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 m/s²)

Concept:

The point in a body or system of bodies at which whole mass of the body or system of bodies is concentrated is called as center of mass.

The force which is responsible for rotational motion of an object is measured in terms of torque. I

t is a vector quantity.

Formula to calculate torque is 

\(\vec τ = \vec r × \vec F = rF\sin \theta (\hat n)\)

where \(τ \) is torque, r is radius and F is force, \(sin θ \) is sine of angle between r and F.

Calculation:

Given that
Mass of rod = 500 g; Length of rod = 200 cm

Rod will be in equilibrium, when net torque about point O will be zero

Torque at point O due to 2 kg mass

\(\vec τ = \vec r × \vec F = rF\sin \theta (\hat n)\)

τ₁ = 20 × 20 × 10^-2 × sin90° \((\hat k)\) =4 N m \((\hat k)\)

Torque due to mass of rod : 

τ₂ = 5 × 60 × 10^-2 × sin90° \((\hat k)\) = 3 N m \(( - \hat k)\)

Net torque about point O will be zero

So \({{\vec τ }_1} + {{\vec τ }_2} + {{\vec τ }_3} = 0\)

⇒ 4 - 3 - 12m = 0

⇒ 12m = 1

\(m = \frac{1}{{12}}kg\)