Maximum height a waitress can push on a glass without it tipping

In summary, the problem involves a waitress trying to push a glass of water without tipping it on a dining table. The maximum height at which the glass can be pushed without tipping is 10 cm above the table top, based on the coefficient of static friction between the glass and the table's surface. The solution involves calculating the net torque and minimum value of force required for the glass to slide across the surface.
  • #1
jfnn

Homework Statement



I have attached the problem. I will write it out:

A waitress attempts to push a glass of water of heigh 15.o cm and diameter of 7.00 cm on a dining table, as shown in the figure. If the coefficient of static friction between the glass and the table's surface is 0.350, what is the maximum height (h) above the tabletop at which the waitress can push the glass without causing it to tip?

Homework Equations


[/B]
F net = 0
Fx = 0
Fy = 0

Something with momentum (p), maybe?

The Attempt at a Solution


[/B]
I defined east as positive and north as positive.

I said the axis of rotation was at the right corner of the class, and I called this O.

I said the sum of forces in x direction = 0 --> The only forces are F and static friction (fs)

Thus, sum of forces in x direction = F-fs

So F = fs where fs=us*N

So... F = us * N --> Equation 1

Then I said the sum of forces in y direction = 0 --> Where the only two forces in y direction are N and W

So...

N - W = 0
N = w
N = mg --> Equation 2 --> Plug into equation 1

F = us * mg

This is as far as I can get and I become lost... Some help would be greatly appreciated.

Thank you in advance!
 

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  • #2
jfnn said:
I said the axis of rotation was at the right corner of the class, and I called this O.
Rotation suggests adding to your list of relevant equations.
 
Last edited:
  • #3
TSny said:
Rotation suggests adding to your list of relevant equation.

Okay, I am going to say that the glass is in equilibrium, so it has to have a net torque equal to zero. Clockwise is negative, cow is positive.

Then I am going to say, that if I defined the axis of rotation at the right end of the glass, the torque of the normal force and static friction is zero because the angle/distance between the force vector and axis vector is 0.

This makes the net torque equal to Torque of weight - torque of F = 0

I will now calculate the torque for the weight and the force.

Tw=rFsin(theta) --> Tw = (0.07m/2)*sin(90)*mg

T force = r*sin(theta) --> T force = (h)*sin(90)*F

Then I can write: 0 = 0.035mg - hF

I get lost here.. is this thought process right/where would I go next?

Thanks again..
 
  • #4
So far so good. If the glass is to slide across the surface, then what is the minimum value for F?

Note: I recommend that you work the entire problem in symbols and plug in numbers at the end.
 
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  • #5
TSny said:
So far so good. If the glass is to slide across the surface, then what is the minimum value for F?

OH!

F= us*mg

Therefore, I will plug this into my equation to get.

0.035mg = h*us*mg

mg cancels

0.035 = h*us

us is given to be = 0.350

h = 0.035 / 0.350

h = 0.100 m

Is this correct?

So the waitress could not push any higher than 0.100 m or 10 cm above the table top without tipping the glass.
 
  • #6
Looks good.
 
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What is the maximum height a waitress can push on a glass without it tipping?

The maximum height a waitress can push on a glass without it tipping depends on several factors, including the size and shape of the glass, the weight of the drink, and the strength and technique of the waitress. As a general rule, it is recommended to not push a glass higher than the height of your shoulder.

What are the potential risks of pushing a glass to its maximum height?

Pushing a glass to its maximum height can result in several risks, including the glass tipping over and causing spills or injuries, the glass breaking, and the contents of the glass splashing onto the waitress or customers. It is important to use caution and proper technique when pushing glasses to their maximum height.

Are there any techniques for pushing a glass to its maximum height without it tipping?

Yes, there are several techniques that can help a waitress push a glass to its maximum height without it tipping. These include using a firm grip on the glass, keeping the glass close to the body, and using a steady and controlled pushing motion instead of jerking or pushing too quickly.

Can the maximum height a waitress can push on a glass be increased with practice?

Yes, with practice and proper technique, a waitress may be able to increase the maximum height they can push on a glass without it tipping. It is important to start with smaller glasses and lighter drinks and gradually work up to larger glasses and heavier drinks to improve strength and technique.

What should a waitress do if a glass does start to tip while they are pushing it?

If a glass starts to tip while a waitress is pushing it, they should immediately stop pushing and try to steady the glass with their hands. If the glass has already tipped over, it is important to carefully clean up any spills and replace the glass to avoid any potential injuries or accidents.

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