Shown in the figure is a rigid and uniform one meter long rod AB held in horizontal position by two strings tied to its ends and attached to the ceiling. The rod is of mass 'm' and has another weight of mass 2m hung at a distance of 75 cm from A. The tension in the string at A is:

Length of uniform rod $$AB$$: $$L = 100\text{ cm} = 1\text{ m}$$

Positions of forces measured from end A:

$$\text{Tension } T_A \text{ at } x = 0\text{ m}$$

$$\text{Weight of uniform rod } mg \text{ at } x = 0.5\text{ m}$$

$$\text{Hung mass } 2mg \text{ at } x = 0.75\text{ m}$$

$$\text{Tension } T_B \text{ at } x = 1\text{ m}$$

Taking torque about end B for rotational equilibrium ($$\Sigma \tau_B = 0$$):

$$T_A(1) - mg(1 - 0.5) - 2mg(1 - 0.75) = 0$$

$$T_A - mg(0.5) - 2mg(0.25) = 0$$

$$T_A - 0.5mg - 0.5mg = 0 \implies T_A = 1mg$$