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(B) The radius of the wheel is $r = 20\, cm = 0.2\, m$.Torque $	au = r \cdot F \cdot \sin 	heta$,where $	heta$ is the angle between the position ve

Vedclass · Accepted Answer

$1.8\, N-m$ (clockwise)

Vedclass · Answer

(B) The radius of the wheel is $r = 20\, cm = 0.2\, m$.Torque $	au = r \cdot F \cdot \sin 	heta$,where $	heta$ is the angle between the position vector and the force vector.$1$. For the $4\, N$ force at $A$: The force is tangential,so $	heta = 90^{\circ}$. It produces an anticlockwise torque: $	au_A = 0.2 	imes 4 	imes \sin 90^{\circ} = 0.8\, N-m$ (anticlockwise).$2$. For the $8\, N$ force at $B$: The angle between the radial line and the force is $30^{\circ}$. It produces a clockwise torque: $	au_B = 0.2 	imes 8 	imes \sin 30^{\circ} = 1.6 	imes 0.5 = 0.8\, N-m$ (clockwise).$3$. For the $6\, N$ force at $C$: The force is directed towards the center,so $	heta = 0^{\circ}$. $	au_C = 0.2 	imes 6 	imes \sin 0^{\circ} = 0\, N-m$.$4$. For the $9\, N$ force at $D$: The force is tangential,so $	heta = 90^{\circ}$. It produces a clockwise torque: $	au_D = 0.2 	imes 9 	imes \sin 90^{\circ} = 1.8\, N-m$ (clockwise).Net torque $	au_{net} = 	au_A - 	au_B - 	au_D = 0.8 - 0.8 - 1.8 = -1.8\, N-m$.The negative sign indicates a clockwise direction. Thus,the net torque is $1.8\, N-m$ (clockwise).

$A$ wheel of radius $20\, cm$ has forces applied to it as shown in the figure. The net torque produced by the forces $4\, N$ at $A$,$8\, N$ at $B$,$6\, N$ at $C$,and $9\, N$ at $D$ at the angles indicated is:

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