Block Toppling on an incline

In summary, the conversation discusses a block on an incline plane with a coefficient of friction of √3. The question is to find the value of theta at which the block begins to topple. The equations used are torque and equilibrium. The normal force and friction force are discussed, as well as the minimum value of the coefficient of static friction required for the block to topple at a given theta. The conversation also touches on the possibility of the block sliding without friction and the role of the normal force and weight in producing torque.
  • #1
Satvik Pandey
591
12

Homework Statement


A block of base ##10cm \times 10cm## and height ##15cm## is kept on an incline plane. The coefficient of friction between them is ##\sqrt { 3 } ##. The inclination of the incline plain is ##\theta##. Find the value of theta for which the block begins to topple.

Homework Equations


1)##\tau =I\alpha ##
2)For equilibrium ##\sum { \tau } =0##

The Attempt at a Solution



I [/B]tried to make free body diagram of the block.
Toppling.png


As the block begins to topple the I think at that moment point A can be treated as Instantaneous Axis of rotation. Can it be?

When the block begins to topple, does the Normal force on block would shift to point A?

If so then, if I take the moment of forces acting on the block then torque of force ##N## and ##f## about A will be ##0##. But that's not correct.
How should I proceed. Sorry if I am missing something very common.
 
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  • #2
In your post I miss the relevant equations...

First thing that strikes me in the picture is that the block isn't resting on its base...

Your question about the action line of the normal force is interesting; intuitively I'm with you, but not quite certain. So perhaps we can try to make it irrelevant.
 
  • #3
BvU said:
In your post I miss the relevant equations...

First thing that strikes me in the picture is that the block isn't resting on its base...

Your question about the action line of the normal force is interesting; intuitively I'm with you, but not quite certain. So perhaps we can try to make it irrelevant.

I have added two equations in my questions. Please have a look over them.

Could you please tell me how to proceed.
 
  • #4
Satvik Pandey said:
does the Normal force on block would shift to point A?

Very good question. To answer it, consider a more general problem: at which point should the normal force be applied for the block to be in equilibrium on an incline?
 
  • #5
Analyse the situation when the block is raised up a little bit from the slope, so that it touches it with the bottom edge only. Releasing, it either returns to its stable position, or topples over.
In that raised --up position, the normal force acts et the edge.
 
  • #6
Time to write down some of the torques around point A ...
 
  • #7
Satvik Pandey said:
When the block begins to topple, does the Normal force on block would shift to point A?

If so then, if I take the moment of forces acting on the block then torque of force ##N## and ##f## about A will be ##0##. But that's not correct.
Why do you say it is not correct?
 
  • #8
Is the value of ##\theta## 33.69?
 
  • #9
Satvik Pandey said:
Is the value of ##\theta## 33.69?
Yes, except that you had to make an assumption to arrive at that. Think about the given information you have not used.
 
  • #10
haruspex said:
Yes, except that you had to make an assumption to arrive at that. Think about the given information you have not used.

I have not used value of friction because the frictional force passes through the point about which I have calculated torque. How should I include that?
Should I calculate torque about CoM of the cube?
 
  • #11
The coefficient of friction is quite big. What do you think, can the block slide on the slope ?
 
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  • #12
ehild said:
The coefficient of friction is quite big. What do you think, can the block slide on the slope ?

Well that depends on the value of ##\theta##. It will not slide unless the value of ##\theta## is equal to the 60.
 
  • #13
If the question asks us to to find the minimum value of ##\mu## (coefficient of static friction) required for block to topple for given ##\theta## then how should I approach to that question.
If I consider moment of forces about point A(please see figure in #post1) then I would not be able to get an expression involving the ##\mu##. Should I consider torques about the CoM of the box?
 
  • #14
OK, so you found 33.69 degrees, or rather ##\arctan{10\over 15}## as the angle at which the thing topples -- provided the friction force is sufficient.
Since this is smaller than 60 degrees (or rather, ##\arctan\sqrt 3 ##), the block will not slide but topple, so that's OK.

At an angle ##\theta## it will slide when the friction force ## \mu_s mg \cos\theta ## is less than ## mg \sin \theta##, in other words, ## \mu_s \lt \ ?\ ##

If I consider moment of forces about point A(please see figure in #post1) then I would not be able to get an expression involving the ##\mu## . Should I consider torques about the CoM of the box?
Not necessary: all you want there is that it doesn't slide and that is about forces, not torques.
 
  • #15
Satvik Pandey said:
If I consider moment of forces about point A(please see figure in #post1) then I would not be able to get an expression involving the ##\mu##.
Quite so, but you're trying to answer whether it would slide, so you need to look at linear forces, not moments.
 
