A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with oil and capped at both ends with tight-fitting pistons. The wider arm of the U-tube has a radius of 18.0 cm and the narrower arm has a radius of The car rests on the piston on the wider arm of the U-tube. The pistons are initially at the same level. What is the initial force that must be applied to the smaller piston in order to start lifting the car? For purposes of this problem, neglect the weight of the pistons. A) 727 N B) 926 N C) 2900 N D) 3330 N E) 1.20 kN i think the answer is E.but my calculations are offcan someone show how to do this step by step?
Mine would be off too, if you don't give a radius for the smaller arm! All you need do is find the cross- sectional area of the 2 sides of the tube (or the pistons, same thing). Then, divide the larger by the smaller area, to find their ratios. Let's say it's 3:1. All you then do, is divide the 12,000N by 3. You will have the small piston force in N to just balance the car. In reality, you would need to exceed that force very slightly, in order to raise the car.
to my know-how F/A for small piston could desire to be equivalent to F/A for great piston area for piston radius x 2 x pi permit tension for small piston x so x/10 pi 20 000 / 30 pi x 6666.7 N
