For a system to be in equilibrium,the torques acting on it must balance. This is true only if the torques are taken about

$A$ metre scale is supported on a wedge at its centre of gravity. $A$ body of weight '$W$' is suspended from the $20 \text{ cm}$ mark and another weight of $25 \text{ g-wt}$ is suspended from the $74 \text{ cm}$ mark to balance it, and the metre scale remains perfectly horizontal. Neglecting the weight of the metre scale, the weight of the body is: (in $\text{ g-wt}$)

$A$ metal bar $70 \; cm$ long and $4.00 \; kg$ in mass is supported on two knife-edges placed $10 \; cm$ from each end. $A$ $6.00 \; kg$ load is suspended at $30 \; cm$ from one end. Find the reactions at the knife-edges. (Assume the bar to be of uniform cross-section and homogeneous.)

State the principle of moments for a lever.

$A$ 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?

Explain the construction and working of an ideal lever and also explain the principle of moments of force.