(a) The bottom surfaces of the piston and plunger are at the same level. N (b) T

Question

(a) The bottom surfaces of the piston and plunger are at the same level.
N

(b) The bottom surface of the output plunger is 1.50 m above that of the input piston.
N

The hydraulic oil in a car lift has a density of 8.30 x 10^-3 m and 0.125 m, respectively. What input force F is needed to support the 26,000-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. N (b) The bottom surface of the output plunger is 1.50 m above that of the input piston. N x 10^2 kg/m^3. The weight of the input piston is negligible. The radii of the input piston and output plunger are 9.20

Explanation / Answer

you baffled me with ?=830 kg/m^3; put it next time like
8.30E2 or 8.3*10^2;
? thus areas of pistons are a=pi*r^2 and A=pi*R^2, where r=7.6E-3 m, R=0.125m;
the pressure in the oil is p=w/A as well as p=F/a, where w=27000 N, F is input force; therefore F=w*a/A for the same level;
F=27000*(7.6E-3/0.125)^2 =99.81 N;
? now suppose they are still at the same level, yet you increase w by ?= ?*A*h*g, where h=1.1m, and A*h is excessive volume of oil;
thus F1= (w + ?*A*h*g)*a/A = F + ?*a*h*g = 99.81 + 1.62= N;

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