The hydraulic oil in a car lift has a density of 8.30 x 10^2 kg/m3. The weight o
ID: 2059905 • Letter: T
Question
The hydraulic oil in a car lift has a density of 8.30 x 10^2 kg/m3. The weight of the input piston is negligible. The radii of the input piston and output plunger are 6.40x10^-3 m and 0.125 m, respectively. What input force F is needed to support the 24000 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 1.40 m above that of the input piston.
Explanation / Answer
In both the parts, the car and the output plunger are being held up by the pressure of the oil underneath. The oil pressure must be sufficient to produce an upward force of 24000N to counteract the weight.
As, Force = pressure × area
24000N = pressure × [(0.125m)2]
Thus, pressure at plunger = 24000N / ((0.125m)²) = 4.89 x 105 Pa
(a) The bottom surfaces of the piston and plunger are at the same level.
In this case, the oil pressure directly under the plunger is the same as the oil pressure at the piston:
pressure at piston = pressure at plunger = 24000N / ((0.125m)²)
Furthermore, pressure at piston = F / (area of piston) = F / ((6.40 × 10-3m)2)
solving for F, we get F = 62.91 N
(b) The bottom surface of the output plunger is 1.40 m above that of the input piston.
In this case, the oil pressure at the piston is not the same as the oil pressure at the plunger because fluid pressure increases with depth.
So in this case,
pressure at piston = (pressure at plunger) + gh
where, = oil's density,
h = difference in height (in this case, 1.40 m).
pressure at piston = (24000N / ((0.125m)²)) + gh = 5 x 105 Pa
Using, pressure at piston = F / (area of piston) = F / ((6.4×10^-3m)²)
F = 64.35 N.
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