Question

Three forces are applied to a wheel of radius 0.350 m

, as shown in the figure (Figure 1). One force is perpendicular to the rim, one is tangent to it, and the other one makes a 40.0∘
angle with the radius.

What is the magnitude of the net torque on the wheel due to these three forces for an axis perpendicular to the wheel and passing through its center?

Answer 1

0.310 Nm, into the page

Step-by-step explanation:

To determine the net torque on the wheel due to the three forces, we'll apply the principle of torque. Torque is calculated using the formula:

`\boxed{ \left \begin{array}{ccc} \text{\underline{Torque:}} \\\\ \tau = Fd\sin(\theta) \\\\ \text{Where:} \\ \bullet \ \tau \ \text{is the torque} \\ \bullet \ F \ \text{is the force applied} \\ \bullet \ d \ \text{is the lever arm distance} \\ \bullet \ \theta \ \text{is the angle between the force vector and the lever arm} \end{array} \right.}`

Given:

• F₁ = 11.9 N, the force perpendicular to the rim
• F₂ = 14.6 N, the force at a 40.0° angle with the radius
• F₃ = 8.50 N, the force tangent to the wheel
• r = 0.350 m, the radius of the wheel
• θ = 40.0°

`\hrulefill`

To calculate the net torque, we can sum up the torques produced by F₂ and F₃. The torque produced by F₁ is zero. This is because the angle between the force and lever arm is 0°, so it does not contribute to the net torque. The net torque can be calculated as follows:

`\vec \tau_{\text{net}}= \vec \tau_1+\vec \tau_2+\vec \tau_3`

`\Longrightarrow \vec \tau_{\text{net}}=0-\vec F_2r\sin(\theta)+\vec F_3r`

Now plug in our given values:

`\Longrightarrow \vec \tau_{\text{net}}=-(14.6 \text{ N})(0.350 \text{ m})\sin(40.0 ^\circ)+(8.50 \text{ N})(0.350 \text{ m})`

`\Longrightarrow \vec \tau_{\text{net}}=-3.28464 \text{ Nm} + 2.975 \text{ Nm}\\\\\\\\\Longrightarrow \vec \tau_{\text{net}} \approx -0.310 \text{ Nm}\\\\\\\\\therefore \big|\big|\vec \tau_{\text{net}}\big|\big| \approx \boxed{0.310 \text{ Nm}, \text{ into the page}}`

Thus, the net torque on the wheel is approximately 0.310 Nm, and since the wheel is rotating clockwise, the direction of the torque is into the page.