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. Assume that F1=11.4N, F2=14.8N, and F3=8.65N. What is the net torque on the wheel due to these three forces for an axis perpendicular to the wheel and passing through its center? Indicate the direction of the net torque by letting counterclockwise torques be positive.

Answer 1

Concepts and reason

The required concepts to solve these questions are torque, and force.

Firstly, calculate the torque for the given forces and perpendicular distance. Calculate the perpendicular distance for all the forces as it is necessary for the calculation of the torque.

Fundamentals

Torque: It is a property of a force to rotate an object about a fulcrum.

The expression for torque is given by,

$\tau = rF\sin \theta$

Here, $F$ is the force and $r$ is the distance.

The equation of the net torque for the given problem is given by,

${\rm{Net Torque = }}{\tau _1} + {\tau _2} + {\tau _3}$

Here, ${\tau _1}$ is the force that is at zero degrees with the axis, ${\tau _2}$ is the tangential force and it has two component, vertical component and horizontal component.

Horizontal component of the force that is having an angle of $40^\circ$ with the radius. ${\tau _3}$ is the tangential force. Sign for the clockwise is taken as negative and anticlockwise is taken as positive.

The free body diagram is given by,

The expression for torque is given by,

$\tau = rF\sin \theta$

From the diagram, net torque is,

$\tau = r{F_1}\sin {\theta _1} + r{F_2}\sin {\theta _2} + r{F_3}\sin {\theta _3}$

Substitute $0.35{\rm{ m}}$ for $r$ , $11.4{\rm{ N}}$ for ${F_1}$ , $11.4{\rm{ N}}$ for ${F_2}$ , $11.4{\rm{ N}}$ for ${F_3}$ , $0$ for ${\theta _1}$ , $40^\circ$ for ${\theta _2}$ and $90^\circ$ for ${\theta _3}$ in the above equation.

$\begin{array}{c}\\\tau {\rm{ = }}(11.4{\rm{ N)(0}}{\rm{.35 m)sin(0}}^\circ ) - (14.8{\rm{ N)}}(0.35{\rm{ m}}\,{\rm{)sin}}\,({\rm{40}}^\circ ) + (8.65{\rm{ N)}}(0.35{\rm{ m}}\,{\rm{)sin}}\,\,9{\rm{0}}^\circ \\\\ = - 3.32{\rm{ N}} \cdot {\rm{m}} + 3.02{\rm{ N}} \cdot {\rm{m}}\\\\ = - 0.30{\rm{ }}\,{\rm{J}}\\\end{array}$

Ans:

The net torque is $- 0.3{\rm{ J}}$ .