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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

Respuesta :

Answer:

[tex]F' = 4F[/tex]

Explanation:

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

[tex]\tau = L F sin90[/tex]

[tex]\tau = LF[/tex]

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

[tex]\tau = \frac{L}{2}F' sin\theta[/tex]

[tex]LF = \frac{L}{2}F' sin30[/tex]

[tex]LF = \frac{F'L}{4}[/tex]

so we have

[tex]F' = 4F[/tex]

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