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

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Homework Help Overview

The problem involves an equilateral triangle positioned on an inclined plane, where the friction is too high to allow sliding. The question seeks to determine the angle of incline necessary for the triangle to begin toppling down the slope.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the concept of the center of mass (CoM) and its role in the toppling process. There is an exploration of how the CoM must shift in relation to the pivot point for toppling to occur. Some participants express uncertainty about the relationship between the angle of incline and the conditions for toppling.

Discussion Status

The discussion is ongoing, with participants exploring the geometric and physical principles involved in the problem. Some guidance has been offered regarding the importance of torque and the CoM's position, but no consensus or resolution has been reached regarding the specific angle at which toppling occurs.

Contextual Notes

Participants are working within the constraints of homework rules that discourage direct solutions. There is an emphasis on understanding the underlying physics rather than simply arriving at an answer.

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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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