$A$ 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:

$A$ $V$-shaped rigid body has two identical uniform arms. What must be the angle between the two arms,so that when the body is hung from one end,the other arm is horizontal?

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

$A$ car weighs $1800 \, kg$. The distance between its front and back axles is $1.8 \, m$. Its centre of gravity is $1.05 \, m$ behind the front axle. The force exerted by the level ground on each back wheel is ....... $N$.

$A$ 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?

Find the value of $T_1$ (in $N$),if the system is in equilibrium and the mass of the rod is $10 \ kg$ and its length is $10 \ m$. $[g = 10 \ m/s^2]$ (in $N$)