Question

A hydraulic press consists of two cylinders of radii 6 cm and 1 cm respectively. The larger

cylinder has to balance a load of 5000 kg. how much force must be applied to the smaller

cylinder ? what is the pressure developed in the oil?​

Answer 1

Answer:

YOUR ANSWER:

Explanation:

1360 N, 4.33 x 106 N/m2

Answer 2

Given:

$Radius, R_{2} = 1cm = 6\times 10^{-2} ,\\Radius, R_{1} = 6cm = 1\times 10^{-2}\\Mass, m = 5000kg\\Acceleration due to gravity, g = 9.8m/s^{2}$

Find: We have to calculate the force applied in the smaller cylinder and the pressure to developed in the oil

Solution:

We know the Pascal's Principle,

So, the expression of the Pascal's Principle,

$\dfrac{F_{1} }{A_{1} } = \dfrac{F_{2} }{A_{2} }$

So, now we easily calculate the Force on the smaller cylinder,

$\dfrac{5000 \times 9.8}{\pi \times (6\times 10^{-2})^{2}} = \dfrac{F_{2} }{\pi \times (1\times 10^{-2})^{2}}$

$\dfrac{49000}{113.04\times 10^{-4}} = \dfrac{F_{2} }{3.14\times 10^{-4}}\\\\43.3474\times 10^{4} = F_{2}\times 0.3184\times 10^{4} \\\\F_{2} = \dfrac{43.3474\times 10^{4}}{0.3184\times 10^{4}} \\\\F_{2} = 136.1413$

So, the Force be applied on the smaller cylinder is 136.1413N

Now, we have to calculate the pressure,

We have the expression of the pressure,

$Pressure = \dfrac{Force}{Area}$

$Pressure = \dfrac{136.1413}{0.3184\times 10^{4}}$

$Pressure = 427.5794\times10^{-4}$

So, the pressure developed in the oil is 0.0427Pa