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.

Hint:

The force multiplication in a hydraulic press is proportional to the square of the ratio of the diameters: $F_2 = F_1 (\frac{D_2}{D_1})^2$.

Answer 1

Correct Answer: 25600

To solve the problem of determining the increased lifting capacity of a hydraulic press when the piston sizes are altered, we need to understand the basic principle behind hydraulic presses, which is Pascal's law. This law states that pressure applied to a confined fluid is transmitted undiminished in all directions throughout the fluid. In terms of a hydraulic press, this implies:

$P=F/A$, where $P$ is pressure, $F$ is force, and $A$ is the area of the piston.

Initially, we have $F\_\{small\}=100\backslash text\{\; kg\}\backslash cdot\; g$ on the larger piston when a mass $m$ is placed on the smaller piston.

The relationship according to Pascal's law can be given by:

$\backslash frac\{F\_\{larger\}\}\{A\_\{larger\}\}=\backslash frac\{F\_\{smaller\}\}\{A\_\{smaller\}\}$

Where:

$A=\backslash frac\{\backslash pi\; d^2\}\{4\}$, using diameter to compute area of a circular piston.

For the initial configuration:

Let $d\_s$ and $d\_l$ be the diameters of the smaller and larger pistons respectively, with $F\_\{small\}=m\backslash cdot\; g$.

The area ratio is:

$\backslash frac\{A\_\{larger\}\}\{A\_\{smaller\}\}=\backslash frac\{d\_l^2\}\{d\_s^2\}$

Increasing the diameter of the larger piston by 4 times and decreasing the smaller piston by 4 times results in new diameters:

$d\_l\text{'}=4d\_l$ and $d\_s\text{'}=\backslash frac\{d\_s\}\{4\}$

The new area ratio becomes:

$\backslash frac\{(4d\_l)^2\}\{(\backslash frac\{d\_s\}\{4\})^2\}=\backslash frac\{16d\_l^2\}\{\backslash frac\{d\_s^2\}\{16\}\}=256\backslash frac\{d\_l^2\}\{d\_s^2\}$

Applying this to our force equation:

$F\_\{new\; larger\}=\backslash frac\{256d\_l^2\}\{d\_s^2\}\backslash cdot\; F\_\{small\}$$=256\backslash cdot\; 100\backslash cdot\; g$

Thus, the mass that can now be elevated on the larger piston is:

$\backslash frac\{F\_\{new\; larger\}\}\{g\}=256\backslash cdot\; 100=25600\backslash text\{\; kg\}$

This computed value, 25600 kg, fits precisely within the given range of [25600, 25600], confirming the solution is correct.