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)

Expert's answer

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=1.2cm2⋅1m2104cm2⋅2700N150N=22⋅10−4m2.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=150N1.2cm2⋅1m2104cm2=1.25⋅106Pa.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=22⋅10−4m2.A_2 = 22 \cdot 10^{-4} m^2.

B) P=1.25⋅106Pa.P = 1.25 \cdot 10^6 Pa.


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