Why do we need extra force to pull an upside down mug out of a bucket?

Question

If I try to pull an upside down mug (full of water) out of a bucket of water, extra force is required to break free from the water surface.

In the case of an inverted mug (rotated by $180^0$ from the previous case), pulling out of water is straight forward i.e. you apply the force equal to the weight of the mug + water.

Have you actually tested that the force needed is not the same for both cases? Or have you read about a study where they have tested it? My first guess would be that the forces are actually equal, and the difference is only psychological, although I might be wrong as well. — JustSaying, Nov 29 '19 at 07:27
Does this answer your question? Why is it so difficult to pull an immersed mug out of the water? — , Jul 05 '20 at 06:50
@FakeMod No. I'm asking about the difference in force between inverted & non-inverted mug. The linked question doesn't talk about that. — user, Jul 05 '20 at 09:15

Hemang Rajvanshy · Answer 1 · 2019-11-30T00:24:48.150

1

Atmospheric pressure is at play here. If the upside-down mug had any amount of water in it, and you start to pull it out of the water, the water has to leave the mug. But since the water leaving the mug needs to be replaced with air, this creates a sort of vacuum. This vacuum is behind the additional force required to break free from the surface (i.e. the force required to create this vacuum).

Edit: One way to think about this is in terms of energy. Creating a vacuum requires energy (think of the created vacuum as a store for this potential energy). Given the relation between potential energy and force, you can see how this would require additional force.

The extra force is mostly still just the weight of the water. As the bottom (closed) of the mug leaves the surface of the water, it pulls the water up with it above the surface of the water because of said vacuum.

edited Nov 30 '19 at 00:24

answered Nov 29 '19 at 04:39

Hemang Rajvanshy

164

Can you expand a bit more on how the vacuum requires that additional force ? – user Nov 29 '19 at 06:28
@user see edit. – Hemang Rajvanshy Nov 30 '19 at 00:25
@HemangRajvanshy Would you say that pulling an upside down mug so as to detach it from the surface of water is similar to pulling a suction cup so as to detach it from the surface that it's adhered to? – Ajay Mohan Nov 30 '19 at 01:34
Similar in that both work based on the pressure difference. Different because suction cups are attached to a rigid surface that can offer an opposing force. In this case, the most force you can get out of it is close to the weight of the water that can fit in the cup. It is very similar to the upside down water cup and card trick – Hemang Rajvanshy Nov 30 '19 at 03:13

Bhavay · Answer 2 · 2020-07-05T04:48:47.803

I was asking the same question to myself the other day and today i finally figured it out.Actually it's not due to vacuum , it is due to the fact that the mug has tiny amount of air trapped inside it.

I am using an approximation that the mug is pulled out slowly i.e without accelerating.

Let the :

Density of water be $\rho_w$
The external force that you apply be F
Atmospheric pressure be $p_o$

The Key thing to notice is that air inside the mug is not exerting the same pressure.

Pressure that trapped air is exerting be $p$
Area of cross section of the mug be $A$

Now since the water is at same horizontal level (at my dashed line ) and the water is at rest . Pressure at the horizontal line is $p_o$.

Now : $p_o=p+\rho_w g h' ====> (p<p_o) $

Let's us say difference in pressure is $\Delta p$

Therefore according my free body diagram we get :

$$F +pA = mg + p_oA $$

$$F = mg + \Delta pA $$

$$F>mg$$

After that, the air enters into the mug and the water spills out.

Why do we need extra force to pull an upside down mug out of a bucket?

2 Answers2