Force must be applied to the small piston to lift the engine?

[AdnamaLeigh]

In a machine shop, a hydraulic lift is used to raise heavy equipment for repairs. The system has a small piston with a cross-sectional area of 0.057m^2 and a large piston with a cross-sectional area of 0.208m^2. An engine weighing 2900N rests on the large piston. What force must be applied to the small piston to lift the engine? If the engine rises 0.235m, how far does the smaller piston move?

I correctly found the force to be 794.712N. I thought that the smaller piston moves the same length as the engine rising, but I could be wrong. I don't know how to figure that out.

 

[andrevdh]

By applying Pascal's principle to the hydraulic jack we get that the change in pressure at the input piston is equal to the pressure change at the output piston , therefore the input force is magnified by the ratio of the areas of the pistons at the output piston. The volume of oil displaced at the input piston will be equal to the volume of oil rising at the output piston as the engine is lifted up. From this you can find the relationship between the areas of the pistons and the distances the two pistons move during the lifting process. Notice what this new relationship obtained this way tells us something about the work done during the lifting process.

 

[quicknote]

I would like to clarify that the force applied to the small piston is necessary to create the necessary pressure to lift the engine. This is because of Pascal's principle which states that pressure applied to a fluid in a closed system is transmitted equally in all directions. In this case, the force applied to the small piston will create a pressure that will be transmitted to the larger piston and eventually lift the engine.

To calculate the force needed to lift the engine, we can use the equation F1/A1 = F2/A2, where F1 is the force applied to the small piston, A1 is the cross-sectional area of the small piston, F2 is the force needed to lift the engine, and A2 is the cross-sectional area of the larger piston.

Substituting the given values, we get F1 = (F2 x A1)/A2 = (2900 N x 0.057 m^2)/0.208 m^2 = 794.712 N. Therefore, a force of 794.712 N must be applied to the small piston to lift the engine.

Regarding the distance the smaller piston moves, it will depend on the displacement of the engine. If the engine rises by 0.235 m, the smaller piston will also move by the same distance. This is because the volume of the fluid displaced by the smaller piston must be equal to the volume of the engine. Therefore, the smaller piston will move 0.235 m to create the necessary pressure to lift the engine.