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A 500-N weight sits on the small piston of a hydraulic machine. The small piston has an area of 2.0 cm2. If the large piston has an area of 40 cm2, Assume the pistons each have negligible weight

(a) how much weight can the large piston support?
(b) Find the pressure applied to the pistons.

Answer :

Khoso123

Answer:

(a) Weight (Uplifting Force)=[tex]10,000N[/tex]

(b) Pressure applied to piston=[tex]250\frac{N}{cm^{2} }[/tex]

Explanation:

Given Data

Force=500 N

A₁(Small Piston Area)=2.0 [tex]cm^{2}[/tex]

A₂(Large Piston Area)=40 [tex]cm^{2}[/tex]

(a) Weight=?

(b) Pressure Applied to Piston=?

Solution

First we have to Solve for part (b) Pressure

Pressure=[tex]\frac{Force}{Small Piston Area}[/tex]

[tex]P=\frac{500}{2}\\ P=250\frac{N}{cm^{2} }[/tex]

For Part (a) Weight

Weight (Uplifting Force)=P×A₂

Weight=250×40

Weight=10,000 N

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