Understanding Physics for the Beer Mug Problem

In summary, the problem involves a bartender sliding a beer mug towards a customer at a certain speed and direction, and the customer misses the grab causing the mug to sail off the end of the frictionless bar. The question is asking for the speed and direction of the mug at impact. Equations are provided to calculate the speed and direction, but more equations are needed to find the x and y components of the final velocity when the mug hits the floor. The conversation also mentions considering the sliding velocity and the x-component of the final velocity without using equations.
  • #1
Let It Be
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1. A bartender slides a beer mug at 1.50m/s towards a customer at the end of a frictionless bar that is 1.2m tall. the customer makes a grab for the mug and misses, and the mug sails of the end of the bar. What are the speed and direction of the mug at impact?


2. Vx=Vocosθ
Vy=Vosinθ
V=√Vx^2+Vy^2
θ=tan-1(Vy/Vx)


3. The correct answer apparently is 5.08m/s and θ=-72.8°. I didn't get that...
 
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  • #2
The equations you give are OK to find the speed and direction, but you need some more equations to find Vx and Vy when the mug hits the floor.
 
  • #3
AlephZero said:
The equations you give are OK to find the speed and direction, but you need some more equations to find Vx and Vy when the mug hits the floor.

I don't know how to solve the problem with the equations I do have though...
 
  • #4
How do you think vx and the sliding velocity compare?

Think about it, don't try use the equations!
 
  • #5
217 MeV said:
How do you think vx and the sliding velocity compare?

Think about it, don't try use the equations!

I know I need to find Vf for the speed when the mug hits the floor. And then direction idk
 
  • #6
Well, start with the x-component of v_f. What will that be?

Remember there's no friction whilst the mug is sliding, and I assume there is no air resistance.
 

FAQ: Understanding Physics for the Beer Mug Problem

What is the "Beer Mug Problem" in physics?

The "Beer Mug Problem" is a thought experiment used to demonstrate the concept of buoyancy in physics. It involves a beer mug filled with liquid being submerged in a bucket of water and the resulting change in weight.

How does understanding physics help in solving the Beer Mug Problem?

Understanding the principles of physics such as buoyancy, density, and Archimedes' principle can help in accurately predicting the change in weight of the beer mug when submerged in water. It also allows for the calculation of the buoyant force acting on the beer mug.

What factors affect the weight of the beer mug when submerged in water?

The weight of the beer mug when submerged in water is affected by its volume, density, and the density of the liquid it is filled with. The weight of the displaced water also plays a role in the overall weight of the beer mug.

How is the buoyant force related to the weight of the beer mug when submerged in water?

The buoyant force, which is the upward force exerted by a fluid on an object, is equal to the weight of the displaced fluid. In the case of the Beer Mug Problem, the buoyant force is equal to the weight of the water that is displaced by the submerged beer mug.

Can the Beer Mug Problem be applied to real-life scenarios?

Yes, the concept of buoyancy and the Beer Mug Problem can be applied to real-life scenarios such as the design of ships and submarines, understanding the behavior of hot air balloons, and even the flotation of objects in water. It is a fundamental principle in fluid mechanics and has many practical applications in engineering and everyday life.

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