(In a hydralic system)
If a weight of 1500 N (Fout) must be lifted, how much force (Fin) must be applied to the input end ?
Area in= 3m^2
Area out= 60m^2
2007-08-04 05:23:43 · 5 answers · asked by ted 1 in Science & Mathematics ➔ Physics
In a hydraulic system, the pressure of the hydraulic fluid is the same everywhere. That's the crucial fact.
At the output point: pressure = Fout/(Area out)
At the input point: pressure = Fin/(Area in)
Since these two pressures are equal, we have:
Fout/(Area out) = Fin/(Area in)
Use algebra to solve for Fin.
2007-08-04 05:37:33 · answer #1 · answered by RickB 7 · 0⤊ 0⤋
Area out is 20 times larger than area in.
so..
weight needed to lift... 1500 N
since Area out is 20 times larger than area in so force needed to apply...
1500/20 = 75N
2007-08-04 12:41:34 · answer #2 · answered by Anindit 1 · 0⤊ 0⤋
Pressure is Force/Area
P=F/A
The pressure is the same on either side, so
F{1}/A{1}=F{2}/A{2}
The car has 20 times the area lifting it, so it has 20 times the force.
2007-08-04 12:40:44 · answer #3 · answered by 2 meter man 3 · 0⤊ 0⤋
Fa/Fb = Aa/Ab a & b should be subscripts
Its a relation that can be proved.
Therefore,
F/1500 = 3/60
or F= 75N
Simple! Isn't it?
2007-08-04 12:40:06 · answer #4 · answered by shubhopriyo 2 · 0⤊ 0⤋
75N
2007-08-04 12:34:19 · answer #5 · answered by andy t 6 · 0⤊ 0⤋