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Question
The average mass that must be lifted by a hydraulic press is 80 kg. If the radius of the larger piston is five times that of the smaller piston , what is the minimum force that must be applied?
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Solution
Mass to be lifted 80 kg . therefore force needed to lift it up= 80*9.8 = 784
Radius of the larger piston is 5 times the radius of the smaller piston therefore if area of the smaller piston is pie*x*x therefore the area of the larger piston= pie*5x*5x = pie*25x^2 therfore area of larger piston is 25 times greater than the area of the smaller piston since pressure applied will be the same due to pascals law but area of larger piston is larger therefore water will exert more force on the larger piston because p =f/a since area of larger piston is 25 times larger therefore minimum force we need to apply on the smaller piston is 1/25 times the force required therefore minimum force required= 1/25*784 or 31.36 newtons.
Radius of the larger piston is 5 times the radius of the smaller piston therefore if area of the smaller piston is pie*x*x therefore the area of the larger piston= pie*5x*5x = pie*25x^2 therfore area of larger piston is 25 times greater than the area of the smaller piston since pressure applied will be the same due to pascals law but area of larger piston is larger therefore water will exert more force on the larger piston because p =f/a since area of larger piston is 25 times larger therefore minimum force we need to apply on the smaller piston is 1/25 times the force required therefore minimum force required= 1/25*784 or 31.36 newtons.
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