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A hydraulic press for compacting powdered samples has a large cylinder which is 10.0 cm in diameter, and a small cylinder with a diameter of 2.0 cm. A lever is attached to the small cylinder as shown in The sample, which is placed on the large cylinder, has an area of 4.0 cm2.

What is the pressure on the sample if F = 270 N is applied to the lever?

Verified Answer :


F small cylinder = 2.F
2.F/s = p = (F sample/S)
p sample = F sample/S sample
▬▬▬
> p sample = 2.F.S/s/S sample = 2.F.(D/d)²/S sample
> p sample = 2280(10/2)^2/4E-4 = 3,50E+07 u.S.I.
> p sample = 35,0 MPa = 345 atm

The hydraulic pressure on sample is \(3.35*10^{7} N/m^{2}\).

What is hydraulic pressure?

Hydraulic pressure is the force imparted per unit area of a liquid on the surfaces which it has contact.

The magnitude of force acting on the small cylinder is equal to \(F_{N}\).

In static equilibrium, all the torques should add up to zero:\(F_{N}l - 270(2l)= 0\\F_{N} = 270(2)\\F_{N} = 540 N\)

The pressure applied to the fluid (through the small cylinder):

\(P= \frac{540}{\pi (\frac{2*10^{-2} }{2} )^2}\)

\(P = 1.71*10^{6} Pa\)

The force of fluid on the large cylinder:

\(F= P\pi R^{3} \\F= 1.71*10^{6} \pi (0.05^{2} )\\F= 1.34*10^{4} N\)

The pressure on the sample:

\(P_{S} =\frac{F}{A} \\P_{S}= \frac{1.34*10^{4} }{4*10^{-4} } \\P_{S}= 3.35*10^{7} N/m^{2}\)

To learn more about Hydraulic pressure here

https://brainly.com/question/857286

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