Fluids - I wish the answer would flow to me

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Fluids - I wish the answer would "flow" to me

I have two problems to do and I seem to suck at fluids...:-p
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The drawing shows a hydraulic system used with disc brakes. The force F is applied perpendicularly to the brake pedal. The brake pedal rotates about the axis shown in the drawing and causes a force to be applied perpendicularly to the input piston (radius = 9.53 10^-3m) in the master cylinder. The resulting pressure is transmitted by the brake fluid to the output plungers (radii = 2.14 10^-2m), which are covered with the brake linings. The linings are pressed against both sides of a disc attached to the rotating wheel. Suppose that the magnitude of F is 9.18 N. Assume that the input piston and the output plungers are at the same vertical level and find the force applied to each side of the revolving disc.
Diagram of a hydraulic disc brake: a brake pedal pivots on an axis, pushing a master cylinder; brake fluid lines lead to plungers with linings pressing on an edge-view rotating disc attached to a wheel. Labels: 0.100 m, 0.0500 m, F, Axis, Brake pedal, Master cylinder, Brake fluid, Plunger, Lining.

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A solid cylinder (radius = 0.150 m, height = 0.120 m) has a mass of 6.60 kg. This cylinder is floating in water. Then oil ( = 725 kg/m3) is poured on top of the water until the situation shown in the drawing results. How much of the height of the cylinder is in the oil?
Diagram of a solid cylinder floating at the interface of oil above and water below in a clear container, with the oil layer labeled Oil and the water layer labeled Water.


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I tried everything I could think of but just don't get it, could I have some clues? Thanks. :smile:
 
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the force applied to each side of the revolving disc.

That :smile:
 
Forgive my double post o:)
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I got the third problem so no help needed there :smile:
 
i really cannot get these and my grade depends on these :rolleyes:

No clue on #1

But for 2nd I try to find sum of forces and set equal to the weight of the cylinder, but it isn't right...
 
The hint that gravitational potential is the same is a great simplificiation. You can focus on the equivalence of forces. Assume that the fluid is incompressible.

You exert a pressure of F/A. The same pressure appears at the two sides of the output plungers. The force would be again F/A. That help?
 
mezarashi said:
The hint that gravitational potential is the same is a great simplificiation. You can focus on the equivalence of forces. Assume that the fluid is incompressible.

You exert a pressure of F/A. The same pressure appears at the two sides of the output plungers. The force would be again F/A. That help?

Tried that and got 46.29

But wrong... so what did I do?