Here's a diagram, BUT use 591 lb for the weight (NOT 500 lb):
http://www.webassign.net/sf/p9_24.gif
Piston 1 has a diameter of 0.25 in. and is attached to a lever arm a distance of 2 in from the pivot point. Piston 2 has a diameter of 1.4 in. An external force F acts on the lever arm at a distance 10 in from piston 1 as shown in the diagram. Assume the height difference between piston 1 and piston 2 is negligible. In the absence of friction, find the force F necessary to support the 591 lb weight (in units of lb.)
2006-12-01 09:16:05 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
3.769 lbs is incorrect
2006-12-01 09:24:48 · update #1
Area1 = pi*(.25/2)^2 = .049 in^2
Area2 = pi*(1.4/2)^2 = 1.54 in^2
Pressure is the same throughout the hydraulic fluid - up, down, and sideways. Pressure = Force/area.
P2 = F2/area2
To support the 591 lb weight, F2 must be 591 lb.
P1 = F1/area1
And P2 = P1 (up, down, and sideways) Solve for F1 and then apply the 5 to 1 mechanical advantage.
2006-12-01 09:45:15 · answer #1 · answered by sojsail 7 · 0⤊ 0⤋
The tension on the two pistons is figured as tension instances the piston section. tension, P, is equivalent during the hydraulic gadget, and the gadget is in equilibrium so F1 + F2 = 0 P*pi*r1^2 + P*pi*r2^2 = 0
2016-12-29 18:49:22 · answer #2 · answered by ? 3 · 0⤊ 0⤋
3.769 pounds.
2006-12-01 09:21:51 · answer #3 · answered by campbelp2002 7 · 0⤊ 0⤋