Three forces are applied to a wheel of radius 0.290 m, as shown in the figure be

Question

Three forces are applied to a wheel of radius 0.290 m, as shown in the figure below. One force (F1 = 12.0 N) is perpendicular to the rim, one (F2 = 9.05 N) is tangent to it, and the other one (F3 = 13.3 N) 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? (Let counterclockwise torques be positive.)

Three forces are applied to a wheel of radius 0.290 m, as shown in the figure below. One force (F1 = 12.0 N) is perpendicular to the rim, one (F2 = 9.05 N) is tangent to it, and the other one (F3 = 13.3 N) 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? (Let counterclockwise torques be positive.)

Explanation / Answer

torque =r x f T1 = 0 x F1 = 0 T2 = 9.05 X .29 = 2.6245 Nm T3 = 13.3 x Sin40 x.29 = 2.47923 Nm not torque = T2 - T3 =2.6245- 2.47923 = 0.14527 Nm

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