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

In summary: Okay, so then what do I do with that?Okay, so then what do I do with that?Torque = Force * perpendicular distance to the CoMAnd you know the force, since it's just mg. So you just need to find the perpendicular distance to the CoM for this force. How long is this line?You're close. Remember, you're looking for the perpendicular distance to the center of mass, so you need to find the length of the line from the point of rotation to the line connecting the center of mass and the point of rotation. This length is the perpendicular distance. So, what is the length of this line in terms of a, b, and theta?It's b/2
  • #1
SignaturePF
112
0

Homework Statement


#24
https://aapt.org/physicsteam/2010/upload/2009_F-maSolutions.pdf

Homework Equations


T = rfsinθ


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.
 
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  • #2
Hello SignaturePF!

This seems like a popular problem nowadays :wink::

https://www.physicsforums.com/showthread.php?t=667319

SignaturePF said:

Homework Statement


#24
https://aapt.org/physicsteam/2010/upload/2009_F-maSolutions.pdf

Homework Equations


T = rfsinθ

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?

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?

SignaturePF said:
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.

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
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
SignaturePF said:
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.

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

SignaturePF said:
And what's wrong with my torque equation.
I said that the force was mgcosθ and the moment is b/2.

Okay, that's fine. I misread it.

SignaturePF said:
How can I incorporate a,b?

What's the next step?

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
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
SignaturePF said:
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?

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
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
SignaturePF said:
I said that the force was mgcosθ and the moment is b/2.
I'm probably missing something, but isn't the moment arm [itex]\sqrt{a^2+b^2}/2[/itex]?
 
  • #9
tms said:
I'm probably missing something, but isn't the moment arm [itex]\sqrt{a^2+b^2}/2[/itex]?

Maybe. I'm not paying too close attention, since it's irrelevant to finding the answer.
SignaturePF said:
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.

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

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  • #11
SignaturePF said:
Here's my diagram: Is this what you're saying?

That is, in fact, exactly what I am saying. So what is theta, in terms of a and b (using trigonometry)?
 
  • #12
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
SignaturePF said:
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?

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.
 

Related to 2009 F=MA # 24 (A block tipping down an incline)

1. What is the purpose of the 2009 F=MA #24 experiment?

The purpose of this experiment is to demonstrate the relationship between force, mass, and acceleration, as described by Newton's second law of motion.

2. How is the block tipped down the incline in the 2009 F=MA #24 experiment?

The block is tipped down the incline by applying a force to it, either by pushing it or by allowing it to slide down the incline due to the force of gravity.

3. What factors affect the acceleration of the block in the 2009 F=MA #24 experiment?

The acceleration of the block is affected by the force applied to it, the mass of the block, and the angle of the incline. The force and mass directly affect the acceleration, while the angle of the incline affects the component of the force acting on the block in the direction of motion.

4. How can the acceleration of the block be calculated in the 2009 F=MA #24 experiment?

The acceleration of the block can be calculated by dividing the net force acting on the block by its mass. This can be represented by the equation a = F/m, where a is acceleration, F is force, and m is mass.

5. What real-world applications can be demonstrated using the results of the 2009 F=MA #24 experiment?

The results of this experiment can be applied to various real-world situations, such as understanding the motion of objects on inclined planes, calculating the acceleration of objects in free fall, and designing efficient pulley systems. This experiment also helps to reinforce the concept of Newton's second law and its application in everyday life.

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