Question

The area of pistons in hydraulic machine are 5cm and 12 cm². What force on
smaller piston will support a load of 1000 N on the larger piston.​

Answer 1

Answer :

The required force on the smaller piston is 416.7 N (approx)

Explanation :

Given :

The area of pistons in hydraulic machine are 5 cm² and 12 cm²

To find :

the force on  the smaller piston which will support a load of 1000 N on the larger piston.​

Solution :

Let F₁ be the force on smaller piston

Also, let

• A₁ = 5 cm² (area of smaller piston)
• A₂ = 12 cm² (area of larger piston)
• F₂ = 1000 N (force on the larger piston)

By the principle of Hydraulic machine, ( works on the principle of Pascal's law - which states that the pressure applied to any part of the enclosed liquid will be transmitted equally in all directions through the liquid. )

Pressure on smaller piston = pressure on larger piston

   P₁ = P₂

  $\sf \dfrac{F_1}{A_1}=\dfrac{F_2}{A_2} \\\\ \sf F_1=\dfrac{F_2}{A_2} \times A_1 \\\\ \sf F_1=\dfrac{1000 N}{12 cm^2} \times 5 cm^2 \\\\ \sf F_1=\dfrac{5000}{12} N \\\\ \sf F_1 \simeq 416.7N$

Therefore, The force on the smaller piston is approximately 416.7 N