Respuesta :
F small cylinder = 2.F
2.F/s = p = (F sample/S)
p sample = F sample/S sample
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> p sample = 2.F.S/s/S sample = 2.F.(D/d)²/S sample
> p sample = 2*280*(10/2)^2/4E-4 = 3,50E+07 u.S.I.
> p sample = 35,0 MPa = 345 atm
The hydraulic pressure on sample is [tex]3.35*10^{7} N/m^{2}[/tex].
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 [tex]F_{N}[/tex].
In static equilibrium, all the torques should add up to zero:[tex]F_{N}l - 270(2l)= 0\\F_{N} = 270(2)\\F_{N} = 540 N[/tex]
The pressure applied to the fluid (through the small cylinder):
[tex]P= \frac{540}{\pi (\frac{2*10^{-2} }{2} )^2}[/tex]
[tex]P = 1.71*10^{6} Pa[/tex]
The force of fluid on the large cylinder:
[tex]F= P\pi R^{3} \\F= 1.71*10^{6} \pi (0.05^{2} )\\F= 1.34*10^{4} N[/tex]
The pressure on the sample:
[tex]P_{S} =\frac{F}{A} \\P_{S}= \frac{1.34*10^{4} }{4*10^{-4} } \\P_{S}= 3.35*10^{7} N/m^{2}[/tex]
To learn more about Hydraulic pressure here
https://brainly.com/question/857286
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