  • #16
BvU said:
the thing topples -- provided the friction force is sufficient

Why is the latter required? As far as I can see, if the block topples, it shall do so even if friction is absent.
 
  • #17
voko said:
Why is the latter required? As far as I can see, if the block topples, it shall do so even if friction is absent.
Without friction, it will slide down the plane. What happens when it gets to the bottom we don't know. Maybe it slides right off, so won't topple.
 
  • #18
@voko: strong point. I imagine a gradual increase of theta until ... etc. There is no solid base for that perception, I agree. However, with ##\theta > \arctan{10\over 15}## and no friction, where can the torque to rotate around the c.o.m. come from ?

Now what about post 14 ? Makes things murky for me: no toppling unless ##\theta > \arctan{10\over 15}## and ##\mu## is irrelevant then ? Doesn't feel good.

[edit]Haru is on the same line I think.
 
  • #19
haruspex said:
Without friction, it will slide down the plane.

Even without friction, the normal force and the weight can produce a non-zero couple when the angle is sufficiently great.
 
  • #20
Don't see a rotation around the c.o.m. coming up ...
 
  • #21
BvU said:
Don't see a rotation around the c.o.m. coming up ...

Can you answer the question I asked in #4?
 
  • #22
That's indeed the key one. What's the line of action for the normal force. My idea: any normal force that would cause a rotation is immediately offset by a shift of the line of action. If N acts at Satvik's point A, that represents a clockwise torque (in combination with ##mg\cos\theta\, ##). All that torque can do is shift the line of action of N upwards.

Would love to trade this gut feeling for a thorough bit of physics, though ... :)
 
  • #23
In equilibrium, the sum of all moments is zero. At some point X within the base of the block, the normal force is exerted. Taking moments about X, the moments of the normal force and friction are zero. So the moment of gravity must also be zero. Clearly this is possible only when X is at the intersection of the vertical line from the C. o. M. with the base (this is a well known "toppling" stability criterion). Equally clear is that at certain angles this is no longer possible, because the vertical line from the C. o. M. will be outside the base, regardless of friction.
 
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  • #24
Toppling.png

I found the answer in this way

Finding moments of forces about point A

Block start toppling if

##7.5mg sin\theta=5mg cos\theta##

or ##tan\theta=\frac{5}{7.5}##

In this there was no use of friction force.

So can we say that even if the friction force is zero the block will topple.

Voko is this what you were saying in #post 19?
 
  • #25
What will happen if ##\theta >60##. Will it topple and slide together?
 
  • #26
voko said:
Even without friction, the normal force and the weight can produce a non-zero couple when the angle is sufficiently great.
Sure, but if it's on the verge of toppling then the normal force acts through the lowest point of the block (A). Its moment about the c.o.m. opposes toppling, so as BvU says there is no moment about the c.o.m. to cause toppling. If you are taking moments about A then bear in mind that a moment is required to account for the acceleration down the plane. Toppling will only occur if the moment that way exceeds that required by the acceleration.
Once sliding, I believe the critical value of ##\mu_s## for toppling is block length / block height.
 
  • #27
Satvik Pandey said:
So can we say that even if the friction force is zero the block will topple.
Please see my post #26.
 
  • #28
Satvik Pandey said:
What will happen if ##\theta >60##. Will it topple and slide together?
If ##\tan(\theta) > \mu_k > \mu_s > \frac{block length}{block height}## then it will topple while sliding.
Correction: I have k and s crossed over there.
 
Last edited:
  • #29
haruspex said:
Sure, but if it's on the verge of toppling then the normal force acts through the lowest point of the block (A). Its moment about the c.o.m. opposes toppling, so as BvU says there is no moment about the c.o.m. to cause toppling.

Then that means that the block will not topple without friction because friction was providing a torque for rotation.
haruspex said:
If you are taking moments about A then bear in mind that a moment is required to account for the acceleration down the plane. Toppling will only occur if the moment that way exceeds that required by the acceleration.
Once sliding, I believe the critical value of ##\mu_s## for toppling is block length / block height.

Are you talking about the torque due to pseudo force?
 
  • #30
haruspex said:
If ##\tan(\theta) > \mu_k > \mu_s > \frac{block length}{block height}## then it will topple while sliding.

From the figure in #post24.
Finding torque about the com

## N\frac{l}{2}=\mu N \frac{h}{2}##

So ##\mu=\frac{l}{h}## (Let ##h## and ##l## be the height and length of the cube)

Did you find the value of ##\mu## in this way?
How did you come to know that at this value of ##\mu## the block will topple and slide simultaneously.
 
