TPRH Hydraulic Jack

[4563728]

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Let W1 be the work don by the force pushing down on Piston #1, and let W2 be the work done by Piston #2 as it raises the load it supports. Calculate the ratio W1/W2.

What I did was set up proportions : ((F1*d1)/A1) =((F2*d2)/A2) with A1=5cm2 and A2=500cm2 (given in the passage). Then I rearranged it so it looked like (F1*d1)/(F2*d2)=A1/A2 and this got me 1/100.

But the book says the ratio is 1 but I don't understand...what was wrong with my calculation? What is the logic behind this?

 

[inasensegone]

Let W1 be the work don by the force pushing down on Piston #1, and let W2 be the work done by Piston #2 as it raises the load it supports. Calculate the ratio W1/W2.

What I did was set up proportions : ((F1*d1)/A1) =((F2*d2)/A2) with A1=5cm2 and A2=500cm2 (given in the passage). Then I rearranged it so it looked like (F1*d1)/(F2*d2)=A1/A2 and this got me 1/100.

But the book says the ratio is 1 but I don't understand...what was wrong with my calculation? What is the logic behind this?

A hydraulic jack doesn't increase the amount of work... this is impossible. Remember, work is a metric of energy. Energy in = energy out.

A hydraulic jack increases the force. In order to keep work constant however, it decreases the distance over which this force is applied:

A force F1 is exerted on the left. This is applied over a small area. This produces a pressure (force/area). The pressure remains constant throughout the fluid of the closed system, so on the right side of the diagram, this pressure becomes: P = F2/A2. Well, the area on the right is larger than the area on the left so the force must also be larger in order for the pressure to be equal. Again, what actually changes is that the distance over which this force is applied is smaller. In other words, if you push down on the left a given distance, the right side is going to rise less distance if it has a larger area.