Question

A hydraulic lift raises a 2000 kg automobile when a 500 N force is applied to the smaller piston. If the smaller piston has an area of 10 cm2 , what is the cross-sectional area of the larger piston

Answer 1

Answer:

The cross-sectional area of the larger piston is 392cm ^{2}[/tex]

Explanation:

To solve this problem we apply the following formula:

Pascal principle: F=P*A   Formula (1)

F=Force applied to the piston

P: Pressure

A= Piston area

Nomenclature:

Fp= Force on the primary piston= 500N

W= car weight =m*g=2000kg*9.8m/s2= 19600N

Fs= Force on the secondary piston= W = 19600N

$Ap= Primary piston area=10cm^{2} =10*10^{-4}m^{2}$

As= Secondary piston area=?

The pressure acting on one side is transmitted to all the molecules of the liquid because the liquid is incompressible.

In the equation (1)

P=F/A

Pp=Ps

$\frac{Fp}{Ap} = \frac{Fs}{As}$

$As= \frac{Fs*Ap}{Fp}$

$As=\frac{19600*10*10^{-4} }{500}$

$As=0.0392m^{2} =0.0392*10^{4}cm^{2}$

$As=392cm ^{2}$

Answer 2

The cross-sectional area of the larger piston is 392 cm².

To calculate the cross-sectional area of the larger piston, we use the formula below.

Formula:

• F/A = f/a................ Equation 1

Where:

• F = Weight of the automobile
• A = cross-sectional area of the automobile
• f = Force applied to the smaller piston
• a = Area of the smaller piston

Make A the subject of the equation

• A = Fa/f............... Equation 2

From the question,

Given:

• f = 500 N
• a = 10 cm²
• F = mg = (2000×9.8) = 19600 N

Substitute these values into equation 2

• A = (19600×10)/500
• A = 196000/500
• A =  392 cm²

Hence, The cross-sectional area of the larger piston is 392 cm².

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