Question

A uniform rod has mass m and length L. One end of the rod is attached to a fixed point O by a hinge and an additional force F is applied to the other end of the rod in the vertical direction(upwards). Given that the rod is in mechanical equilibrium, what is the magnitude of the applied force F, expressed as a numerical mutiple of mg, where g is the magnitude of the acceleration due to gravity? (To 3 significant figures).

Answer 1

Question#20363

A uniform rod has mass $m$ and length $L$. One end of the rod is attached to a fixed point $O$ by a hinge and an additional force $F$ is applied to the other end of the rod in the vertical direction (upwards). Given that the rod is in mechanical equilibrium, what is the magnitude of the applied force $F$, expressed as a numerical multiple of $mg$, where $g$ is the magnitude of the acceleration due to gravity? (To 3 significant figures).

Solution:

The rod is in mechanical equilibrium its means that (according to the law of the laver):

$$\frac{F}{L} = \frac{mg}{L/2}, \frac{F}{L} = \frac{2mg}{L}$$ $$\frac{F}{L} = \frac{2mg}{L}$$ $$F = 2mg$$

Answer:

$$F = 2mg$$