Question #87906
Pistons are free to move in two cylinders which are connected by a hydraulic fluid. The area of the smaller piston is 1.2 cm2. A force of 150 N on this piston will move a weight of 2700 N on the larger piston. (Assume that friction is negligible.)
A) What is the area of the large piston?
B) What is the pressure in the fluid? (Give answer in Pascals)
1
Expert's answer
2019-04-12T09:43:20-0400

A) We can find the area of the large piston from the hydraulic press formula:


F1A1=F2A2,\dfrac{F_1}{A_1} = \dfrac{F_2}{A_2},

here, F1F_1 is the force that acts on the smaller piston, F2F_2 is the force that acts on the larger piston, A1A_1 is the area of the smaller piston, A2A_2 is the area of the larger piston.

Then, from this formula we can calculate the area of the large piston:


A2=A1F2F1=1.2cm21m2104cm22700N150N=22104m2.A_2 = A_1 \dfrac{F_2}{F_1} = 1.2 cm^2 \cdot \dfrac{1m^2}{10^4 cm^2} \cdot \dfrac{2700 N}{150 N} = 22 \cdot 10^{-4} m^2.

B) We can find the pressure in the fluid from the formula:


P=F1A1=150N1.2cm21m2104cm2=1.25106Pa.P = \dfrac{F_1}{A_1} = \dfrac{150 N}{1.2 cm^2 \cdot \dfrac{1m^2}{10^4 cm^2}} = 1.25 \cdot 10^6 Pa.

Answer:

A) A2=22104m2.A_2 = 22 \cdot 10^{-4} m^2.

B) P=1.25106Pa.P = 1.25 \cdot 10^6 Pa.


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