Question #184954

A hydraulic press has a small piston of diameter 2 cm. If a force of 60 N is applied to the smaller piston, a force of 300 N is exerted on the larger piston. What is the diameter of the larger piston?

Expert's answer

Pascal's Principle: "F1\/A1=F2\/A2"

F_{1}=60N r_{1}=(2÷2)1/100

r_{1}=0.01m

F_{2}=300N r_{2}=?

A=(πr^{2})

A_{1}=π(0.01^{2})m^{2}

A_{2}=π(r^{2})m^{2}

(60N/π(0.01)^{2})=300N/π(r^{2})

r^{2}/0.01^{2}=5

√ r^{2}=√0.0005

r=0.0224m

r= (0.224x100) cm

r=2.24cm

Diameter= (2.24cm x2)

D=4.48 cm

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