They call it the Mighty Mug—and the idea is that it isn't easily knocked over. Of course, you can tip just about anything over if you try hard enough. But really, how does this work? Let me be clear: I haven't actually played with a Mighty Mug—so most of this is just physics-based speculation.
It seems the key component of this mug is some type of suction cup on the bottom. When you put the cup down, a rubber seal forms with a smooth table surface. If someone knocks the cup and the rubber bottom moves up a tiny bit, the interior air pressure will slightly decrease to a value smaller than the atmospheric pressure. This atmospheric pressure on the top results in a force that can keep the cup down.
But how do you pick it up then? Honestly, I'm not quite sure. If I had to guess, there must be some type of valve that lets air get back under the rubber seal when you lift it. That's my super-short explanation.
Now is a good time to review the physics of a suction cup. They don't actually suck at all. The force from a suction cup comes from the atmosphere. Let's imagine the atmosphere as a bunch of tiny balls bouncing around.
These tiny balls are moving around, and as they travel, some of them collide with a surface. Since the tiny ball changes momentum when it collides, it exerts a small force on the area. The bigger the area, the more collisions and the greater the force. On a bigger scale, if you know the pressure and the area you can calculate the force.
If you also have air (or balls in the tiny balls model of air) below the surface, there will be two forces. One force pushing down from the collisions above and a force pushing up from the collisions below. With equal pressures, these two forces would be equal, giving a net force of zero. In real life, the pressure above and below the object are not quite the same (at least in a constant gravitational field). This will produce a very tiny upward net force that we call buoyancy. This buoyancy force doesn't really do anything significant to the Mighty Mug, but it's fun to mention.
Getting back to the suction cup. There is a net downward force on the suction cup because there is less air underneath the cup (and less pressure) than there is above the cup. However, this means that suction cups don't work if there is no air. Here is a – video showing what happens if you put a suction cup in a vacuum bell and remove the air. But in the end, this cup probably stays up with suction cups.
Now for a good physics estimation. Suppose I push the cup near the top, like this.
Yes, a lot of stuff is going on here. To start, of course, there is the gravitational force and the force from the push. The table pushes up as would be true for any cup. In this case, the table is only pushing up on the right side of the cup. I have put the force from the suction cup as a downward force on the left side of the cup; if you push on the top of the cup and to the right then only the left side of the suction would pull down. The last force is the frictional force, which prevents the cup from sliding.
If I want to solve for this maximum pushing force, I need to use two ideas. First, that the net force is zero. This means that the forces in the vertical direction have to add to zero as well as the forces in the horizontal direction adding to zero. Second, the total torque is zero. These two ideas together mean that the cup is in equilibrium (or not falling over).
OK, I'm going to skip over most of the details in this calculation, so I will just give a few starting notes.
- I need to estimate the mass and size of the cup. I'm going to use a mass of 400 grams with a height of 15 cm and a diameter of 8 cm (mostly just from guessing).
- The tricky force to calculate is the suction force. I am going to assume the pressure inside the suction cup is half of an atmosphere and the contact area is half of the area of the bottom of the cup (since the other side pushes up not down).
- When calculating the torque, it's not just the force that matters but also the location of that force. For the suction force, I am going to use a location that's the same as the center of mass of a half circle.
- The frictional force is as big as it needs to be so that the cup doesn't slide.
- I guess I should also say that I will assume the cup plus liquid has a uniform density (so the center of mass is in the center).
Using that, I can write the following equation for the sum of the vertical forces.
Calculating the magnitude of Fsuck, I get approximately:
The ΔP is the difference in pressure above and below the suction cup—or 0.5 atmospheres. With these values, I get a sucking force of (yes, I know it doesn't really suck) 127 Newtons. Now I can solve for the force of the table, but I can get by without it.
Now for the total torque. If the net torque is zero about some point on the object, the net torque is zero about any point on the object. I am going to sum the torques about a point on the right edge of the cup. Assuming the upward table force acts on this corner as well as the frictional force, then both of these forces produce zero torque about this point (to make things easier). My torque equation now becomes (if you want more details on torque calculations, check out this post on the mass of Darth Vader):
I decided to just push the cup at the top (because it's simpler) and I am calling x the distance from the edge of the cup to the location of the suck force. Putting in my values and solving for the push force, I get a maximum value of 49 Newtons (11 pounds). That's pretty impressive. Probably a normal tap would just be around 2 to 3 Newtons, but I'm just guessing. Oh, also it should be clear that this is just a rough calculation and not to be used for mission critical events involving the non-spillage of beverages.
One final note. If you want a very awesome method to prevent drink spills, check out the physics of the no-spill drink tray. It's a great demo that you can (and should) try yourself.
