How can I determine the instant the block overcomes static friction?

  • #1
Thermofox
144
26
Homework Statement
A sphere is connected via a spring, attached to its center, to a block. Initially the block and the sphere are resting on a rough plane, then a constant force F=20N is applied on the sphere. This causes the sphere to start a pure rolling motion.

I need to determine the angular velocity and how much does the spring moves from its resting point when the block starts moving. What I've difficulty with is finding this moment.

ms=3Kg ; mb= 2Kg
R= 30cm
k= 200N/m
μs= 0.5 ; μd= 0.3
Relevant Equations
Fs= -kx
ΣF= ma ; Στ= Iα
I know that the block will move only if the force that pushes him is greater than μs
Screenshot 2024-06-10 163319.png
mb g.

The only force that can act on the block is the elastic force (Fs) generated by the expansion of the spring, caused by the rolling sphere, that rolls because a force F is acting on the sphere.

Then can I say that Fs > μs mb g? I've determined this equation by analysing the forces acting on the block and found that:

Fs- Fsf ( static friction force) = 0. I've considered the right direction as positive and put the equation equal to zero because the block is not moving => it has no acceleration.

I don't know how to proceed forward from this point. How can I find the instant, when the block starts moving, if my condition is greater than something? Because all I can find now is that, assuming x is how much did the spring expanded, x >(μs mb g)/k.
 
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  • #2
As I see it, you might not need to find the time at which it starts moving (though that can be done considering the SHM of the sphere up to that point).
Try writing an energy balance equation as well as the force balance equation.
 
  • #3
Energy balance equation:

energy when block starts moving - Initial energy = work done by friction
=> mbvb2/2 + Icω2/2 + msvs2/2 + kx2/2 - 0 = 0

Force balance equation of the sphere:
F + Ff,s - Fsp= msac
Ns= msg
τF + τFf,s - τFsp = Ic α

Force balance equation of the block:
Nb= mbg
Fsp- Ff,b = 0

I don't know if I made any mistakes.
 
  • #4
Let x(t) represent the displacement of the sphere at time t. Assuming the sphere does not slip relative to the ground, the angular velocity of the sphere is $$\omega(t)=\frac{1}{R}\frac{dx}{dt}$$and its angular acceleration is $$\alpha=\frac{1}{R}\frac{d^2x}{dt^2}$$Let f be the frictional force of the ground on the sphere. What is the moment balance on the sphere?
 
  • #5
You have a coefficient of static friction (I believe), so you can determine the displacement of the spring (center of sphere) just prior to onset of motion from the sum of forces on the block. Then utilize Work-Energy to find ##x(t)## over that period. Don't forget ( as you appear to have done so in post #3 ) that the constant force ##F=20 ~\text{N} ## is doing work on the sphere over displacement ##x## and the spring is doing negative work on the sphere. Also, over this period ask yourself if the block is moving?

Finally, see LaTeX Guide to format mathematics here. I think you'll learn it in very little time.
 
  • #6
erobz said:
You have a coefficient of static friction (I believe), so you can determine the displacement of the spring (center of sphere) just prior to onset of motion from the sum of forces on the block. Then utilize Work-Energy to find ##x(t)## over that period. Don't forget ( as you appear to have done so in post #3 ) that the constant force ##F=20 ~\text{N} ## is doing work on the sphere over displacement ##x## and the spring is doing negative work on the sphere. Also, over this period ask yourself if the block is moving?

Finally, see LaTeX Guide to format mathematics here. I think you'll learn it in very little time.
Hi, thanks for answering!
I need to find the displacement of the spring right after the block has started moving. Therefore I don't understand why I should determine the displacement of the spring before the block moves. As for the energetic balance you are right, I forgot that ##F## does work. Then the equation should be $$...= W_F - W_{Fsp}?$$ Moreover can I say that the displacement ##x## is equal to the displacement of the spring? I don't understand how I should determine ##x(t)##, could you elaborate a bit more?

Btw you were right, learning LaTeX is fun.
 
