An iron sphere weighing $10 \, N$ rests in a $V$-shaped smooth trough whose sides form an angle of $60^o$ as shown in the figure. Then the reaction forces are:

$P, Q$ and $R$ are three coplanar forces acting at a point and are in equilibrium. Given $P = 1.9318 \, kg \, wt$,$\sin {\theta _1} = 0.9659$,the value of $R$ is (in $kg \, wt$):

$A$ mass $M \text{ kg}$ is suspended by a weightless string. The horizontal force required to hold the mass at $60^{\circ}$ with the vertical is

$A$ book of mass $5 \,kg$ is placed on a table and it is pressed by $10 \,N$ force. Then,the normal force exerted by the table on the book is ......... $N$.

$A$ body of mass $m$ is suspended by two strings making angles $\theta_1$ and $\theta_2$ with the horizontal ceiling with tensions $T_1$ and $T_2$ respectively. If $T_1 = \sqrt{3} T_2$,then the angles $\theta_1$ and $\theta_2$ are:

As shown in the figure,the two sides of a step ladder $BA$ and $CA$ are $1.6 \; m$ long and hinged at $A$. $A$ rope $DE$ of length $0.5 \; m$ is tied halfway up. $A$ weight of $40 \; kg$ is suspended from a point $F$,$1.2 \; m$ from $B$ along the ladder $BA$. Assuming the floor to be frictionless and neglecting the weight of the ladder,find the tension in the rope and the forces exerted by the floor on the ladder. (Take $g = 9.8 \; m/s^2$)