Two spherical objects each of radii $R$ and masses $m_1$ and $m_2$ are suspended using two strings of equal length $L$ as shown in the figure $(R << L)$. The angle $\theta$ which mass $m_2$ makes with the vertical is approximately
$\frac{m_1 R}{\left(m_1+m_2\right) L}$
$\frac{2 m_1 R}{\left(m_1+m_2\right) L}$
$\frac{2 m_2 h}{\left(m_1+m_2\right) L}$
$\frac{m_2 R}{\left(m_1+m_2\right) L}$
$A$ right triangular plate $ABC$ of mass $m$ is free to rotate in the vertical plane about a fixed horizontal axis through $A$. It is supported by a string such that the side $AB$ is horizontal. The reaction at the support $A$ is:
A non-uniform bar of weight $W$ is suspended at rest by two strings of negligible weight as shown in 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.
Consider the situation shown in the figure. Uniform rod of length $L$ can rotate freely about the hinge $A$ in vertical plane. Pulleys and string are light and frictionless. If therod remains horizontal at rest when the system is released then the mass of the rod is :
Two light vertical springs with equal natural lengths and spring constants $k_1$ and $k_2$ are separated by a distance $l$. Their upper ends are fixed to the ceiling and their lower ends to the ends $A$ and $B$ of a light horizontal rod $AB$. $A$ vertical downwards force $F$ is applied at point $C$ on the rod. $AB$ will remain horizontal in equilibrium if the distance $AC$ is
When it is said that body is in mechanical equilibrium ?