What is the necessary force to support a 500 lb weight with a hydraulic lift?

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To determine the necessary force to support a 500 lb weight with a hydraulic lift, the relationship between the forces and areas of the pistons is crucial, expressed by the equation F1=(A1/A2)F2. The user initially calculated F1 as 12.8 but questioned the accuracy, considering the height of the fluid column in the larger piston. Further discussion emphasized the importance of torque in the calculation, suggesting that the torque should be evaluated around the point next to piston 2. The correct approach involves analyzing the torque created by the force exerted and the pressure force acting on the pistons. Accurate calculations must account for both the forces and the geometry of the hydraulic system.
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Homework Statement


Piston 1 in Figure P9.24 has a diameter of 0.32 in.; Piston 2 has a diameter of 2.0 in. In the absence of friction, determine the force, , necessary to support the 500 lb weight.

Picture
http://www.webassign.net/sf/p9_24.gif

Homework Equations



F1=(A1/A2)F2
A=pi*r^2

The Attempt at a Solution


I solved for F1 using the above equation and got 12.8. THIS IS WRONG Is this because the column of fluid is higher in the larger piston? So i have to factor in the weight of the higher water as well? Can someone please point me in the correct direction?

thanks
 
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you got the force that the piston at 1 exerts upwards... what force does the hand need to exert to keep the piston from moving.. use torque.
 
torque=rF

would the radius be from the hand to the center of piston 1?
 
Izmad said:
torque=rF

would the radius be from the hand to the center of piston 1?

you want the torque about the point next to the piston 2. so torque due to the F force is F*12.0in clockwise... what is the torque due to the pressure force...
 

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