A uniform rod of length 200 cm and mass 500 g is balanced on a wedge placed at 40 cm mark. A mass of

Question

A uniform rod of length 200 cm and mass 500 g is balanced on a wedge placed at 40 cm mark. A mass of 2 kg is suspended from the rod at 20 cm and another unknown mass ' $m$ ' is suspended from the rod at 160 cm mark as shown in the figure. Find the value of ' $m$ ' such that the rod is in equilibrium. $\left(g=10 \mathrm{~m} / \mathrm{s}^{2}ight)$\

Professor PC Thomas Classes · Answer

(a) : Here, $A D=20 \mathrm{~cm}$\ $A E=160 \mathrm{~cm}$\ $g=10 \mathrm{~m} / \mathrm{s}^{2}$\ Take moments about $C$\ Clockwise moment $=$ Anticlockwise moment $$ \begin{array}{ll} & 2 	imes g 	imes D C=0.50 	imes g 	imes C C_{1}+m 	imes g 	imes C E \ 	ext { or } & 2 	imes(40-20)=0.500(100-40)+m 	imes(60+60) \ 	ext { or } & 40-30=120 \mathrm{~m} \ 	ext { or } & m=\frac{1}{12} \mathrm{~kg} \end{array} $$