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3 years ago

5

What force (in N) must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2400 kg car (a large car

) resting on the slave cylinder? The master cylinder has a 1.50 cm diameter, while the slave has a 24.0 cm diameter.

1 answer:

makvit [3.9K]3 years ago

8 0

Answer:

Fm= 91.88 N

Explanation:

Pascal principle

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

The pressure is definited like this:

P=F/A

Where:

P: Pressure in pascals (Pa)

F: Force acting in the area (N)

A : Area where the force acts (m²)

Pascal principle

Pm=Ps

Fm/ Am= Fs/ As Formula (1)

Where :

Pm : Pressure on the master piston

Ps : Pressure on the slave piston

Fm : Force on the master piston (N)

Fs: Force on the slave piston ((N)

Am: master piston area (m²)

As: slave piston area (m²)

Area Formula (A)

A= π*R²

R : piston radius

Calculation of the weight of the car (W)

W= m*g= 2400 kg*9.8m/s²= 23520 N

W = Fs

Data

Fs = 23520 N

Dm = 1.5 cm

Ds = 24 cm

Rm = 0.75 cm

Rs = 12 cm

Am = π*Rm² = π*(0.75)²

As = π*Rs² = π*(12)²

Force exerted on the master cylinder

We replace data in the formula (1)

$\frac{F_{m} }{A_{m} } = \frac{F_{s} }{A_{s} }$

$F_{m} = \frac{F_{s}*A_{m} }{A_{s}}$

$F_{m} = \frac{(23520 N)*(\pi *(0.75)^{2})(cm^{2})}{(\pi *(12)^{2})(cm^{2})}$

$F_{m} = (23520 N)*\frac{(0.75)^{2} }{(12)^{2} }$

Fm= 91.88 N

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