Physics Piston question, the picture works now :)?

Question

Piston 1 in Figure P9.24 has a diameter of 0.30 in. Piston 2 has a diameter of 1.3 in. In the absence of friction, determine the force, , necessary to support the 500 lb weight.

Here is the link to Figure P9.24: http://i136.photobucket.com/albums/q200/physics2006/piston.gif

thanks a bunch guys!

Answer 1

From looking at the picture, it is clear that this is both a piston problem and a lever problem. So the previous answers were only partially correct.

So we know that the pressure at both pistons must be the same, so:

P(1) = P(2) or F(1)/A(1) = F(2)/A(2)

But we also know that the torques at the lever must also be equal:

Let d1 = 10 inches and d2 = 2 inches, then

F(2)d2 = F(lever)d1

So what we want to solve for is F(lever):

F(lever) = F(2)d2/d1

and F(2)=F(1)A(2)/A(1)

F(lever) = F(1)A(2)d2/A(1)d1 = mgA(2)d2/A(1)d1

where m = 500 lbs and g is Earth's gravity (32.17405 ft/s²)

Remember that most of the inputs are in lbs and inches, so if you want the answer in Newtons, you'll need to convert to the metric system.

Answer 2

The force on the 2 pistons is figured as
Pressure times the piston Area.
Pressure, P, is equal throughout the hydraulic system, and the system is in equilibrium so
F1 + F2 = 0
P*pi*r1^2 + P*pi*r2^2 = 0

Answer 3

I think all you have to do is to sum the torques around the small piston. Since the system is stationary, then the sum of the torques is zero.

In that case (taking clockwise as positive)

F*(10) - 500*(2) = 0

Answer 4

Do your own homework!