Question

Multiple-Concept Example 8 presents an approach to problems of this kind. The hydraulic oil in a car lift has a density of 8.30 ???? 10 2 kg/m3. The weight of the input piston is negligible. The radii of the input piston and output plunger are 7.70 ???? 10????3 m and 0.125 m, respectively. What input force F is needed to support the 24 500-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 1.30 m above that of the input piston?

Answer 1

Answer:  

a. = 93 N

b. =  94.9 N

Explanation:

Answer:

Explanation:

from the question we were given the following:

density of oil (ρ) = 8.3 x $10^{2}$  $\frac{kg}{m^{3} }$

radius of the output plunger (R) = 0.125 m

radius of the input piston (r) =  7.70 x $10^{-3}$  m = 0.0077 m

output force (F2) = 24500 N

input force (F1) = ?

acceleration due to gravity (g) = 9.8 $\frac{m}{s^{2} }$

difference in height of the plunger and piston (h) = 1.3 m

• we can find the required input force when the piston and the plunger are at the same height using the equation below

$\frac{F2}{A2}$  = $\frac{F1}{A1 }$

F1 = (  $\frac{F2}{A2 }$  ) x A1

where A1 is the area of the input piston and A2 is the area of the output plunger

Area = π x $radius^{2}$

A2 = π x $0.125^{2}$ = 0.049

A1 = π x $0.0077^{2}$ = 0.000186

recall that from above F1 =  (  $\frac{F2}{A2 }$  ) x A1

F1 = (  $\frac{24500}{0.049 }$  ) x 0.000186

F1 = 93 N

• we can find the required input force when the height of the piston and the plunger are 1.3 m apart using the equation below

P2 = P1 + ρgh

where P = pressure = $\frac{F}{ A }$

therefore the equation above now becomes

$\frac{F2}{ A2 }$ = $\frac{F1}{ A1 }$  + ρgh

F2 = ( $\frac{F1}{ A1 }$  + ρgh ) x A2

F2 =  ( $\frac{24500}{ 0.000186 }$  + ( 8.3 x $10^{2}$  [tex] x 9.8 x 1.3  ) ) x 0.049

F2 = 94.9 N