In the fig., the areas of the plunger A and cylinder B are 38.7 cm2 and 387 cm2, respectively, and the weight of B is 4500 N. The vessel and the connecting passages are filled with oil of specific gravity 0.75. What force F is required for equilibrium, neglecting the weight of A?

Problem Statement

In the figure, the areas of the plunger A and cylinder B are \( 38.7 \, \text{cm}^2 \) and \( 387 \, \text{cm}^2 \), respectively, and the weight of B is \( 4500 \, \text{N} \). The vessel and the connecting passages are filled with oil of specific gravity \( 0.75 \). What force \( F \) is required for equilibrium, neglecting the weight of A?

Solution

Given:

  • Area of plunger A (\( A_A \)) = \( 38.7 \, \text{cm}^2 \)
  • Area of cylinder B (\( A_B \)) = \( 387 \, \text{cm}^2 \)
  • Weight of B (\( W_B \)) = \( 4500 \, \text{N} \)
  • Specific weight of water (\( \gamma \)) = \( 9810 \, \text{N/m}^3 \)
  • Specific weight of oil (\( \gamma_{\text{oil}} \)) = \( 0.75 \times 9810 = 7357.5 \, \text{N/m}^3 \)

Pressure Equation:

Using the pressure balance equation:

\( P_A + \gamma_{\text{oil}} h_{\text{oil from A to X_L}} = \frac{W_B}{A_B} \)

Substitute the values:

\( P_A + 7357.5 \times 4.8 = \frac{4500}{387 \times 10^{-4}} \)

Simplify:

\( P_A + 35316 = 116242.63 \)

\( P_A = 116242.63 – 35316 \)

Final Value:

\( P_A = 80963 \, \text{N/m}^2 \)

Force \( F \):

The force required for equilibrium is:

\( F = P_A \times A_A \)

Substitute the values:

\( F = 80963 \times 38.7 \times 10^{-4} \)

Final Value:

\( F = 313 \, \text{N} \)

Explanation

This problem uses hydrostatic principles to determine the force required for equilibrium:

  1. The pressure at A includes contributions from the oil column height and the pressure needed to balance the weight of B.
  2. The areas of the plunger and cylinder influence the pressure distribution, as seen in the equations.
  3. The resulting force is calculated using the pressure at A and the area of the plunger.

Physical Meaning

  1. Force Equilibrium: The calculated force \( F \) ensures equilibrium in the system by balancing the weight of B and the pressure contributions.
  2. Specific Weight: The specific weight of oil plays a critical role in determining the pressure changes due to the fluid column height.
  3. Hydrostatic Pressure: The pressure distribution in the system is governed by the principles of hydrostatics, accounting for fluid properties and geometry.

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