$A$ metal rod of weight $W$ is supported by two parallel knife-edges $A$ and $B$. The rod is in equilibrium in a horizontal position. The distance between the two knife-edges is $r$. The centre of mass of the rod is at a distance $x$ from $A$. The normal reaction on $A$ is

$A$ 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?

$A$ small ball of mass $m$ is suspended from the ceiling by a string of length $L$. The ball moves along a horizontal circle with constant angular velocity $\omega$, as shown in the figure. The torque about the centre $(O)$ of the horizontal circle is

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

$A$ 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:

$A$ rigid body in the shape of a $V$ has two equal arms made of uniform rods. What must the angle between the two rods be so that when the body is suspended from one end,the other arm is horizontal?