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Question: A uniform rod is 4m long and weight 10kg. If it is supported on a knife edge at one meter from the e...

A uniform rod is 4m long and weight 10kg. If it is supported on a knife edge at one meter from the end, what weight placed at that end keeps the rod horizontal.

A

8kg

B

10kg

C

12kg

D

15kg

Answer

10kg

Explanation

Solution

The problem involves a uniform rod in rotational equilibrium. For an object to be in rotational equilibrium, the net torque acting on it about any point must be zero. We will choose the knife edge (support) as our pivot point.

  1. Identify the properties of the rod:

    • Length of the rod (L) = 4m
    • Weight of the rod (W_rod) = 10kg (This is effectively the mass, and we can use it directly in torque calculations as long as we are consistent, as 'g' will cancel out).
    • Since the rod is uniform, its weight acts at its geometric center. The center of the 4m rod is at 2m from either end.

  2. Locate the pivot point:

    • The knife edge is placed at 1m from one end. Let's call this End A. So, the pivot is at 1m from End A.

  3. Identify the forces and their distances from the pivot:

    • Weight of the rod (W_rod): Acts at the center of the rod (2m from End A).
      • Distance of W_rod from the pivot: The pivot is at 1m from End A, and the center of the rod is at 2m from End A. So, the distance is |2m - 1m| = 1m.
      • This force tends to rotate the rod clockwise about the pivot.
    • Added weight (W_added): This weight is placed at "that end", which refers to End A (the end from which the 1m support distance was measured). So, W_added is at 0m from End A.
      • Distance of W_added from the pivot: The pivot is at 1m from End A, and W_added is at 0m from End A. So, the distance is |0m - 1m| = 1m.
      • This force tends to rotate the rod counter-clockwise about the pivot.
    • Normal force from the knife edge: This force acts at the pivot point itself, so it produces no torque about the pivot.

  4. Apply the condition for rotational equilibrium: For the rod to remain horizontal, the sum of clockwise torques must be equal to the sum of counter-clockwise torques about the pivot. Torque (τ) = Force × Perpendicular distance from pivot

    • Clockwise Torque (τ_cw) due to W_rod: τ_cw = W_rod × (distance of W_rod from pivot) τ_cw = 10 kg × 1 m

    • Counter-clockwise Torque (τ_ccw) due to W_added: τ_ccw = W_added × (distance of W_added from pivot) τ_ccw = W_added × 1 m

  5. Equate the torques: τ_ccw = τ_cw W_added × 1 m = 10 kg × 1 m W_added = 10 kg

Therefore, a weight of 10kg placed at that end keeps the rod horizontal.