A rigid mass less rod of length L is hinged at its one end and a weight W is hung at a distance l(

Question

A rigid mass less rod of length L is hinged at its one end and a weight W is hung at a distance l( < L) from this end. What a force P should be applied upwards at the other end so that the rod remains in equilibrium horizontally? (A) (W(L-l)/L) (B) (W(L-l)/l) (C) (W l/L) (D) (W L/l)

Tardigrade Admin · Accepted Answer

Taking torque about hinge A. P L-W l=0 ⇒ P=(Wl/L)

Q. A rigid mass less rod of length $L$ is hinged at its one end and a weight $W$ is hung at a distance $l (< L)$ from this end. What a force $P$ should be applied upwards at the other end so that the rod remains in equilibrium horizontally?

$\frac{W ( L - l )}{L}$

$\frac{W ( L - l )}{l}$

$\frac{W l}{L}$

$\frac{W L}{l}$

Taking torque about hinge $A$ .
$P L - W l = 0 \Rightarrow P = \frac{W l}{L}$

Q. A rigid mass less rod of length L is hinged at its one end and a weight W is hung at a distance l(<L) from this end. What a force P should be applied upwards at the other end so that the rod remains in equilibrium horizontally?

LW(L−l)​

lW(L−l)​

LWl​

lWL​