Question

Suppose that you bring your car to an auto mechanic. The auto mechanic, to get a better look at what’s broke, puts your car on a hydraulic lift, shown below in which there is an incompressible fluid of density € ρ = 900 kg m3 . A force € F1 is applied to a piston of area € A1 . This supports the weight of the car € F2 on the second piston of area € A2. Assume that both of the pistons have negligible masses. a. If the car has a mass of € m = 500kg what magnitude of force € F1 is needed to support the car so that the bottoms of each piston are at the same height? Assume that the radii of piston 1 and piston 2 are € r 1 = 1 8 m and € r2 = 1 2 m respectively and that each piston is negligibly thin.

Suppose that you wanted to lift the car up so that you could see under it. If you wanted to lift the car by a distance € d2 = 2mabove the car’s initial height, how far down € d1 below piston 1’s starting height, would you have to push piston 1? Assume that piston 1 can move as far down as you would like.

How much extra force would be required to push piston 1 down by distance € d1?

Answer 1

F pascal's law we con Α. Using crite | A pressure at Point P,= P₂ The refore, F Ulf Fi- Ai Az x F₂ Tri xa ri Ez are Now , F = Wear = mg SO, fi 18² x 500 x 9.8 12° | 1025 N Now if the car moves 2m above volume distance . water lifted = 2 Az volume of Then move - Suppose water d'. A piston shifting down liquid is in compressit since the २.2 ď A 2 at 2A₂ AI 2x 12 2x Tre To" 0889 180 دی ۸ suppose Fi extra force is applied on Piston P so, work done W - (Eit Å) d' Piston Now, 1 moves down Piston 2 ď dist ona (1) and is by anco So, liquid by change in potential energy ΔΟι + PA₂ 2 2 change in pe of car, Auc= M, g x 2 2 pa, dhe Now m. gx2 W = au e touc (, + F' Dxd's -Phone + P, A, 2 + (F. +fi'). -PAS -Pad + P Az x2 + meg? d' 900 x 1 18"x0.8889 + qoox Tx/²x2 0.8889 2 So8x9.8 x 2 0-8899 519963 N So, F F- 519963 - 11025 508938 N