Question

The small piston of a hydraulic lift has diameter of 8.0 cm, and its large piston has a diameter of 40 cm. The lift raises a load of 15,000 N. What is the force that must be applied to the small piston?

Answer 1

Answer:

$\bigstar{\bold{Force\:on\:small\:piston\:=\:600\:N}}$

Explanation:

$\Large{\underline{\underline{\it{Given:}}}}$

• Diameter of small piston (d) = 8cm=0.08m
• Radius of small piston(r) = 0.04m
• Diameter of big piston (D) = 40cm= 0.4m
• Radius of big piston(R) = 0.2m
• Force on the big piston (F) = 15000 N

$\Large{\underline{\underline{\it{To\:Find:}}}}$

• Force on the small piston (f)

$\Large{\underline{\underline{\it{Solution:}}}}$

→ The area of small piston (a) = πr²=3.14×0.04×0.04 = 5.024×10⁻³m²

→ The area of big piston (A) = πR²= 3.14×0.2×0.2=0.1256m²

→ Force on the small piston is given by the formula

  f = F×(a/A)

→ Substituting the given datas we get,

  f = 15000×(5.024×10⁻³/0.1256)

  f = 600 N

  $\boxed{\bold{Force\:on\:small\:piston\:=\:600\:N}}$

$\Large{\underline{\underline{\it{Notes:}}}}$

• The working principle of hydraulic lift is based on Pascal's law
• Pascal's law state that the pressure exerted at any point on an enclosed liquid is transmitted equally in all directions.
• Hydraulic lift is used to lift heavy bodies by applying small force.

Answer 2

The force that must be applied to the small piston is 600 N

• A Hydraulic lift generally works on the Principle of Pascal's Law.
• According to the Pascal's Law, "pressure exerted at any point on an enclosed liquid is transmitted equally in all directions."

Given,

Diameter of the small piston (d₁) = 8 cm = 0.08 m

Diameter of the large piston (d₂) = 40 cm = 0.4 m

Load on the large piston (F₂) = 15,000 N

Suppose, force applied on the small piston = F₁

we know, from the Pascal's Law,

F₁ / A₁ = F₂ / A₂   [F₁ = load on small piston, F₂ = load on large piston

                              A₁ = Area of small piston, A₂ =  Area of large piston]

⇒ F₁ / d₁² = F₂ / d₂² [∵ Area = πd²/4, d= diameter]

⇒ F₁ = 15,000 × $\frac{(0.08)^2}{(0.4)^2}$

⇒ F₁ = 15,000 × 0.04

⇒ F₁ = 600

So, The force that must be applied to the small piston is 600 N.