What are the physics behind balancing drinks on a tray?

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The discussion centers on the physics of balancing drinks on a tray, emphasizing the importance of torque and center of gravity. Torque is defined as the product of the weight of the drink and its distance from the tray's center, requiring that the sum of all torques equals zero for balance. Additionally, the height of the glass affects stability, with taller glasses having a higher center of gravity, making them more prone to tipping. The conversation provides a practical example involving a 10 Newton object and a 25 Newton object to illustrate torque balance calculations.

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Okay, you know how waiters carry those trays with lots of lots of drinks on them. What determines how drinks can fall of the tray. Like I realized that the odds of the drink falling is proportional to the drink height. This all had to do with aerodynamics. Could you guys give me some equations to play around with or some useful info about carries trays and not allowing things to fall of with physics?
 
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Well, first and foremost, your torques have to be balanced.
Check out this picture I made. It's a top-down view of a tray with four drinks on it. Imagine the center is where the waiter's hand is.

Torque, if you didn't know, is a "twisting force". The torque that a glass of water exerts on a tray is equal to the weight of the drink, times the distance it is away from the center.
In order for this tray to be balanced, all four torques have to add up to zero.
torque_waiting.jpg


As a simple example, if you have a 10 Newton object, 40 cm away from the fulcrum of a balance beam, where should you place a 25 Newton object in order to balance it? To see it as an equation, (10*40)+(25*x)=0. If you solve for x, you find out that x= -16, meaning that you should place the 25 Newton object 16 cm away from the fulcrum, on the opposite side.

The picture above is a little more complicated, because you're balancing torque in more than one direction. All you have to do, is break each distance vector into its x and y components, multiply each component vector by the weight of its glass, and do the balance beam problem, once in the x direction, and once in the y direction.

torque_waiting2.jpg



Another important factor is the shape of the glass itself. As you suggested in your post, a taller glass is harder to keep standing. This is because it has a higher center of gravity than a shorter glass. A glass will start to fall over once its center of gravity is tipped passed the base of the glass.

glasses.jpg


The glass on the left only has to be tipped 15° before it starts to fall over.
The glass on the right has to be tipped a whole 45° before it starts to fall over.
 

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