Homework Help: 2009 F=MA # 24 (A block tipping down an incline)

1. Jan 29, 2013

SignaturePF

1. The problem statement, all variables and given/known data
#24

2. Relevant equations
T = rfsinθ

3. The attempt at a solution
FBD:
Axes: (along the incline is x), perpendicular to that is y:
Down: mgcosθ
Up: N
Right: mgsinθ
Left: Ff (friction)

If the box is to tip over, it will rotate around one vertex, namely the bottom right.
We choose that as PoR and do net torque.
(b/2)mgcosθ = Torque, isn't that the only torque?
Or is it:
b/2mgcosθ = aμmgcosθ

If the box isn't slipping down the plane, then
Ff = mgsinθ
μmgcosθ = mgsinθ
tanθ = μ

I don't really know what to do beyond that.

2. Jan 29, 2013

cepheid

Staff Emeritus
Hello SignaturePF!

This seems like a popular problem nowadays :

Yeah, the only torque is from gravity. mgcosθ is a force, not a torque, so you need to multiply this by the correct lever arm. But as I hinted to the other poster in the thread I linked to above, what is more important, for toppling, than the magnitude of the torque, is its direction. Look at the picture given in your exam. What do you notice about the vertical line along which the force acts (which is called the "line of action" of the force)? Where is it located in relation to the PoR? It's offset to the left of it. Given that, what is the direction of the torque that it produces? I.e. in what direction does it want to make the box rotate? Is this the toppling direction? What must be true about the line of action of the force, relative to the PoR, for the box to begin toppling?

Just a minor nitpick. If the box ISN'T slipping, then this means that there is MORE friction than there is force trying to pull the thing down the plane. So it should be:

Ff > mgsinθ
μmgcosθ > mgsinθ
tanθ < μ

3. Jan 29, 2013

SignaturePF

Ahh I see what you're saying. What I need to do is adjust the position of the force such that its moment of action hits or is to the right of the PoR.

How can I incorporate a,b?
And what's wrong with my torque equation.
I said that the force was mgcosθ and the moment is b/2.

What's the next step?

4. Jan 29, 2013

cepheid

Staff Emeritus
Yeah, and you do this by making the hill steeper.

Okay, that's fine. I misread it.

Suppose that you increased theta until the line of action of the force passed right through the PoR. This is the critical point. Any steeper and it will topple. What is the relationship between a, b, and θ at this point? You need to draw a picture and look at the geometry.

5. Jan 29, 2013

SignaturePF

I drew a picture, I just don't see the relationship.
Here's what I was thinking:
The square is at a right angle to the ramp, so no use of theta there. I thought of drawing diagonals/parallel lines, couldn't find how to relate theta either. Could you give me a hint?

6. Jan 29, 2013

cepheid

Staff Emeritus
When you're at the critical angle, the force passes through the CoM and the PoR. Consider a line connecting these two points. What is the angle between this line and the wall of the box? Hint: this line is vertical, and the wall of the box is perpendicular to the incline. You know that theta is the angle between the horizontal and a line parallel to the incline. To get from the horizontal to the vertical, you just rotate by 90 degrees. To get from parallel to the incline to perpendicular to the incline, you also just rotate by 90 degrees. So, if you just rotated both lines by 90 degrees, has the angle between them changed?

7. Jan 29, 2013

SignaturePF

No, the angle has not changed. that angle is still theta. Ok so L, where L is the length we have discussed, is simply theta degrees above the vertical. So Lcostheta = a and Lsintheta = b
Putting that back into the equation
Lsintheta/2*mgcostheta = Torque.

8. Jan 29, 2013

tms

I'm probably missing something, but isn't the moment arm $\sqrt{a^2+b^2}/2$?

9. Jan 29, 2013

cepheid

Staff Emeritus
Maybe. I'm not paying too close attention, since it's irrelevant to finding the answer.

No, much simpler than this. Just geometry. You have a line joining the CoM and the PoR. It's at an angle of theta from the wall of the box. So, it forms a right triangle with a section of the side of the box of length ____, and a section parallel to the bottom of the box of length ____. Based on this right triangle, you have a relation between theta, a, and b.

10. Jan 29, 2013

SignaturePF

Here's my diagram: Is this what you're saying?

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11. Jan 29, 2013

cepheid

Staff Emeritus
That is, in fact, exactly what I am saying. So what is theta, in terms of a and b (using trigonometry)?

12. Jan 29, 2013

SignaturePF

tanθ = b/a

Ahh, now I can solve the problem.

I found before that Ff > mgsinθ
umgcosθ > mgsinθ
u > tanθ
u > b/a

This is correct.
Thanks so much. Not only were you EXTREMELY helpful, but you helped me in a very timely fashion. Your help is greatly appreciated.

Ok so the main concepts in the problem are that
1) If the block isn't to slip
Ff > x-component force
2) Finding a simple ratio of sides of a,b using trigonometry at a critical point where the torque causes toppling in the INTENDED direction.
3)Substitution of the ratio into the expression from 1)

PS. If you're willing, could you take a look at my other questions?

13. Jan 29, 2013

cepheid

Staff Emeritus
You're very welcome. I'm glad that I could help. Yeah, the main concept is that as you increase θ from 0° to 90°, you increase tanθ from 0 to infinity. If tanθ > b/a, the block will tip over. If tanθ > μ, the block will slip. So, if you want the block to tip first, before slipping, it must be true that you reach the tipping angle before you reach the slipping angle. Therefore it must be true that b/a < μ

I have time to look at maybe one or two more threads.