One force is perpendicular to the rim, one is tangent to it, and the other one m

Question

One force is perpendicular to the rim, one is tangent to it, and the other one makes a 40.0(^circ) angle with the radius. Assume that ( exttip{F_{ m 1}}{F_1} = 11.4{ m N}), ( exttip{F_{ m 2}}{F_2} = 15.6{ m N}), and ( exttip{F_{ m 3}}{F_3} = 8.70{ m N}). Part A 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. Express your answer in newton meters to three significant figures. ( au) = ({ m N cdot m}) SubmitMy AnswersGive Up Provide FeedbackContinue Figure 1 of 1

Explanation / Answer

Let radius = 0.350 m

You need to calculate the tangential net force first. The perpendicular force would have zero force along the tangent. The tangential force, of course, would be 100% tangential. The third force would be scaled according to the cosine of the angle between it and the tangent line, or the sine of the angle between it and the radius:
8.7*sin(radians(40)) = 6.48 N

Now, the NET torque depends on the DIRECTION of the two forces that have an effect. If they are both in the same direction, say clockwise, they add up: 15.6+6.48 = 22.08 N, but if they are opposed they subtract: 15.6-6.48 = 9.12 N, that is, 9.12 N in the direction to the tangential force.

Finally, to calculate the net torque, we multiply the net tangential force with the radius of the wheel:
22.08 * 0.350 = 7.728 Nm
or...
9.12 * 0.350 = 3.192 Nm

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