Chapter 8, Problem 24P

To determine

To Find: The net torque based on the axle of the wheel.

Answer to Problem 24P

The net torque about the axle of the wheel is $1.4\text{N}\cdot \text{m}$ .

Explanation of Solution

Given information:

Assume that a friction torque of $0.40\text{N}\cdot \text{m}$ opposes the motion.

  

Formula used:

  $\tau =rF\sin \theta$

Here,

  $\tau$ is the torque.

  $L$ is the distance.

  $\theta$ is the angle.

  $F$ is the force.

Calculation:

Torque is defined as the product of the force and the perpendicular distance between the rotation axis and the point of application of the force,

  $\begin{array}{l}\tau =rF\sin \theta \\ \tau =r_{\perp }F\end{array}$

Here,

F is the force.

  $r_{\perp }$ is the perpendicular distance.

The wheel is acted upon by different forces.

Hence, the force is at an angle $45°$ to the face of door is $29\text{N}\cdot \text{m}$ .

  

In the above figure, three forces of different magnitudes are being acted on the wheel at different distances from the axis of rotation Let the forces of magnitude $18\text{N},28\text{}\text{}\text{}\text{N}$ and $35\text{N}$ be $F_1,F_2\text{and}{F}_3$ .

Convert the distances from the axis of rotation from cm to m.

  $\begin{array}{l}{r}_1=\left(12\text{cm}\right)\left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)\\ r_1=0.12\text{m}\\ r_2=\left(\text{24}\text{cm}\right)\left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)\\ r_2=0.24\text{m}\end{array}$

Assume that counter clockwise torques as positive and clockwise torques as negative.

Calculate the torque on the wheel about the axle of the wheel due to three applied forces.

  $\tau =-F{\times }_1{r}_2+F_2\times \left(r_1\right)-F_3\times \left(r_1\right)$

Substituting the values in equation (1) as shown below,

  $\begin{array}{c}\tau =-F{\times }_1{r}_2+F_2\times \left(r_1\right)-F_3\times \left(r_1\right)\\ =-\left(18\text{N}\right)\left(0.24\text{m}\right)+\left(28\text{N}\right)\left(0.24\text{m}\right)-\left(35\text{N}\right)\left(0.12\text{m}\right)\\ =-4.32\text{N}\cdot \text{m}+6.72\text{N}\cdot \text{m}-4.2\text{N}\cdot \text{m}\\ =-1.8\text{N}\cdot \text{m}\end{array}$

Negative sign indicates that the torque due to the applied forces is in the clock wise direction.

Frictional always opposes the motion of the object. Since, the net torque on the object is ln the clockwise motion, the frictional torque $f_{\text{torque}}$ will be counter clockwise.

Calculate the net torque about the axle of the wheel.

  ${\tau }_{net}=-\tau +f_{\text{torque}}$

Substituting the values as shown below,

  $\begin{array}{l}{\tau }_{net}=-\tau +f_{\text{torque}}\\ =-1.8\text{N}\cdot \text{m}+\text{0}\text{.40}\text{N}\cdot \text{m}\\ {\tau }_{\text{net}}=-1.4\text{N}\cdot \text{m}\end{array}$

Negative sign indicates that the net torque on the wheel is in the clock wise direction.

Therefore, the net torque about the axle of the wheel is $1.4\text{N}\cdot \text{m}$ .

Conclusion:

The net torque about the axle of the wheel is $1.4\text{N}\cdot \text{m}$ .