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Centre of mass: Toppling

  1. Mar 21, 2009 #1
    1. The problem statement, all variables and given/known data
    A uniform square metal plate of side 10 cm has a square of side 4cm cut away from one corner.

    (a) Find the position of the centre of mass of the remaining plate.

    It is now placed on a rough inclined plane as shown in the diagram. The place is inclined to the horizontal at an angle Θ.

    (b) Calculate the maximum possible angle of inclination of the plane for the lamina to be in equilibrium.

    Diagram:
    http://img23.imageshack.us/img23/7048/p2103091901small.th.jpg [Broken]

    2. Relevant equations

    x bar = [tex]\frac{\sum mx}{\sum m}[/tex]

    3. The attempt at a solution
    I managed to do part a) but I do not know how to do part b)

    a)
    x bar = (100*5 - 16*2)/84
    x bar = 5.57
    y bar = (100*5 - 16*2)/84
    y bar = 5.57
    [tex]\sqrt{10^{2}+10^{2}}[/tex]-[tex]\sqrt{5.57^{2}+5.57^{2}}[/tex] = 6.26 cm

    b)
    I have no idea how to work out the maximum angle if the lamina is not rectangular. Haven't been taught and can't think of where I would start. Can anyone teach me the steps in order to work it out?

    Thanks in advance!
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 21, 2009 #2

    lanedance

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    Hi Ulti

    The plate will start to topple when a line drawn vertically through the centre of mass falls outside the base support

    This is equivalent to calculating the net torque on the plate due to the gravitational & reaction forces. When a line drawn vertically down through the centre of mass falls inside the base support, the net torque will sum to zero (ie it is in static equilibrium).

    When a line drawn vertically through the centre of mass falls outside the base support, the net torque is no longer zero and the plate is unstable & will move.
     
    Last edited: Mar 21, 2009
  4. Mar 21, 2009 #3
    Thanks for the answer, I've done the rest of the questions which is the same but with a rectangular lamina and I just drew a rectangle from it and used the inverse of tan to calculate it.

    I'm not too sure on how to calculate the moments (I assume you mean that by "torque") in this kind of situation... Am I missing something very fundamental here?
     
  5. Mar 21, 2009 #4

    LowlyPion

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    You don't need to calculate the torque. It is enough to know the angle of inclination tolerated by observing what angle a line from the down incline corner of the base through the center of mass (x,y) makes with the vertical.

    The center of mass requires a support along the line that is in the direction of gravity.
     
  6. Mar 21, 2009 #5
    If I draw a line from the centre of mass to the direction of gravity I still don't see how I can work out the angle. I believe I can use trigonometry somewhere but I fail to see where I can apply it...
     
  7. Mar 21, 2009 #6

    LowlyPion

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    What is the coordinate of the base corner that would be furthest down the incline?

    Then what is the coordinate of the Center of Mass?

    The difference in the x coordinate divided by the y-height of the center of mass is the tan-1 of the angle.
     
  8. Mar 21, 2009 #7
    That would be (0, 4)

    (5.57, 5.57)

    The difference in the x coordinate divided by the y-height of the center of mass is the tan-1 of the angle.[/QUOTE]

    tan-1 [tex]\frac{5.57-4}{5.57}[/tex] = 16 = Right answer!

    Thanks! One thing I still don't understand is how this gave me the right answer though...
     
  9. Mar 21, 2009 #8

    lanedance

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    hi Ulti

    what don?t you understand? if you re-read the posts I think its been explained (i made some corrections to my earlier one wher i had some typos as well)

    the angle is calculated when the centre of mass is directly above the left base corner, which is what you calculated...
     
    Last edited: Mar 21, 2009
  10. Mar 21, 2009 #9

    LowlyPion

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    Since you seemed resistant to figuring the angle with the object rotated in the horizontal plane with gravity vertical, I had you calculate it based on rotating gravity and viewing the object in the plane of the incline with the line that the center of mass acts through passing through the left hand corner.
     
  11. Mar 23, 2009 #10
    Ah I understand now! Thanks for all the help! Now I can do the rest of this exercise with ease. Thanks once again!
     
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