The hydraulic oil in a car lift has a density of 8.30 102 kg/m3. The weight of t
ID: 2017638 • Letter: T
Question
The hydraulic oil in a car lift has a density of 8.30 102 kg/m3. The weight of the input piston is negligible. The radii of the input piston and output plunger are 6.50 10-3 m and 0.125 m, respectively. What input force F is needed to support the 25500 N combined weight of a car and the output plunger under the following conditions?(a) The bottom surfaces of the piston and plunger are at the same level.
1 N
(b) The bottom surface of the output plunger is 1.10 m above that of the input piston.
2 N
Explanation / Answer
weight Fout = 25500 Nradii of the output piston rout = 0.125 m
radii of the input piston rin = 6.5 x 10-3 m we know that , force F = pressure*area therefore , F A from this equation , F1/F2 = A1/A2
a) from above relation , Fin / Fout = Ain / Aout
Fin = [Fout](rin / rout)2
= [25500 N]( 6.5 x 10-3 m / 0.125 m)2
= 68.95 N
b) the bottom surface of the output plunger rout = 1.1 m
the output piston movens a smaller distance than the input piston work done = Fi * di = 68.95*1.1 m = 75.845 J F0do = Fidi Fi = F0d0/di = (68.95 N ) / 1.1m = 62.68 N here F0 is the force which are at same height . weight Fout = 25500 N
radii of the output piston rout = 0.125 m
radii of the input piston rin = 6.5 x 10-3 m we know that , force F = pressure*area therefore , F A from this equation , F1/F2 = A1/A2
a) from above relation , Fin / Fout = Ain / Aout
Fin = [Fout](rin / rout)2
= [25500 N]( 6.5 x 10-3 m / 0.125 m)2
= 68.95 N
b) the bottom surface of the output plunger rout = 1.1 m
the output piston movens a smaller distance than the input piston work done = Fi * di = 68.95*1.1 m = 75.845 J F0do = Fidi Fi = F0d0/di = (68.95 N ) / 1.1m = 62.68 N here F0 is the force which are at same height .
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