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  1. Problems
  2. /A hydraulic jack has pistons 2.0 cm and 25 cm across…
MediumPhysicsHigh School (11–12)Mechanics › Pressure and FluidsNumericalNEETBoard ExamAP

A hydraulic jack has pistons 2.0 cm and 25 cm across…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A hydraulic jack has pistons 2.0 cm2.0\ \text{cm}2.0 cm and 25 cm25\ \text{cm}25 cm across. What force on the small piston holds up a 2000 kg2000\ \text{kg}2000 kg van? How far must the small piston move in all to raise the van by 10 cm10\ \text{cm}10 cm? (g=9.8 m/s2g = 9.8\ \text{m/s}^2g=9.8 m/s2.)

Check your answer

Work it out on paper, then type your number. The answer itself stays hidden.

N
m
Need a reminder? The key ideaShowHide

Pressure is force per unit area, p=FAp = \dfrac{F}{A}p=AF​, in pascal. Below the surface of a liquid, p=p0+ρghp = p_0 + \rho ghp=p0​+ρgh; the gauge pressure is ρgh\rho ghρgh. Pascal's law: a change of pressure in an enclosed fluid reaches every part of it, so a hydraulic machine gives F2=F1A2A1F_2 = F_1\dfrac{A_2}{A_1}F2​=F1​A1​A2​​.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. Area ratio. (25 cm2.0 cm)2=156.25\left(\dfrac{25\ \text{cm}}{2.0\ \text{cm}}\right)^2 = 156.25(2.0 cm25 cm​)2=156.25.

Step 2. Force. 2000 kg×9.8 m/s2156.25=125.4 N\dfrac{2000\ \text{kg} \times 9.8\ \text{m/s}^2}{156.25} = 125.4\ \text{N}156.252000 kg×9.8 m/s2​=125.4 N.

Step 3. Distance. 0.10 m×156.25=15.6 m0.10\ \text{m} \times 156.25 = 15.6\ \text{m}0.10 m×156.25=15.6 m, in many short strokes.

Answer: 125.4 N125.4\ \text{N}125.4 N; 15.6 m15.6\ \text{m}15.6 m of pumping in all.

Check. Work in: 125.4 N×15.6 m=1960 J125.4\ \text{N} \times 15.6\ \text{m} = 1960\ \text{J}125.4 N×15.6 m=1960 J; work out: 19 600 N×0.10 m=1960 J19\,600\ \text{N} \times 0.10\ \text{m} = 1960\ \text{J}19600 N×0.10 m=1960 J.

Watch out. Square the ratio of the diameters.

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