Hydraulic lift has two pistons; small one of cross-sectional area of 4cm^2 and large one cross-sectional area 250cm^2. Designed to raise 3500kg car, what minimum force must be applied to the small piston?
Also, if the force is applied through compressed air, what must be the minimum air pressure applied to the small piston?
I know the answers are: 549N and 1.37x10^6 Pa. Just need to know how to get there.
- 1 decade agoFavorite Answer
How do you know what the answers are if you don't know how to get there? If you wanted some one to confirm your work that I could understand, but your answers are correct, so odds are you did it right
the force on the piston is easy 3500kgf = 34,300N, 34,300N * (4 / 250) = 548.8N or 549N
Another answerer gave a correct equation to find the pressure in kgfcm^2 ; 4/250 x 3500/4 kg
since the 4cm^2 area of the small piston cancels, I'd use 34,300N / 250cm^2 = 137.2Ncm^2 to get Pa multiply by 1x10^4, therefore 1.37 X 10^6Pa
- MadhukarLv 71 decade ago
Force = 3500 x 9.8 x (4/250) N = 5488 N
Pressure = 5488 / 4 x 10^(-4) Pa = 1.22 x 10^(-2) Pa.
- Anonymous1 decade ago
4/250 x 3500kg
4/250 x 3500/4 kg
- MurtazaLv 61 decade ago
i think 70 N