$A$ uniform rod of length $1\, m$ and mass $4\, kg$ is supported on two knife-edges placed $10\, cm$ from each end. $A$ $60\, N$ weight is suspended at $30\, cm$ from one end. The reactions at the knife edges are:

Two uniform rods of equal length but different masses are rigidly joined to form an $L-$ shaped body,which is then pivoted as shown in the figure. If in equilibrium the body is in the shown configuration,the ratio $M/m$ will be

$A$ rod of weight $W$ is supported by two parallel knife edges $A$ and $B$ and is in equilibrium in a horizontal position. The knives are at a distance $d$ from each other. The centre of mass of the rod is at distance $x$ from $A.$ The normal reaction on $A$ is

$A$ uniform rod of mass $250 \ g$ having length $100 \ cm$ is balanced on a sharp edge at $40 \ cm$ mark. $A$ mass of $400 \ g$ is suspended at $10 \ cm$ mark. To maintain the balance of the rod,the mass to be suspended at $90 \ cm$ mark is: (in $g$)

Write the conditions for the equilibrium of a rigid body.

$A$ rod of mass $M = 1 \ kg$ and length $L = 1 \ m$ is suspended horizontally by two ideal strings as shown in the figure. The strings are attached at a distance of $1/3 \ m$ from each end. First, a mass $m_1$ is suspended from the left end while keeping the rod horizontal. Then, a second mass $m_2$ is suspended from the right end, again keeping the rod horizontal. What is the maximum total mass $(m_1 + m_2)$ that can be suspended in this way while maintaining the horizontal orientation of the rod?