Question

A rod of length L is pivoted about its left end and has a force F applied perpendicular to the other end. The force F is now removed and another force F' is applied at the midpoint of the rod. If F' is at an angle of 30° with respect to the rod, what is its magnitude of the resultant torque is the same as when F was applied

Answer 1

Answer:

$F' = 4F$

Explanation:

When force applied to the end of the rod in perpendicular direction then net torque on the rod is given as

$\tau = L F sin90$

$\tau = LF$

Now another force is applied at mid point of the rod at an angle of 30 degree with the rod

so new value of torque is given as

$\tau = \frac{L}{2}F' sin\theta$

$LF = \frac{L}{2}F' sin30$

$LF = \frac{F'L}{4}$

so we have

$F' = 4F$