presents one method for modeling this problem. Multiple-Concept Example 8 also p
ID: 1701719 • Letter: P
Question
presents one method for modeling this problem. Multiple-Concept Example 8 also presents an approach to problems of this kind. The hydraulic oil in a car lift has a density of 8.17 x 102 kg/m3. The weight of the input piston is negligible. The radii of the input piston and output plunger are 6.44 x 10-3 m and 0.179 m, respectively. What input force F is needed to support the 22700-N combined weight of a car and the output plunger, when (a) the bottom surfaces of the piston and plunger are at the same level, and (b) the bottom surface of the output plunger is 1.10 m above that of the input plunger?Explanation / Answer
The density of the oil is = 8.17*10^2 kg/m^3
The radius of the piston is r1 = 6.44*10^-3 m
The radius of the plunger is r2 = 0.179 m
The combeined weight of car and the plunger is F2 = 22700 N
1)
The force applied on the piston will be
F1 = F2(A1/A2)
A2 and A1 are the reas of the plunger and piston respectively
F1 = F2(r1^2/r2^2)
F1 = 22700*(6.44^2*10^-6/0.179^2)
F1 = 29.38 N
2)
The distance between the piston and the plunger is d = 1.1 m
The change in pressure due to this distance is P = gd
P = 8.17*10^2 * 9.8* 1.1
P = 0.88 Pa
The extra force on piston due to this pressure is
F' = PA2
F' = 0.028 N
The total force on piston is
F1 = F2(A1/A2) + F'
F1 = 29.38+0.028
F1 = 29.41 N
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