A force of `36N` is applied to a particle located at `0.15 m` from the axis of rotation. What is the magnitude of the torque about this axis, if the angle between the direction of the applied force and radius vector is (a)`120^(@) (b) 45^(@)`?

A force of `36N` is applied to a particle located at `0.15 m` from the

1.	A force of `36N` is applied to a particle located at `0.15 m` from the axis of rotation. What is the magnitude of the torque about this axis, if the angle between the direction of the applied force and radius vector is (a)`120^(@) (b) 45^(@)`?
Answer» Here, `F = 36N, r = 0.15 m, tau = ?` (a) `theta = 120^(@)` `tau = rF sin theta` `= 0.15 xx 36 sin 120^(@) = 0.15 xx 30 xx (sqrt(3))/(2) = 4.67 N-m` (b) `theta = 45^(@)` `tau = rF sin theta` `= 0.15 xx36 sin 45^(@) = 0.15 xx 36 xx (1)/(sqrt(2)) = 3.81 N-m`

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A force of `36N` is applied to a particle located at `0.15 m` from the axis of rotation. What is the magnitude of the torque about this axis, if the angle between the direction of the applied force and radius vector is (a)`120^(@) (b) 45^(@)`?

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