Page 38 - ebook fluid mechanics_finalize

FLUID MECHANICS

Thus, the small force F can raise the larger load W because the jack has a mechanical

advantage of A/a.

-4
-4
2
2
Putting F = 650 N, a = 15x10 m , A = 150x10 m
F = W
a A

150x 10 − 4
So that W = F  A = 650  − = 6500 N
a 15x 10 4

b. if the large piston is 0.65 m below the smaller piston

p = p + 
gh
1
2
F 650
4
2
p = = = 43 3 .  10 N / m
1
a 15 10 − 4
3
2
Putting ρ 10 kg= 3 / m , h = 0.65 m and g = 9.81 m/s
p = p + 
gh
2
1
3
4
p 2 = 43 3 .  10 + ( 10  . 9 81 ) 0 . 65

4
3 .
= 43  10 + 6376 5 .
2
= 439.38kN/m and

W = p 2 A

−
4
3
= 439 . 38 10  150  10
6590= 7 . N

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