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

AI Thread Summary
An equilateral triangle on an inclined plane will not slide due to high friction but can topple if the slope is steep enough. The critical angle for toppling is determined by the position of the center of mass (CoM) relative to the pivot point, which is one of the triangle's vertices in contact with the slope. For the triangle to begin toppling, the CoM must be horizontally offset from the pivot point, creating a net torque in the toppling direction. The correct angle for this to occur is 60 degrees, as established through geometric reasoning. Understanding the relationship between the incline angle and the CoM offset is essential for solving this problem.
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Homework Statement



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
toppling?
(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

Homework Equations


t = r x F
t = I(alpha)

The Attempt at a Solution


Not sure, I'm completely lost. Maybe something to do with torque?
 
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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?
 
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.
 
SignaturePF said:
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.

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:
So why is it 60 degrees?
 
SignaturePF said:
So why is it 60 degrees?

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

cepheid said:
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.

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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