The small piston of a hydraulic lift has diameter of 8.0 cm, and its large piston has a diameter of 40 cm. The lift raises a load of 15,000 N. What is the force that must be applied to the small piston?
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Sm = small piston
la = large piston
P=F/A
P=15000/(20^2)π
F of sm = PA
= (75/2π)•((8^2)π)
= (75•64)/2
= 4800/2
= 2400N
We already know the pressure but giving it in an approximate decimal form, to two significant figures (since that's what your supplied precision is at):
la = large piston
P=F/A
P=15000/(20^2)π
F of sm = PA
= (75/2π)•((8^2)π)
= (75•64)/2
= 4800/2
= 2400N
We already know the pressure but giving it in an approximate decimal form, to two significant figures (since that's what your supplied precision is at):
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