A uniform rod of length 100 cm and mass 3 kg is suspended horizontally by two vertical strings attached at its ends (at the 0 cm and 100 cm marks). A block of mass 6 kg is placed on the rod at a distance x cm from the 0 cm end. If the tension in the string at the 0 cm mark is twice the tension in the string at the 100 cm mark, the value of x is ________.

Step-by-step solution

Let T₁ and T₂ be the tensions in the strings at the 0 cm and 100 cm marks, respectively. Given that T₁ = 2T₂ . For translational equilibrium, the total upward force equals the total downward force: T₁ + T₂ = 3g + 6g = 9g 2T₂ + T₂ = 9g 3T₂ = 9g T₂ = 3g For rotational equilibrium, the net torque about the 0 cm end must be zero. The forces producing torque are the weight of the rod (acting at 50 cm ), the weight of the block (acting at x cm ), and the tension T₂ (acting at 100 cm ). (3g 50) + (6g x) = T₂ 100 150g + 6g