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

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SUMMARY

The problem involves determining the angle at which an equilateral triangle will begin to topple on an inclined plane with high friction. The correct angle for toppling is 60 degrees, as established through analysis of torque and the center of mass (CoM) of the triangle. The triangle pivots around the vertex in contact with the slope, requiring a horizontal offset of the CoM from this pivot point to generate the necessary torque for toppling. Understanding the geometry of the situation is crucial for solving the problem.

PREREQUISITES
  • Understanding of torque and its calculation (t = r x F)
  • Knowledge of center of mass (CoM) concepts
  • Familiarity with rotational dynamics (t = I(alpha))
  • Basic geometry skills for visualizing angles and offsets
NEXT STEPS
  • Study the principles of torque and equilibrium in rigid bodies
  • Learn about the center of mass and its role in stability and motion
  • Explore the geometry of inclined planes and their effects on objects
  • Practice problems involving rotational dynamics and toppling scenarios
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for illustrative examples of torque and stability in rigid bodies.

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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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