In a hydraulic lift, the diameters of the pistons are 4.96 cm and 20.2 cm. A car weighing W = 10.4 kN is to be lifted by the force of the large piston. (a) What force Fa must be applied to the small piston? (b) When the small piston is pushed in by 10.4 cm, how far is the car lifted? (c) Find the mechanical advantage of the lift, which is the ratio W/Fa.
Any help would be appreciated... I'm just looking to understand the problem!
2007-07-17 08:40:26 · 3 answers · asked by Kelly 1 in Science & Mathematics ➔ Physics
all the ratio's you need come from the ratio of the area of the cylinder cross section,
areas are: pi*r^2
4.96cm = 19.31cm^2
20.2cm=320.31cm^2
so the force needed is 10.4*19.31/320.31 = .626kN or 626 N
and the distance traveled is computed the same way, so its .626cm
the mechanical advantage is the large area over the small area 320/19.3=16.5
2007-07-17 08:49:16 · answer #1 · answered by Anonymous · 1⤊ 0⤋
The ratio of forces on each piston is proportional to the ratio of areas.
So, if one piston has 3 times the diameter of another, it has 9 times the surface area, because the surface area of a circle goes as the square of diameter. Therefore, 1 N applied to the small piston generates 9 N of force at the large piston.
However, the distance traveled by each piston goes as the INVERSE of the piston area. So, in my example, if you pushed the small piston down by 18 inches, the larger piston would move by only 2 inches (one ninth as far).
2007-07-17 08:50:53 · answer #2 · answered by lithiumdeuteride 7 · 0⤊ 0⤋
LOOK FOR AN ONLINE ENGINEERING GENERATOR
2007-07-17 09:05:21 · answer #3 · answered by manunitedk 3 · 0⤊ 0⤋