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

`F' = 4F`

Step-by-step 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 = (L)/(2)F' sin\theta`

`LF = (L)/(2)F' sin30`

`LF = (F'L)/(4)`

so we have

`F' = 4F`