A uniform meter scale of mass 0.24 kg is made of steel. It is kept on

A uniform meter scale of mass 0.24 kg is made of steel. It is kept on two wedges, W₁ and W₂, in a horizontal position. W₁ is at a distance of 0.2 m from one of its ends, while W₂ is at distance of 0.4 m from the other end. If the force on the scale is N₁ due to W₁ and N₂ due to W₂, then : (take g = 10.0 m s⁻²)

N₁ = 1.6 N and N₂ = 0.8 N
N₁ = 0.8 N and N₂ = 1.6 N
N₁ = 0.6 N and N₂ = 1.8 N
N₁ = 1.8 N and N₂ = 0.6 N
This question was previously asked in
UPSC NDA-1 – 2024
The correct option is C. Calculating the forces using equilibrium conditions yields N₁ = 0.6 N and N₂ = 1.8 N.
For a uniform meter scale of mass 0.24 kg, its weight (W = mg = 0.24 * 10 = 2.4 N) acts at its center of mass, which is at the 0.5 m mark from either end. W₁ is at 0.2 m from one end (say, the 0m mark), and W₂ is at 0.4 m from the other end (the 1m mark), placing it at 1.0 – 0.4 = 0.6 m from the 0m mark. For equilibrium, the sum of upward forces equals the sum of downward forces (N₁ + N₂ = W = 2.4 N), and the sum of torques about any point is zero. Taking torques about the point W₁ (at 0.2m), the weight W (at 0.5m) creates a clockwise torque W * (0.5 – 0.2) = 2.4 * 0.3 = 0.72 Nm. The force N₂ (at 0.6m) creates an anticlockwise torque N₂ * (0.6 – 0.2) = N₂ * 0.4 Nm. Setting the sum of torques to zero: 0.72 – 0.4 N₂ = 0, which gives N₂ = 0.72 / 0.4 = 1.8 N. Substituting N₂ into the force equation: N₁ + 1.8 = 2.4, which gives N₁ = 2.4 – 1.8 = 0.6 N.
This problem is a classic example of applying the conditions for static equilibrium: zero net force and zero net torque. Choosing the pivot point for calculating torques at one of the unknown force locations simplifies the calculation by eliminating that force from the torque equation.
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