A hydraulic press can lift 100 kg when a mass m is placed on the smaller piston. It can lift ......

Q: Why do we use the square of the diameter for the area?

The area of a circle is $\pi r^2$ or $\pi (D/2)^2$. When you take the ratio of two areas, the constants $\pi$ and $1/4$ cancel out, leaving the ratio of the squares of their diameters. So, if diameter increases by a factor of 4, area increases by a factor of $4^2 = 16$.

Q: Why doesn't the value of $g$ matter in this problem?

Because $g$ appears on both sides of the equation ($F = mg$). When we equate the pressures, $g$ cancels out. We are essentially comparing masses directly.

Q: What happens to the mass $m$ on the smaller piston?

The problem states that the same mass $m$ is kept on the smaller piston in the final state. This is a crucial constant that allows us to link the initial and final states by dividing their respective equations.

Q: Is energy conserved in a hydraulic press?

Absolutely! While a hydraulic press multiplies force (Mechanical Advantage), it does not multiply work. The smaller piston must move a much greater distance downwards to lift the heavier load on the larger piston upwards by a small amount. Work in equals work out.

Q: Why did we divide the two equations instead of substituting?

Dividing equations is a powerful algebraic trick when you have multiplicative constants that you want to eliminate. Here, dividing immediately cancelled out $m$, $g$, $A_1$, and $A_2$, leaving only the numbers and our target variable $M$.

Question

A hydraulic press can lift 100 kg when a mass $m$ is placed on the smaller piston. It can lift ...... kg when the diameter of the larger piston is increased by 4 times and that of the smaller piston is decreased by 4 times keeping the same mass $m$ on the smaller piston.

Sigma Percentile · Accepted Answer

### The Magic of Pascal's Law

Imagine you are standing in a garage, looking at a massive car being lifted effortlessly by a mechanic pushing a small lever. How is this possible? Is it magic? No, it is **Pascal's Law** in action!

Pascal's Law states that when pressure is applied to an enclosed fluid, it is transmitted undiminished to every part of the fluid and the walls of its container. This simple yet profound principle is the heart of a hydraulic press.

In our problem, we have a hydraulic press with two pistons: a smaller one where we apply our effort, and a larger one that lifts the heavy load. Because the pressure must be equal on both sides, we can write our master equation:

$$ P_1 = P_2 $$

$$ \frac{F_{	ext{large}}}{A_{	ext{large}}} = \frac{F_{	ext{small}}}{A_{	ext{small}}} $$

### Analyzing the Initial Setup

Let's break down the first scenario. We are told that the press can lift a $100 	ext{ kg}$ load when a mass $m$ is placed on the smaller piston.

Let the area of the larger piston be $A_1$ and the area of the smaller piston be $A_2$. The force exerted by the load is its weight, $100g$, and the force exerted by the effort mass is $mg$. Substituting these into our master equation gives us:

$$ \frac{100g}{A_1} = \frac{mg}{A_2} \quad \dots 	ext{(Equation 1)} $$

This equation perfectly captures the initial state of our hydraulic press.

### The Power of Scaling Diameters

Now, the problem introduces a twist. We are going to modify the press by changing the diameters of the pistons.

The diameter of the larger piston is increased by $4$ times. But wait, how does this affect the area? Remember that the area of a circular piston is given by $A = \frac{\pi D^2}{4}$. This means the area is directly proportional to the **square** of the diameter ($A \propto D^2$).

If the diameter increases by a factor of $4$, the new area $A_1'$ becomes $4^2 = 16$ times the original area:

$$ A_1' = 16A_1 $$

Simultaneously, the diameter of the smaller piston is decreased by $4$ times. Following the same logic, its new area $A_2'$ becomes $\left(\frac{1}{4}ight)^2 = \frac{1}{16}$ of the original area:

$$ A_2' = \frac{A_2}{16} $$

### Setting Up the Final State

With our new, highly modified hydraulic press, we want to find out the new mass $M$ it can lift, assuming we keep the exact same mass $m$ on the smaller piston.

Let's plug our new areas and masses back into Pascal's Law:

$$ \frac{Mg}{A_1'} = \frac{mg}{A_2'} $$

Substituting the expressions for the new areas:

$$ \frac{Mg}{16A_1} = \frac{mg}{\left(\frac{A_2}{16}ight)} $$

The fraction in the denominator on the right side looks a bit messy. Let's simplify it by bringing the $16$ up to the numerator:

$$ \frac{Mg}{16A_1} = \frac{16mg}{A_2} \quad \dots 	ext{(Equation 2)} $$

### The Final Calculation

We now have two beautiful equations describing the initial and final states. We need to find $M$, and the smartest algebraic move here is to divide Equation 1 by Equation 2. This will elegantly eliminate all the unknown variables like $m$, $g$, $A_1$, and $A_2$.

$$ \frac{\left(\frac{100g}{A_1}ight)}{\left(\frac{Mg}{16A_1}ight)} = \frac{\left(\frac{mg}{A_2}ight)}{\left(\frac{16mg}{A_2}ight)} $$

Watch how everything cancels out! The $g$, $A_1$, $m$, and $A_2$ all vanish, leaving us with a clean, simple ratio:

$$ \frac{100 	imes 16}{M} = \frac{1}{16} $$

Now, it's just basic arithmetic. Cross-multiplying gives us:

$$ M = 100 	imes 16 	imes 16 $$

$$ M = 100 	imes 256 $$

$$ M = 25600 	ext{ kg} $$

### The Way Forward

Take a moment to appreciate what just happened. By simply tweaking the diameters of the pistons, we increased the lifting capacity of the press from a mere $100 	ext{ kg}$ to a massive $25600 	ext{ kg}$! This is the incredible power of **Mechanical Advantage**.

However, nature always demands a trade-off. While we multiplied our force drastically, energy remains conserved. To lift that $25600 	ext{ kg}$ load even a tiny fraction of a millimeter, the smaller piston will have to be pushed down a significantly larger distance. Work input always equals work output!

Enter Numerical Value:

Visualized Solution

$P a sc a l^{'} s L a w in H y d r a u l i c P r ess$

$P r ess u r e E q u a t i o n$

$I ni t ia l S t a t e$

$A r e a an d D iam e t er R e l a t i o n$

$N e w A r e a s$

$F ina l S t a t e S e t u p$

$S im pl i f y in g F ina l S t a t e$

$D i v i d in g E q u a t i o n s$

$F ina l C a l c u l a t i o n$

$T h e W a y F or w a r d$

The Sigma Insight: Fluid Pressure and Pascal's Law

The Magic of Pascal's Law

Analyzing the Initial Setup

The Power of Scaling Diameters

Setting Up the Final State

The Final Calculation

The Way Forward

Similar Questions

A U-shaped tube contains a liquid of density $ρ$ and it is rotated about the line as shown in the figure. Find the difference in the levels of liquid column.

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Pascal′s Law in Hydraulic Press

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Pressure Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial State

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Area and Diameter Relation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } New Areas

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final State Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying Final State

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dividing Equations

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward