Question

The hydraulic oil in a car lift has a density of 8.04 x 10² kg/m³. The weight of the input piston is negligible. The radii of the input piston and output plunger are 7.63 x 10^-3 m and 0.158 m, respectively. What input force F is needed to support the 26400-N combined weight of a car and the output plunger, when (a) the bottom surfaces of the piston and plunger are at the same level, and (b) the bottom surface of the output plunger is 1.60 m above that of the input plunger?

Answer 1

In both cases, the car & 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 26400 N to counteract the weight. Use the formula:

Force = pressure * area

26400 N = pressure * (pi * (0.158m)²)

Solve for pressure:

pressure at plunger = 26400N / (pi(0.158m)²)

P1 = 336790.71 N/m^2

(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 = 26400N / (pi*(0.158m)²).......(2)

pressure at piston = F / (area of piston) = F / (pi* (7.63*10^-3)²) ......(3)

Combine (2) and (3), and solve for F

F = 336790.71 * pi*( 7.63*10^-3)^2

F = 61.57 N

(b) The bottom surface of the output plunger is 1.10 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.

This is because fluid pressure increases with depth. So in this case:

pressure at piston = (pressure at plunger) + pgh .....(4)

where "p" is the oil's density, and "h" is the difference in height (in this case, 1.0 m). So combine (1) and (4):

pressure at piston = (26400N / (pi(0.158m)²)) + pgh ....(5)

Combine (3) and (5), and solve for "F"

F / (pi* (7.63*10^-3)²) = 316790.71 + 8.04*10^2 * 9.8*1.6

F = 60.24 N