Fluids - I wish the answer would flow to me

[stuplato]

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

I have two problems to do and I seem to suck at fluids...
.
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.

.
.
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?

.
.
I tried everything I could think of but just don't get it, could I have some clues? Thanks.

 

[sarujin]

What are we solving for in the first problem?

 

[stuplato]

the force applied to each side of the revolving disc.

That

 

[stuplato]

Forgive my double post
.
I got the third problem so no help needed there

 

[stuplato]

i really cannot get these and my grade depends on these

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...

 

[mezarashi]

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?

 

[stuplato]

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?

 

[mezarashi]

Maybe you should try writing out your calculations.