Operation of a Hydraulic Jack Consider a hydraulic jack being used in a car repair shop, as in Fig. 3–12. The pistons have an area of A1 = 0.8 cm2 and A2 = 0.04 m2. Hydraulic oil with a specific gravity of 0.870 is pumped in as the small piston on the left side is pushed up and down, slowly

Question

Operation of a Hydraulic Jack

Consider a hydraulic jack being used in a car repair shop, as in Fig. 3–12. The pistons have an area of [latex]A_{1}=0.8 cm ^{2} ext { and } A_{2}=0.04 m ^{2}[/latex]. Hydraulic oil with a specific gravity of 0.870 is pumped in as the small piston on the left side is pushed up and down, slowly raising the larger piston on the right side. A car that weighs 13,000 N is to be jacked up.

(a) At the beginning, when both pistons are at the same elevation (h = 0), calculate the force [latex]F_{1}[/latex] in newtons required to hold the weight of the car.

(b) Repeat the calculation after the car has been lifted two meters (h = 2 m). Compare and discuss.

Accepted Answer

We are to determine the force required to lift a car with a hydraulic jack at two different elevations.

Assumptions 1 The oil is incompressible. 2 The system is at rest during the analysis (hydrostatics). 3 Air density is negligible compared to oil density.

Analysis (a) When h = 0, the pressure at the bottom of each piston must be the same. Thus,

[latex]P_{1}=\frac{F_{1}}{A_{1}}=P_{2}=\frac{F_{2}}{A_{2}} ightarrow F_{1}=F_{2} \frac{A_{1}}{A_{2}}=(13,000 N ) \frac{0.8 cm ^{2}}{0.0400 m ^{2}}\left(\frac{1 m }{100 cm } ight)^{2}=26.0 N[/latex]

At the beginning, when h = 0, the required force is thus [latex]F_{1}[/latex] = 26.0 N.

(b) When h ≠ 0, the hydrostatic pressure due to the elevation difference must be taken into account, namely,

[latex]\begin{aligned}P_{1}=& \frac{F_{1}}{A_{1}}=P_{2}+ ho g h=\frac{F_{2}}{A_{2}}+ ho g h \F_{1}=& F_{2} \frac{A_{1}}{A_{2}}+ ho g h A_{1} \=&(13,000 N ) \frac{0.00008 m ^{2}}{0.04 m ^{2}} \&+\left(870 kg / m ^{3} ight)\left(9.807 m / s ^{2} ight)(2.00 m )\left(0.00008 m ^{2} ight)\left(\frac{1 N }{1 kg \cdot m / s ^{2}} ight)= 2 7 . 4 N\end{aligned}[/latex]

Thus, after the car has been raised 2 meters, the required force is 27.4 N.

Comparing the two results, it takes more force to keep the car elevated than it does to hold it at h = 0. This makes sense physically because the elevation difference generates a higher pressure (and thus a higher required force) at the lower piston due to hydrostatics.

Discussion When h = 0, the specific gravity (or density) of the hydraulic
fluid does not enter the calculation—the problem simplifies to setting the two
pressures equal. However, when h ≠ 0, there is a hydrostatic head and therefore
the density of the fluid enters the calculation. The air pressure on the right
side of the jack is actually slightly lower than that on the left side, but we have neglected this effect.

Question 3.2: Operation of a Hydraulic Jack Consider a hydraulic jack bein...

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