Three forces are applied to a wheel of radius 0.350 m, as shown in ?Fig. E10.4?. 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 net torque on the wheel due to these three forces for an axis perpendicular to the wheel and passing through its center?
Solution 4E We have to calculate the net torque on the wheel due to the given forces. Mathematically, Torque = Force X Perpendicular distance between the force and the axis. Let us have a look at the following figure. Force 11.9 N The perpendicular distance of this force from the axis is zero. Therefore, this force can not produce a torque. So, torque produced by 11.9 N force is zero. = 0 1 Force 8.50 N The perpendicular distance of this force from the axis of rotation is 0.350 m. Moreover, this force acts counterclockwise. Hence the torque produced by it will be positive. Torque by the force 8.50 N, 2 = 8.50 N × 0.350 m = 2.97 N.m Force 14.6 N This force has two components that has been shown in the figure above. The 0 0 x-component is 14.6 cos 40 and the y-component is 14.6 sin 40 . The x-component has a perpendicular distance of zero from the axis of rotation. Therefore, the torque produced by it is zero. The y-component has a perpendicular distance of 0.350 m from the axis. But this component will act clockwise and hence the torque produced by it will be negative. Therefore, torque produced by this force, = 14.6 sin 40 N × 0.350 m 3 = 3.28 N.m 3 Therefore, the net torque on the wheel due to all the force is, = +1 + 2 3 = 0 + 2.97 N.m 3.28 N.m = 0.31 N.m The net torque is -0.31 N.m. The negative sign indicates that it will be clockwise.