A hydraulic press has a ram of 30 cm diameter and a plunger of 5 cm diameter. Find the weight lifted by the hydraulic press when the force applied at the plunger is 400 N.

Hydraulic Press Weight Calculation

Problem Statement

A hydraulic press has a ram of 30 cm diameter and a plunger of 5 cm diameter. Find the weight lifted by the hydraulic press when the force applied at the plunger is 400 N.

Given Data

  • Diameter of ram, \(D = 30 \, \text{cm} = 0.3 \, \text{m}\)
  • Diameter of plunger, \(d = 5 \, \text{cm} = 0.05 \, \text{m}\)
  • Force applied at plunger, \(F = 400 \, \text{N}\)

Solution

1. Calculate Piston Areas

First, we calculate the area of the plunger (\(a\)) and the ram (\(A\)).

$$ a = \frac{\pi}{4} d^2 $$ $$ a = \frac{\pi}{4} (0.05 \, \text{m})^2 $$ $$ a \approx 0.0019635 \, \text{m}^2 $$
$$ A = \frac{\pi}{4} D^2 $$ $$ A = \frac{\pi}{4} (0.3 \, \text{m})^2 $$ $$ A \approx 0.0706858 \, \text{m}^2 $$

2. Apply Pascal’s Law

According to Pascal’s Law, the pressure exerted on the plunger is transmitted equally to the ram. Let \(W\) be the weight lifted by the ram.

$$ P_{\text{plunger}} = P_{\text{ram}} $$ $$ \frac{F}{a} = \frac{W}{A} $$

3. Solve for the Weight Lifted (W)

Rearrange the formula to solve for \(W\) and substitute the known values.

$$ W = F \times \frac{A}{a} $$ $$ W = 400 \, \text{N} \times \frac{0.0706858 \, \text{m}^2}{0.0019635 \, \text{m}^2} $$ $$ W = 400 \times 36 $$ $$ W = 14400 \, \text{N} $$

The weight can also be expressed in kilonewtons (kN).

$$ W = \frac{14400}{1000} \, \text{kN} = 14.4 \, \text{kN} $$
Final Result:

The weight lifted by the hydraulic press is \( 14400 \, \text{N} \) or \( 14.4 \, \text{kN} \).

Explanation of Pascal’s Law

Pascal’s Law is a fundamental principle in fluid mechanics that states a pressure change at any point in a confined incompressible fluid is transmitted throughout the fluid such that the same change occurs everywhere. In a hydraulic system, this means the pressure created by the small force on the plunger (\(P = F/a\)) is the same pressure that acts on the large ram.

Because this pressure (\(P\)) acts on a much larger area (\(A\)) at the ram, the resulting upward force (\(W = P \times A\)) is much larger than the initial force (\(F\)). This phenomenon is known as force multiplication.

Physical Meaning

The result demonstrates the powerful advantage of a hydraulic press. A relatively small input force of 400 N (roughly the weight of a 40 kg object) applied to the 5 cm plunger is magnified to lift a massive weight of 14,400 N (roughly the weight of a 1468 kg object, like a car).

The force multiplication factor is equal to the ratio of the areas (\(A/a\)), which in this case is 36. This means the output force is 36 times greater than the input force, showcasing how hydraulic systems provide a significant mechanical advantage.

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top