How can one solve this Physics problem about pistons?

Question

Piston 1 in Figure P9.24 has a diameter of 0.25 in.; Piston 2 has a diameter of 1.1 in. In the absence of friction, determine the force, , necessary to support the 500 lb weight.
_________ lb
http://www.webassign.net/sf/p9_24.gif

Answer 1

The force exerted at 1 is proportional to the surface at 1 multiplied by the pressure. The pressure is equal to the weight at 2 divided by the area at 2.

So

F1 = 500 lbs * 0.25^2/1.1^2 = 25.8 lbs

To get this force a torque of
T = 25.8lbs*2in = 51.6 lbs.in is required at 2

At the end of the lever the force required is
T = F.r so
F = T/r = 51.6/12 = 4.30 lbs
______
Note: this method is equivalent to the first answerer's method except I avoided computing the surfaces at 1 and 2, thus I get a better precision.
The second answer is wrong.

Answer 2

The area of piston 2 is approximately 0.95 sq-in. So, the pressure of the fluid is 500/0.95=526.3 psi.

Since the fluid is considered to be incompressible, the pressure will be the same in piston 1. The area of piston 1 is about 0.05 sq-in. So, the force being applied to the piston is 526.3*0.05 = 26.3 lb.

The force on piston 1 is acting 2 in from the pivot, so the moment of that force is 26.3*2=52.6 in-lb. To get the balancing force, you divide this by the moment arm of the applied force, 12 in. So, the applied force is 52.6/12 = 4.4 lb.

Hope this helps.

Answer 3

Piston 2 has a force of 500lb acting above it.
Piston 2 must have the same force acting below it to support the weight.
Piston 1 therefore has to transfer this force to piston 2.
Area of piston 2 = 0.55 x 0.55 x 3.142 = 0.951in²
= 500 ÷ 0.951 = 525.8psi
Area of piston 1 = 0.125 x 0.125 x 3.142 = 0.049in²
525.8 ÷ 0.049 = 25.8lbs force required on piston 1.

(edited)

Answer 4

For extremely everybody else, the suited answer has a mistake indoors the Torque calculation Torque = tension of friction * radius the previous poster seems to have used the diameter somewhat than the radius, which does no longer yield the suited suited answer.