A wheel of radius R with an axle of radius R/ | vedclass

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(B) The net torque $	au_{	ext{net}}$ is the sum of the torques produced by each force about the central axis.$1$. Torque due to force $F$ at the out

Vedclass · Accepted Answer

$3.2$

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(B) The net torque $	au_{	ext{net}}$ is the sum of the torques produced by each force about the central axis.$1$. Torque due to force $F$ at the outer rim (at $45^{\circ}$): The tangential component of this force is $F \cos(45^{\circ})$. The torque is $	au_1 = R \cdot F \cos(45^{\circ}) = R \cdot F \cdot \frac{1}{\sqrt{2}} \approx 0.707 RF$.$2$. Torque due to force $F$ at the inner axle: This force is applied tangentially at radius $R/2$. The torque is $	au_2 = \frac{R}{2} \cdot F = 0.5 RF$.$3$. Torque due to force $2F$ at the outer rim: This force is applied tangentially at radius $R$. The torque is $	au_3 = R \cdot 2F = 2 RF$.All these torques act in the same rotational sense (counter-clockwise). Therefore,the net torque is:$	au_{	ext{net}} = 	au_1 + 	au_2 + 	au_3 = 0.707 RF + 0.5 RF + 2 RF = 3.207 RF$.Rounding to the nearest value,the magnitude of the net torque is nearly $3.2 FR$.

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