  • #31
Satvik Pandey said:
Then that means that the block will not topple without friction because friction was providing a torque for rotation.
That is the conclusion BvU and I have come to.
Are you talking about the torque due to pseudo force?
Maybe. Never been one for pseudo forces, but as I understand it the pseudo force associated with the linear acceleration down the slope would be equal and opposite to the resultant of the actual forces. Similarly, a pseudo torque. The pseudo force here has a pseudo torque about A. I believe adding that into the torque equation about A will show there's no toppling without friction.
 
  • #32
haruspex said:
Maybe. Never been one for pseudo forces, but as I understand it the pseudo force associated with the linear acceleration down the slope would be equal and opposite to the resultant of the actual forces. Similarly, a pseudo torque. The pseudo force here has a pseudo torque about A. I believe adding that into the torque equation about A will show there's no toppling without friction.

I have a confusion:

If we choose our axis of rotation at the com of the body then we do not include pseudo force even if the c.o.m is accelerating. Right?

But if we choose our axis of rotation (other than c.o.m) on the body and if that point the body(through which the axis is passing) is accelerating then we have to consider pseudo force.

Where does the the pseudo force act on the body? Does it act on the c.o.m of the body?

Please reply to my #post30.:D
 
  • #33
Satvik Pandey said:
From the figure in #post24.
Finding torque about the com

## N\frac{l}{2} = \mu N \frac{h}{2}##

So ##\mu=\frac{l}{h}## (Let ##h## and ##l## be the height and length of the cube)

Did you find the value of ##\mu## in this way?
Yes
How did you come to know that at this value of ##\mu## the block will topple and slide simultaneously.
As you've just shown, if it is sliding then the toppling criterion is ##\mu_k > \frac{l}{h}##. (I had k and s crossed over before. I often do that. 'k' feels like 'stuck', while 's' feels like 'slipping'.) In order to slide we must have ##\tan(\theta) > \mu_s##. And necessarily ##\mu_s >= \mu_k##.
 
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  • #34
Satvik Pandey said:
Where does the the pseudo force act on the body? Does it act on the c.o.m of the body?
The pseudo force is the equal and opposite of ma, not just as a vector but in all respects, so necessarily it acts through the c.o.m.
 
  • #35
haruspex said:
And necessarily ##\mu_s >= \mu_k##.


Why should ##\mu_s >= \mu_k##?
Sorry if I am missing some thing obvious?
 
<h2>What is block toppling on an incline?</h2><p>Block toppling on an incline is a phenomenon in which a block or object placed on an inclined surface starts to slide or topple due to the influence of gravity.</p><h2>What factors affect block toppling on an incline?</h2><p>The factors that affect block toppling on an incline include the angle of the incline, the mass and shape of the block, the coefficient of friction between the block and the incline, and the force of gravity.</p><h2>How is the angle of the incline related to block toppling?</h2><p>The steeper the incline, the greater the force of gravity acting on the block, which increases the likelihood of toppling. As the angle of the incline decreases, the force of gravity decreases and the block is less likely to topple.</p><h2>What is the role of friction in block toppling on an incline?</h2><p>Friction between the block and the incline can either help prevent or cause toppling. If the coefficient of friction is high, it can provide enough resistance to keep the block from sliding or toppling. However, if the coefficient of friction is low, it can contribute to the block's movement and increase the likelihood of toppling.</p><h2>What are some real-world applications of studying block toppling on an incline?</h2><p>Understanding block toppling on an incline is important in fields such as engineering and construction, where it is necessary to consider the stability of objects placed on inclined surfaces. It is also relevant in sports and recreational activities, such as rock climbing, where the angle of the incline can affect the safety and difficulty of the activity.</p>

What is block toppling on an incline?

Block toppling on an incline is a phenomenon in which a block or object placed on an inclined surface starts to slide or topple due to the influence of gravity.

What factors affect block toppling on an incline?

The factors that affect block toppling on an incline include the angle of the incline, the mass and shape of the block, the coefficient of friction between the block and the incline, and the force of gravity.

How is the angle of the incline related to block toppling?

The steeper the incline, the greater the force of gravity acting on the block, which increases the likelihood of toppling. As the angle of the incline decreases, the force of gravity decreases and the block is less likely to topple.

What is the role of friction in block toppling on an incline?

Friction between the block and the incline can either help prevent or cause toppling. If the coefficient of friction is high, it can provide enough resistance to keep the block from sliding or toppling. However, if the coefficient of friction is low, it can contribute to the block's movement and increase the likelihood of toppling.

What are some real-world applications of studying block toppling on an incline?

Understanding block toppling on an incline is important in fields such as engineering and construction, where it is necessary to consider the stability of objects placed on inclined surfaces. It is also relevant in sports and recreational activities, such as rock climbing, where the angle of the incline can affect the safety and difficulty of the activity.

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