Question

2. A hydraulic lift is used to lift a car. The small piston has a radius of 5 cm and the large piston has a radius of 50 cm. If a driver applies a force of 88 N to the small piston, what is the weight of the car the large piston can support? A) 880 N B) 8800 N C) 8.8 N D) 88000 N

Answer 1

Answer:

8800N

Explanation:

a=7.85×10^-3

A=0.785

F=88×0.785/7.85×10^-3

W=8800N

Answer 2

Answer:

Option B (8800N) is the right option.

See the attachment given below,

By Pascal's Law we can state,

Input pressure on one end of the hydraulic is equal to the output pressure at the other end,

P(in)=P(out)

$\frac{F(in)}{A1}=\frac{F(out)}{A2}$

Radius of the pistons are,

Radius of small piston=r₁=5cm

Radius of large piston=r₂=50cm

A₁=πr₁²

A₂=πr₂²

$F(out)=\frac{A2}{A1}*F(in) \\ F(out)=\frac{\pi*50^{2} }{\pi5^{2} }*88\\ F(out)=100*88$

F(out)=8800N

Hence, Large piston can support a car of  8800N weight.