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Homework Help: F = MA Exam 2012 #3 - (Triangle toppling down a plane)

  1. Dec 28, 2012 #1
    1. The problem statement, all variables and given/known data

    3. An equilateral triangle is sitting on an inclined plane. Friction is too high for it to slide under any circumstance,
    but if the plane is sloped enough it can “topple” down the hill. What angle incline is necessary for it to start
    (A) 30 degrees
    (B) 45 degrees
    (C) 60 degrees ← CORRECT
    (D) It will topple at any angle more than zero
    (E) It can never topple if it cannot slide

    2. Relevant equations
    t = r x F
    t = I(alpha)

    3. The attempt at a solution
    Not sure, I'm completely lost. Maybe something to do with torque?
  2. jcsd
  3. Dec 28, 2012 #2


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    Staff: Mentor

    Consider where the center of mass of the triangle is located. If an object is sitting on one of its sides, what has to happen to the center of mass in order for it to tip over?
  4. Dec 28, 2012 #3
    I think that originally, the CoM is in the geometric center of the equilateral triangle. Are you suggesting that for the triangle to topple, the CoM must relocate to one side? I'm not seeing where to go from here.
  5. Dec 28, 2012 #4


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    If the thing is going to topple over, what this really means is that it rotates around some pivot point (which is a point where the object makes contact with the surface). In this case, the pivot point is one of the two vertices of the triangle that is in contact with the slope (the "downhill" one). If this rotation is to occur, there must be a net torque around that pivot point. This torque comes from gravity, but in order for it to be non-zero, the CoM (which is where gravity acts) must be horizontally offset from the pivot point in such a way as to make the torque be in the toppling direction. The size and direction of this offset depends on how inclined the plane is. Draw a picture. :wink:
    Last edited: Dec 28, 2012
  6. Dec 28, 2012 #5
    So why is it 60 degrees?
  7. Dec 28, 2012 #6


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    Well, we aren't going to do your homework for you. I told you the relevant physics already:

    From this point, it's just geometry: draw a picture and work it out. Figure out what inclination angle is the critical one (where the torque switches to being in the correct direction to topple the object).

    To gain some intuition for the problem: draw one picture with the triangle on a very shallow slope. Draw another with the triangle on a very steep slope. What is the difference between the two (in terms of HOW the CoM is offset from the pivot point)?
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