Hydraulic Press 50 kn

In a hydraulic lift, the radii of the pistons are 2.50 cm and 9.4 cm. A car weighing W = 8.6 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 8.6 cm, how far is the car lifted? (c) Find the mechanical advantage of the lift, which is the ratio W / Fa.

EDIT: It is the Area of the pistons: a) Fa =Ari/Ar2 x F load = pi x (r1^2) / pi x (r2^2) x 8.6 kN = (2.5^2)pi / (19.6 cm^2 / 283.5 cm^2 x 8.6 kN = .069 zx 8.8 kN =.594 kN or 594 Newtons. EDIT: Not sure of (b)--is it distance x radius =d x r, or Area r of piston x d = area x dist (that is, mech. advantage/distance)? b) r1 d1 =r2d2 so, d2=- r1d1/r2 = 2., cm x 8.6 cm /9.4 cm = 2.128 cm. c) W/ fa = 8.6 kN/.594 kN = 14.48 : 1 .

Easy to use and powerful enough for most press work, choose from the bench top 6 ton hydraulic press or the 12 ton floor model. These guys can handle installing or removing bearings, gears, U-joints and can be used for straightening or bending jobs as well. The tables can be quickly adjusted for height and each includes table plates. Size: 20"W x 17"D x 37"H Aprox shipping weight 70 lbs

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The small piston of a hydraulic lift has a cross-sectional area of 2.50 cm2, and its large piston has a cross-sectional area of 199 cm2. What force must be applied to the small piston for the lift to raise a load of 15.4 kN?

Force is proportional to area. The ratio of the areas is 199 cm² / 2.5 cm² = 79.6 That means the force goes down by that ratio, 15400 N / 79.6 = 193 newtons .

The small piston of a hydraulic lift has a cross-sectional area of 2.50 cm2, and its large piston has a cross-sectional area of 196 cm2. What force must be applied to the small piston for the lift to raise a load of 15.4 kN? (In service stations, this force is usually exerted by compressed air.) In Newtons

F/A (big)= F/A (small) 15.4/ 196 = F /2.5 = 0.19643 KN =196.43 N

In a hydraulic lift, the radii of the pistons are 2.50 cm and 10.0 cm. A car weighing W = 10.0 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 10.0 cm, how far is the car lifted?

The pressures of each piston face must be equal (or greater to actually cause lift) Pressure is Force divided by the area so the pressure on the large piston would be: 10/(pi*0.1^2) =318 kPa roughly The force then is the pressure multiplied by the smaller area 318*(pi*0.025^2) =0.625 kN The volume change for each piston is equal so divide the volume of the small piston's change by the the area of the large piston to find the total height changed. (pi*0.025^2*0.1)/(pi*0.1^2) =6.25mm

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