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 (g = 10 ms⁻²).

By Shivank Goel|Updated : August 17th, 2022

A uniform rod of length 200 cm and mass 500 g

  1. 1/2 kg
  2. ⅓ kg
  3. ⅙ kg
  4. 1/12 kg

We know that

The measure of the force which causes an object to rotate about an axis is the torque. The force causes an object to accelerate in linear kinematics. In a similar way, torque causes an angular acceleration. Therefore, torque is defined as the rotational equivalent of linear force.

The forces acting on the rod are shown in the figure below

A uniform rod of length 200 cm and mass 500 g_1

For the rod to remain in equilibrium

𝛕net = 0

Let us calculate the net torque of the wedge

2g x 20 = 0.5g x 60 + mg x 120

By further calculation

m = 0.5/ 6 kg = 1/12 kg

Therefore, the value of m is 1/12 kg.

Summary:

A uniform rod of length 200 cm and mass of 500 g is balanced on a wedge placed at the 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 the 160 cm mark as shown in the figure. Find the value of m such that the rod is in equilibrium (g = 10 ms⁻2).

A uniform rod of length 200 cm and mass of 500 g is balanced on a wedge placed at the 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 the 160 cm mark as shown in the figure. The value of m is 1/12 kg such that the rod is in equilibrium (g = 10 ms⁻2).

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