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

  1. Aug 7, 2017 #1
    1. The problem statement, all variables and given/known data

    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?

    2. Relevant equations

    F net = 0
    Fx = 0
    Fy = 0

    Something with momentum (p), maybe?

    3. The attempt at a solution

    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!
     

    Attached Files:

  2. jcsd
  3. Aug 7, 2017 #2

    TSny

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    Rotation suggests adding to your list of relevant equations.
     
    Last edited: Aug 7, 2017
  4. Aug 7, 2017 #3
    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..
     
  5. Aug 7, 2017 #4

    TSny

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    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.
     
  6. Aug 7, 2017 #5
    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.
     
  7. Aug 7, 2017 #6

    TSny

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    Gold Member

    Looks good.
     
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