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Friction Questions.

  • Thread starter SheldonG
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  • #1
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Homework Statement



A block slides with constant velocity down an inclined plane with slope angle [itex] \theta [/itex]. The block is then projected up the same plane with an initial speed [itex] v_o [/itex]. How far up the plane will it move before coming to rest, and after coming to rest, will it slide down the plane again?

Homework Equations


F = ma.
[tex] f = \mu mg [/tex]

The Attempt at a Solution



The block will travel a distance of [tex] v^2_0/(4g\sin\theta) [/tex]. But the second part of this problem is not clear to me. I am inclined to say there isn't enough information to answer this, but if pressed, I would say that it would probably slide back down. I would say this because the coefficient of kinetic friction in this problem is [tex] \tan\theta [/tex]. If the box would not move, then it would seem that [tex] \mu_s = \mu_k [/tex], which contradicts the usual situation where [tex] \mu_s > \mu_k [/tex]. The book answer is no.

I would appreciate any insights offered.

Thank you,
Sheldon
 
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Answers and Replies

  • #2
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If the box would not move, then it would seem that [tex] \mu_s = \mu_k [/tex],
I can see no reason for that. Since the kinetic friction exactly cancels gravity and static friction will be bigger.
 
  • #3
Hootenanny
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I would say this because the coefficient of kinetic friction in this problem is [tex] \tan\theta [/tex]. If the box would not move, then it would seem that [tex] \mu_s = \mu_k [/tex], which contradicts the usual situation where [tex] \mu_s > \mu_k [/tex]. The book answer is no.
If the box isn't moving then there is no kinetic friction, only static friction. What is of vital importance is the following information,
A block slides with constant velocity down an inclined plane.
Which means, as you correctly say, that:

[tex]\mu_k = \tan\theta[/tex]

Now once the box has stopped, there are only two forces acting parallel to the incline: static friction and a component of the block's weight. We know that the maximum static frictional force is given by,

[tex]f_s = \mu_sN[/tex]

And as you correctly say we know that [itex]\mu_s>\mu_k[/itex], which means,

[tex]\mu_s > \tan\theta[/tex]

Now let's multiply both sides by [itex]mg\cos\theta[/itex] (i.e. the component of the block's weight normal to the incline),

[tex]\mu_smg\cos\theta > mg\sin\theta[/tex]

Notice that the LHS is the maximum static frictional force and the RHS is the component of the block's weight acting parallel to the incline.

Does that help?
 
  • #4
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If the box isn't moving then there is no kinetic friction, only static friction. What is of vital importance is the following information,

Which means, as you correctly say, that:

[tex]\mu_k = \tan\theta[/tex]

Now once the box has stopped, there are only two forces acting parallel to the incline: static friction and a component of the block's weight. We know that the maximum static frictional force is given by,

[tex]f_s = \mu_sN[/tex]

And as you correctly say we know that [itex]\mu_s>\mu_k[/itex], which means,

[tex]\mu_s > \tan\theta[/tex]

Now let's multiply both sides by [itex]mg\cos\theta[/itex] (i.e. the component of the block's weight normal to the incline),

[tex]\mu_smg\cos\theta > mg\sin\theta[/tex]

Notice that the LHS is the maximum static frictional force and the RHS is the component of the block's weight acting parallel to the incline.

Does that help?
I'm still a bit confused, unless we are both saying the same thing in a different way. Since the block travels at constant velocity down the incline, we have

[tex] m(0) = mg\sin\theta - \mu_k mg\cos\theta [/tex]

which leads to
[tex] \mu_k = \tan\theta [/tex]

If the block stops and does not slide back down:

[tex] 0 = mg\sin\theta - \mu_s mg\cos\theta [/tex]

which gives
[tex] \mu_k = \mu_s = \tan\theta [/tex]

Unless I have mucked up my force equations (very possible), then the correct answer to the question is not "the block will not slide" as the textbook states, but the block will slide, since we need [tex] \mu_k > \tan\theta [/tex] so a smaller angle would be required to keep the block from sliding back.

Or perhaps, "not enough information" to decide would be the right answer.

Where am I going astray here?
 
  • #5
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I can see no reason for that. Since the kinetic friction exactly cancels gravity and static friction will be bigger.
Why assume this? The problem doesn't say what happened prior to the constant velocity of the block down the incline. Perhaps the block was placed on the incline, wouldn't move, and then someone gave it a push. On the other hand, perhaps the block was placed on the incline, wouldn't move, and someone increased the angle until it slid at constant v down the incline. In this second case, if you projected the block up the incline again, it should eventually slide back.

I don't see any way from the information given in the problem to conclude that the block must stop. If you actually calculate the coefficients from the information in the problem, you end up with the same value for both of them, unless I made a mistake (which I admit is very possible).
 
  • #6
Gokul43201
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If the block stops and does not slide back down:

[tex] 0 = mg\sin\theta - \mu_s mg\cos\theta [/tex]
That's where the error is.

[itex]\mu _sN[/itex] is not the force of static friction, it is the maximum force of static friction.

i.e., [itex]F_{stat} \leq \mu_sN[/itex]
 
  • #7
Doc Al
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I'm still a bit confused, unless we are both saying the same thing in a different way. Since the block travels at constant velocity down the incline, we have

[tex] m(0) = mg\sin\theta - \mu_k mg\cos\theta [/tex]

which leads to
[tex] \mu_k = \tan\theta [/tex]
OK. Now you know [itex]\mu_k[/itex]--what can you immediately deduce about [itex]\mu_s[/itex]?

If the block stops and does not slide back down:

[tex] 0 = mg\sin\theta - \mu_s mg\cos\theta [/tex]

which gives
[tex] \mu_k = \mu_s = \tan\theta [/tex]
What it tells you is that the minimum value of [itex]\mu_s[/itex] to prevent sliding is equal to what you just found for [itex]\mu_k[/itex]. Is that criterion met?

Unless I have mucked up my force equations (very possible), then the correct answer to the question is not "the block will not slide" as the textbook states, but the block will slide, since we need [tex] \mu_k > \tan\theta [/tex] so a smaller angle would be required to keep the block from sliding back.

Or perhaps, "not enough information" to decide would be the right answer.
You've got all the info you need to answer the question, but you also need to know a basic fact about how maximum static friction compares to kinetic friction.
 
  • #8
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OK. Now you know [itex]\mu_k[/itex]--what can you immediately deduce about [itex]\mu_s[/itex]?


What it tells you is that the minimum value of [itex]\mu_s[/itex] to prevent sliding is equal to what you just found for [itex]\mu_k[/itex]. Is that criterion met?


You've got all the info you need to answer the question, but you also need to know a basic fact about how maximum static friction compares to kinetic friction.
I understand now, thank you so much. I see my mistake now. I was thinking of the situation where the incline was increased to overcome the static friction and cause the block to slide at constant velocity. But after further calculation, I see that this is an impossibility, such an angle does not exist. Any increase in the angle great enough to overcome static friction would cause the block to accelerate. You would have to lower the angle during the slide to get constant velocity, which would of course cause static friction to hold it still when it stopped.

You have all been very patient and so helpful. Thank you very much.
 

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