# What is the pressure on the sample if 300 N is applied to the lever?

1. Nov 11, 2004

### physicsss

A hydraulic press for compacting powdered samples has a large cylinder which is D1 = 13.0 cm in diameter, and a small cylinder with a diameter of D2 = 3.0 cm (Fig. 13-48). A lever is attached to the small cylinder as shown. The sample, which is placed on the large cylinder, has an area of 4.0 cm2. What is the pressure on the sample if 300 N is applied to the lever?

http://www.webassign.net/gianpse3/13-48alt.gif

2. Nov 11, 2004

### cepheid

Staff Emeritus
I think you know how to calculate the pressure on the small cylinder, given the applied force and the area of the cylinder's face. Next, use Pascal's law: Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel. This is probably in your text. If you look up the example of how this applies to a hydraulic lift, then the answer to the problem of the hydraulic press should also be clear.

3. Nov 11, 2004

### physicsss

Do I have to worry about the L at all?

4. Nov 11, 2004

### cepheid

Staff Emeritus
Good point...I never thought of that. If it is true that the same torque around the pivot point is being applied to all points on the lever, then the equivalent force in the centre of the lever would be larger than 300N. But I am not at all sure of this. Could somebody else chime in? How do you determine with what force the cylinder (at the centre of the lever) will push down on the fluid?

5. Nov 13, 2004

### Clausius2

I would say:

$$F_{cylinder}=\frac{300\cdot 2L}{L}$$ as angular equilibrium states.

6. Nov 14, 2004

### cepheid

Staff Emeritus
So let me see if I understand what you are saying: we have an applied torque:

$$\tau_{app} = (300)(2L)$$

and the cylinder must provide a counter-torque? Is that it? In that case:

$$F_{cylinder}(L) = (300)(2L)$$

$$F_{cylinder}=\frac{300\cdot 2L}{L}$$

OK, fine. But why must there be equilibrium at all? After all...the cyliner is moving downward. And at least some part of the cylinder's force should be directed sideways as well as downward, right? The question confuses me...

7. Nov 14, 2004

### Clausius2

Sure, after you push down the cylinder, it is moving downwards. And the pivoting bar is also moving with some angular acceleration.

But just when you push down the cylinder, the system can be considered in rotationally static. Think of it, only a small displacement of the bar of length 2L is possible, because it is not extensible. Such a small angular displacement is enough to provide a pressure wave that reaches the other cylinder surface. It is usual to simplify these types of movements as I have just said to you due to those small displacements, and it is not needed to formulate the dynamic equilibrium of the bar, because it is almost rotationally static.

8. May 2, 2005

### hotmail590

How do we find out the length of L?