Please answer fully to receive credit. My car weighs about 3400 pounds. I would

Question

Please answer fully to receive credit.

My car weighs about 3400 pounds. I would love to have a large hydraulic lift in my garage for hoisting the entire car so I could work on it. A schematic of such a device is shown in Figure 4E.1 below. Let's assume that the diameter of the large hydraulic ram for lifting the car is 12 inches (D2). Let's also assume that I want to be able to step on the smaller hydraulic ram (FI) to elevate the car. If I can comfortably exert 25 pounds (FI), what is the smaller hydraulic ram diameter (Di, in inches) needed to just lift the vehicle? What is the pressure in psig inside the hydraulic system? What are the values for Di and P if D2 = 6 inches? Figure 4E.1. Schematic of Pascal's hydraulic machine (modified) from Cengel and Boles Thermodynamics, 6th Edition online material

Explanation / Answer

Answer :

Principle Used : The pressure inside liquid at all the points is the same, provided there is not much variation in height, and in this case, we assume that there is not much height of the hrdaulic set-up so as to give negligible variations in pressure inside liquid

Given data :

Weight of the car, W = 3400 pounds

Weight of the person, F1 = 25 pounds

D2 = 12 inches

Principle of hydraulic lift :

In equilibrium, pressure exerted by smaller hydraulic ram, Ps = pressure exerted by larger hydraulic ram, Pl

Since pressure = force / area

So, Ps = F1/A1 and Pl = W/A2

Since Ps = Pl, so, F1/A1 = W/A2

A1 = pi*D12/4 and A2 = pi*D22/4

Putting values, we get : D1 = 1.028 inches

When D2 = 6 inches, putting values we get : D1 = 0.514 inches

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