The magnitude of torque on a particle of mass | vedclass

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Question

(A) The formula for the magnitude of torque is given by $	au = rF \sin 	heta$,where $r$ is the position vector magnitude,$F$ is the force magnitude,

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

(A) The formula for the magnitude of torque is given by $	au = rF \sin 	heta$,where $r$ is the position vector magnitude,$F$ is the force magnitude,and $	heta$ is the angle between them.Given: $	au = 2.5\,Nm$,$F = 1\,N$,and $r = 5\,m$.Substituting the values into the formula: $2.5 = 5 	imes 1 	imes \sin 	heta$.This simplifies to $\sin 	heta = \frac{2.5}{5} = 0.5$.Since $\sin 	heta = 0.5$,the angle $	heta = \arcsin(0.5) = \frac{\pi}{6}$ radians.

The magnitude of torque on a particle of mass $1\,kg$ is $2.5\,Nm$ about the origin. If the force acting on it is $1\,N$,and the distance of the particle from the origin is $5\,m$,the angle between the force and the position vector is (in radians)

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