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A gymnast stands on a balance beam, which is 5 meters long and has a mass of 40 kg. The balance beam is supported by two pivot points at each end, and the gymnast is positioned 2 meters away from one end. Determine the mass and position of the gymnast needed to maintain equilibrium on the balance beam.

Answer :

A gymnast should stand at a position 2 meters away from one end of the 5-meter-long balance beam to maintain equilibrium.

To determine the mass and position of the gymnast needed to maintain equilibrium on the balance beam, we can set up an equilibrium condition where the total torque around either pivot point is zero.

Let's break down the steps:

  • The total length of the balance beam (L) = 5 meters.
  • The mass of the balance beam (M) = 40 kg.
  • Let x be the position of the gymnast from one end of the beam.
  • Let m be the mass of the gymnast.

Setting up the Equilibrium Condition:

  • The balance beam is in equilibrium when the sum of torques around any pivot point is zero.
  • Choose one end of the beam (say, the left end) as the pivot point.

Torque Due to the Beam's Weight:

  • The weight of the balance beam acts downward at its center (2.5 meters from either end).
  • Torque due to the beam's weight Mg (where g is the acceleration due to gravity) around the left pivot (chosen pivot point) is (Mg)×2.5 meters.

Torque Due to the Gymnast's Weight:

  • The gymnast's weight (mg) acts downward at a distance x from the left end.
  • Torque due to the gymnast's weight around the left pivot is (mg) × x meters.

Setting Up the Equilibrium Equation:

For equilibrium, the sum of clockwise torques must equal the sum of counterclockwise torques.

  • (Mg) × 2.5 = (mg) × x

Simplifying the equation:

  • M × 2.5 = m × x

Rearrange the equation to solve for m:

  • m = (M × 2.5) / x

Substituting Values:

Substitute: M = 40 kg and x = 2 meters (position of the gymnast)

  • x = (40 × 2.5) / 50
  • x = 2 m.

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