  • #7
Chestermiller said:
Let x(t) represent the displacement of the sphere at time t. Assuming the sphere does not slip relative to the ground, the angular velocity of the sphere is $$\omega(t)=\frac{1}{R}\frac{dx}{dt}$$and its angular acceleration is $$\alpha=\frac{1}{R}\frac{d^2x}{dt^2}$$Let f be the frictional force of the ground on the sphere. What is the moment balance on the sphere?
##τ_F + τ_f - τ_{Fsp} = I_c α?##
 
  • #8
Thermofox said:
##τ_F + τ_f - τ_{Fsp} = I_c α?##
$$fR=I_c\alpha$$
 
  • #9
Thermofox said:
Hi, thanks for answering!
I need to find the displacement of the spring right after the block has started moving. Therefore I don't understand why I should determine the displacement of the spring before the block moves. As for the energetic balance you are right, I forgot that ##F## does work. Then the equation should be $$...= W_F - W_{Fsp}?$$ Moreover can I say that the displacement ##x## is equal to the displacement of the spring? I don't understand how I should determine ##x(t)##, could you elaborate a bit more?

Btw you were right, learning LaTeX is fun.
Maybe I'm not seeing the correct question. I thought the system is at rest, the wheel begins to roll via application of ##F##, the spring begins to stretch, until static friction can no longer hold the connected block of mass ##m_b## from accelerating. I thought the question was how long that would take?

If this isn't it, then I've mis-interpreted and not sure what is being asked.
 
  • #10
Chestermiller said:
$$fR=I_c\alpha$$
Yes, because ##τ_F## and ##τ_{Fsp}## are on the axis of rotation so we only have the moment caused by the friction force, ##fR##
 
  • #11
Thermofox said:
Then the equation should be $$...= W_F - W_{Fsp}?$$ Moreover can I say that the displacement ##x## is equal to the displacement of the spring? I don't understand how I should determine ##x(t)##, could you elaborate a bit more?
We'll whatever is the case the work done by ##F## is responsible for the potential on the spring, and the kinetic energies of the sphere and block.

I was thinking that it was going to lead to a first order ODE with ##\frac{dx}{dt} = \cdots ##, but I guess I have the wrong problem.
 
  • #12
erobz said:
Maybe I'm not seeing the correct question. I thought the system is at rest, the wheel begins to roll via application of ##F##, the spring begins to stretch, until static friction can no longer hold the connected block of mass ##m_b##. I thought the question was how long that would take?

If this isn't it, then I've mis-interpreted and not sure what is being asked.
The situation you described is what its happening. I'm so sorry, I probably used wrong words to describe what the problem is asking. The question is: when the block starts moving how stretched is the spring?
 
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  • #13
What do the two coefficients of friction represent?
 
  • #14
erobz said:
What do the two coefficients of friction represent?
##μs= 0.5## static coefficient of friction , ##μd= 0.3## dynamic coefficient of friction
 
  • #15
Thermofox said:
##μs= 0.5## static coefficient of friction , ##μd= 0.3## dynamic coefficient of friction
I guess you have to figure out whether or not the force ##F## is large enough to free the block, otherwise the sphere is a red herring...

I think you should do as @haruspex suggested and correctly write Work-Energy as a place to start.
 
  • #16
erobz said:
I guess you have to figure out whether or not the force ##F## is large enough to free the block, otherwise the sphere is a red herring...

I think you should do as @haruspex suggested and correctly write Work-Energy as a place to start.
Isn't the one I wrote correct?

mbvb2/2 + Icω2/2 + msvs2/2 + kx2/2 - 0 = ##W_F##
 
  • #17
Thermofox said:
Homework Statement: A sphere is connected via a spring, attached to its center, to a block. Initially the block and the sphere are resting on a rough plane, then a constant force F=20N is applied on the sphere. This causes the sphere to start a pure rolling motion.

I need to determine the angular velocity and how much does the spring moves from its resting point when the block starts moving. What I've difficulty with is finding this moment.

ms=3Kg ; mb= 2Kg
R= 30cm
k= 200N/m
μs= 0.5 ; μd= 0.3
Relevant Equations: Fs= -kx
ΣF= ma ; Στ= Iα

I know that the block will move only if the force that pushes him is greater than μs View attachment 346805mb g.

The only force that can act on the block is the elastic force (Fs) generated by the expansion of the spring, caused by the rolling sphere, that rolls because a force F is acting on the sphere.

Then can I say that Fs > μs mb g? I've determined this equation by analysing the forces acting on the block and found that:

Fs- Fsf ( static friction force) = 0. I've considered the right direction as positive and put the equation equal to zero because the block is not moving => it has no acceleration.

I don't know how to proceed forward from this point. How can I find the instant, when the block starts moving, if my condition is greater than something? Because all I can find now is that, assuming x is how much did the spring expanded, x >(μs mb g)/k.
-This would be difficult if we jerked the wheel and set it oscillating but the statement is that the motion is pure rolling so the answer is very simple. By now much must the spring be stretched to overcome the "stiction" of the block Am I missing something here? $$F=\mu_s m_b g$$ this means $$x\gt \frac {\mu_s m_b g}k$$ The OP has already found this answer. Am I missing something?
 
