theta$

Since the applied force $F$ is applied tangentially to the rim. So, it will make an angle of $90°$ with respect to the radius of the wheel. Therefore, $\theta =90°$.

And, the radial distance from the axis of rotation is

$r=\frac{d}{2}=\frac{40}{2}=20\text{cm}$

Substitute $20\text{ cm}$ for $r$, $30\text{ N}$ for $F$, and $90°$ for $\theta$

$\begin{array}{c}\tau =\left(20\text{ cm}\right)\left(\frac{1\text{ m}}{100\text{ cm}}\right)\left(30\text{ N}\right)\sin 90°\\ =6\text{ N}\cdot \text{m}\end{array}$

Conclusion:

The magnitude of the torque for the given applied force is $6\text{ N}\cdot \text{m}$.