For the system to be in equilibrium,the value of mass $m$ should be .......... $kg$.

Consider the situation shown in the figure. $A$ uniform rod of length $L$ can rotate freely about the hinge $A$ in a vertical plane. Pulleys and string are light and frictionless. If the rod remains horizontal at rest when the system is released,then the mass of the rod is:

$A$ metre scale is balanced on a knife edge at its centre. When two coins,each of mass $10\, g$,are placed one on top of the other at the $10.0\, cm$ mark,the scale is found to be balanced at the $40.0\, cm$ mark. The mass of the metre scale is found to be $x \times 10^{-2}\, kg$. The value of $x$ is

$A$ weightless rod is acted on by upward parallel forces of $2N$ and $4N$ at ends $A$ and $B$ respectively. The total length of the rod $AB = 3m$. To keep the rod in equilibrium,a force of $6N$ should act in the following manner:

Write the condition for rotational equilibrium and translational equilibrium.

$A$ non-uniform bar of weight $W$ is suspended at rest by two strings of negligible weight as shown in the figure. The angles made by the strings with the vertical are $36.9^{\circ}$ and $53.1^{\circ}$ respectively. The bar is $2 \; m$ long. Calculate the distance $d$ of the centre of gravity of the bar from its left end.