  • #18
hutchphd said:
-This would be difficult if we jerked the wheel and set it oscillating but the statement is that the motion is pure rolling so the answer is very simple. By now much must the spring be stretched to overcome the "stiction" of the block Am I missing something here? $$F=\mu_s m_b g$$ this means $$x\gt \frac {\mu_s m_b g}k$$ The OP has already found this answer. Am I missing something?
I'm not sure, my concern was that the force in general could be not large enough to dislodge the mass from its rest position?
 
  • #19
What part of the physics says that??
 
  • #20
Thermofox said:
Isn't the one I wrote correct?

mbvb2/2 + Icω2/2 + msvs2/2 + kx2/2 - 0 = ##W_F##
Is the block moving?
 
  • #21
hutchphd said:
What part of the physics says that??
I don't know @hutchphd , we have half of a problem statement. No need to be upset with me...
 
  • #22
We had an OP statement asking for the time which made no sense, therefore the responses were in kind, but still nonsense. Hence my confusion.
 
  • #23
hutchphd said:
I don't understand the point.
I thought maybe it was a "trick" question. Otherwise the rolling wheel is a red herring.
 
  • #24
hutchphd said:
-This would be difficult if we jerked the wheel and set it oscillating but the statement is that the motion is pure rolling so the answer is very simple. By now much must the spring be stretched to overcome the "stiction" of the block Am I missing something here? $$F=\mu_s m_b g$$ this means $$x\gt \frac {\mu_s m_b g}k$$ The OP has already found this answer. Am I missing something?
I didn't find how stretched the spring is the moment right after the block started moving. I found a range of displacements not a value.
 
  • #25
erobz said:
Is the block moving?
Yes
 
  • #26
Thermofox said:
Yes
Gives us the entire problem statement. Thats counter to what I thought we agreed the actual problem was...
 
  • #27
Can you provide the calculation? I don't see that there will be a range of valuesw
 
  • #28
erobz said:
Gives us the entire problem statement. Thats counter to what I thought we agreed the actual problem was...
Screenshot 2024-06-12 180243.png
 
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  • #29
Thermofox said:
Ok,

In 1) The block is static until the threshold of overcoming static friction is reached. Just before it breaks loose is the maximum deflection of the spring, the block is stationary at that point in the analysis ( i.e it begins its motion with ##v = 0 ##).

In short, the answer I was looking for was No...the "instant" the block breaks free it has no kinetic energy...

Then you can use that deflection and Work-Energy to find the angular velocity in 2).
 
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  • #30
hutchphd said:
Can you provide the calculation? I don't see that there will be a range of valuesw
##μ_s m_b g##= ##9,81N##
##k= 200 N/m##
=> ##x\gt \frac {\mu_s m_b g}k = 0,0491m##, hence ##x\gt 0,0491m##
I didn't found the "x" but a threshold. If x is below that than the block will not move if it is higher it will.
 
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  • #31
yes But the threshold value is a particular value at the low edge of a range. This is merely a semantic quibble.
 
  • #32
erobz said:
Ok,

In 1) The block is static until the threshold of overcoming static friction is reached. Just before it breaks loose is the maximum deflection of the spring, the block is stationary at that point in the analysis ( i.e it begins its motion with ##v = 0 ##).

Then you can use that deflection and Work-Energy to find the angular velocity in 2).
I see the consideration I didn't make:
the displacement of the spring reaches its maximum when the block is about to move. I still don't understand why that's the case, but I managed to solve the problem. Thank you infinitely.
 
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  • #33
Thermofox said:
I still don't understand why that's the case
Feel free to ask about it(what are you unsure of?)! It's more than just solving a particular problem, understanding is the primary goal of your education.
 
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  • #34
erobz said:
Feel free to ask about it(what are you unsure of?)! It's more than just solving a particular problem, understanding is the primary goal of your education.
I understand why the spring has maximum stretch right before the block moves. I don't get why we use this value. Because when the block starts moving this will move by dx, this causes a compression of dx in the spring. Therefore the deflection when the block moves should be ##x_{max}-dx## and not ##x_{max}## as we said.
 
  • #35
Thermofox said:
I see the consideration I didn't make:
the displacement of the spring reaches its maximum when the block is about to move. I still don't understand why that's the case, but I managed to solve the problem. Thank you infinitely.
What then did you get for the angular velocity? What method?
 
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