# Weight of water in different vessels

1. Oct 20, 2011

### Himal kharel

Water is poured to same level in vessels having different shapes. If pressure is the same at the bottom of each vessel, the force experienced by base of each vessel is same. Why then do these different shape vessels have different readings when placed on weighing scale?

2. Oct 20, 2011

### elegysix

Volume may not be the same. Pressure is a function of depth, not volume, the volume is what the weight depends on. That's my guess.

3. Oct 20, 2011

### Staff: Mentor

That's true only if the areas of the bottoms of the vessels are the same and ignores downward forces on any other nonvertical surfaces.

4. Oct 21, 2011

### anigeo

russ watters is certainly right.if the area of the base is less as compared to some other and provided the pressure is kept same, then the weight will be lesser.
but question arises when we consider centre of mass and normal reaction of the vessel with water.

5. Oct 21, 2011

### Naty1

Watters has it right....

Pressure is force per unit area; observed weight is due to the total volume of water...

You can see this if you consider,say, a three foot tall, 1/2" diameter pipe filled with
water (ignore the weight of the container), and some other very different volume
vessel, say a cylinder, a foot in diameter, of the same height.

Pressure at the bottom is the same, P = hd, but not total weight.

6. Oct 21, 2011

### Himal kharel

I guess i didnt explain the question properly. so here is a better explaination.
Consider two cylinders with same base area but one with converging walls and other with diverging walls. let they be filled to same height. now pressure is same at the base as given by P=hdg. So the force on bottom is same as given by force=pressure*area. this suggests that the force on weighing scale given by both cylinders is same. so they must have same readings when placed on scale. but it is not so. Please explain this contradictory result?

7. Oct 21, 2011

Your vessels aren't holding the same amount of water, so they don't weigh the same. Simple. Additionally, your vessels may weigh differently. If you want it in terms of the pressure, it is because the pressure is $\rho g h$ at the bottom, but $h$ is measured from the bottom of the vessel straight up to the top of the water, so with a container that gets narrower at the top, $h$ decreases as you go away from the center of the base leading to a lower overall force. For a vessel that is wider at the top, you are neglecting the vertical component of the force on the walls of the vessel, which are no longer vertical. This leads to a higher overall force. Of course, no matter how you slice it, it comes down to one vessel having more water in it than the other.

8. Oct 21, 2011

### JHamm

I think your confusion comes from the difference between the force from the water on the base of the cylinder and the force of the base of the cylinder on the scale, the first depends on the height of the water, the second on the total amount of water and there is no reason for them to necessarily be the same

9. Oct 21, 2011

### Staff: Mentor

I gave the answer in my first post: the water imparts a force on the nonvertical walls and that force is transfered throught the container, to the scale.

10. Oct 21, 2011

### elegysix

The pressures from the water are contained by the walls of the vessel - no net force, which is why the weight on your scale does not depend on internal pressure, only mass.

11. Oct 21, 2011

### elegysix

Take any aerosol or gas canister, weigh it, cool it or heat it some and weigh it again. definitely changes internal pressure, but i guarantee it'll weigh the same. Don't heat it up a whole lot though, because they tend to explode. :)