Three forces are applied to a wheel of radius 0.350 m, as shown in the figure (F

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 degree angle with the radius. Assume that F1 = 12.0N , F2 = 14.6N , and F3 = 8.25N . 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? What is the direction of the net torque in part A (into the page or out of the page)?

Explanation / Answer

F1 AND COMPONENT of F2 ( F2 COS(40) ) produces zero torque about centre

torque due to F3 = 8.25*0.350 = 2.887 Nm

torque due to F2 sin(40) = 14.6 sin(40)* 0.350 = 3.285 N m

NET TORQUE -------- CLOCKWISE = 0.398 N m

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