Q.
A metre rod of mass 200 g is suspended by two strings at each end as shown in figure. A body of mass 300 g is suspended by a weightless string at 30 cm mark. The tensions in the string 2 is
Laws of Motion
Solution:
The various forces acting on the metre rod are shown in figure.
Net force acting on the rod is zero.
That is, $T_1 + T_2 - Mg - mg = 0$
or $T_1 + T_2 = (M + m) g $ [ $\therefore $ g = 10 m $s^{-1}$
= (0.3 + 0.2) $\times$ 10 N = 0.5 $\times$ 10 N = 5 N ....(1)
$\therefore $ Net torqure acting on the rod = $T_2 \times (1.00 \, m ) - Mg \times (0.3 \, m) - mg \times (0.5 \, m)$
For equilibrium, net torque acting on the body is zero
$\therefore $ $T_2 \times (1.00 \, m ) - Mg \times (0.5) - mg \times (0.5 \, m) = 0$
i.e. $T_2 = mg \times 0.5 +Mg \times 0.3 = 0.2 \times 10 \times 0.5 + 0.3 \times 10 \times 0.3 $
= 1 + 0.9 = 1.9 N
