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B Both Tension and Friction?

  1. Sep 25, 2018 #1
    This is a bit of a general question.

    Say that there's a block on a ramp, with a (massless) rope attaching it to the top of a ramp. If we're given values for the mass of the block, maximum tension, and coefficients of static and kinetic friction, how would we determine the friction or tension?

    My interpretation is that if static friction is enough to counter gravity, then the tension is zero.

    On the other hand, if static friction is not enough, I can't convince myself of what the tension force will equal. Would it be the gravity in the ramp direction minus the maximum static friction? Or, do we have to take into account kinetic friction? In general, I'm really confused :).

    Your help would be appreciated.
     
  2. jcsd
  3. Sep 25, 2018 #2

    jbriggs444

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  4. Sep 25, 2018 #3

    kuruman

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    If the angle is such that the mass on the incline will accelerate downhill starting from rest, then you can consider the following. The downhill acceleration is ##a=g\sin \theta -\mu_k g \cos \theta##. An uphill force ##F_{uphill} = m(g\sin \theta -\mu_k g \cos \theta)## will result in zero acceleration of the mass. This force could be due to any kind force a person pushing from below or a string pulling from above. Arguably, there is no kinetic friction when the block is tied to the string. However, is there a difference in the dynamics of a block sliding down the incline at constant velocity and a block being prevented from accelerating downhill by a string tied to it? The same uphill force is required to have zero acceleration in either case. Is there a hole in my argument?
     
  5. Sep 25, 2018 #4

    Mister T

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    Newton's 2nd law.

    In other words, if the block won't slide when there's no rope attached, then the tension in a rope that is attached would be zero.

    Have you drawn a free-body diagram of the block, which would include an arrow representing each force? Once you do, consider all the components of those forces along a line parallel to the ramp. Compare the sum of all the components directed up the ramp (one of which is the tension force) to the sum of all the components directed down the ramp. Since the block is not moving, the friction force is static.
     
  6. Sep 25, 2018 #5
    @jbriggs444, do you mean that this system is statically indeterminate? I'm still not really sure about what the concept means.

    @kuruman, I agree that the constant velocity, like the sting case, is in mechanical equilibrium. It's just that there's both tension and friction in the string case, as opposed to just friction. I'm mostly confused about the case where without a string, the block would not be in equilibrium (i.e. friction isn't strong enough). In this case, do we think "since this is static friction, there has to be a minimum additional upward force of ____ to ensure that the block has zero acceleration?"
     
  7. Sep 25, 2018 #6

    jbriggs444

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    That depends on the process used to bring the rock to its static configuration. See #2 above.
     
  8. Sep 25, 2018 #7

    jbriggs444

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    It means that the answer is unknown and unknowable. The provided information is not sufficient to determine the answer. The set of equations is under-determined.

    A similar thing occurs with a four-legged, perfectly rigid ideal table with legs just touching a perfectly rigid ideal planar floor. It is not possible to determine the force on each leg. Multiple answers are consistent with the given facts. Three legged tables are easy to solve. Four legged ideal tables are impossible [unless they are uneven so that the table wobbles onto just three legs]

    If you evenly soften the floor just a bit the table situation becomes easy to solve. If you make the string just a bit elastic and specify how the block was brought to rest, the block on plane may become solvable as well.
     
    Last edited: Sep 25, 2018
  9. Sep 25, 2018 #8

    Dale

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    I agree with @jbriggs444 the situation is statically indeterminate.

    Since the friction up the ramp is more than enough to counter gravity then you could have zero tension. But then you could tighten the rope a little bit, and you would need less friction up the ramp. And you could tighten it more and more until you got less friction force up the ramp and then even friction down the ramp.
     
  10. Sep 25, 2018 #9

    kuruman

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    @jbriggs444 and @Dale I a afraid you missed the point of what I was asking. My questions are,

    Given that the block would accelerate downhill from rest if it were not restrained by the string,


    1. How is the string tension needed to keep the block in place different from the force exerted by some other agent required to keep the sliding block at zero acceleration?
    2a. If there is no difference, then why can one not say that the tension in the string is equal to the force exerted by some other agent required to keep the sliding block at zero acceleration?
    2b. If there is a difference, what is it and why?

    Thanks.
     
  11. Sep 25, 2018 #10
    I think I see what you mean. So in the case where static friction is sufficient, if we pull with a force less than the maximum static friction, the tension just 'feeds off of' the static friction force, while still maintaining this zero acceleration.

    How about in the case where static friction isn't sufficient? If the mass will slide when there's no string and the string is stationary, what would we consider the friction and tension forces to be?
     
  12. Sep 25, 2018 #11

    Dale

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    Let’s not derail the OP’s thread. Let this thread focus on his/her question. If you have other questions please open a separate thread.
     
  13. Sep 25, 2018 #12

    Dale

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    It is still statically indeterminate. However in that case the minimum tension is greater than zero.
     
  14. Sep 25, 2018 #13

    kuruman

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    Far be it from me to derail the thread. In post # 1, OP wrote,
    I thought I was addressing exactly this when I posted
    as a preamble before asking my questions, not to derail but to focus on OP's question and present a line of reasoning for getting around the static indeterminacy in this particular instance.
     
  15. Sep 25, 2018 #14

    Dale

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    Oops, my apologies. You are exactly correct. I had missed that part of the OP. Your post was on topic, I should have been more careful.
     
  16. Sep 26, 2018 #15

    jbriggs444

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    This seems confused. If we have an inextensible, breakable string attached to a fixed point then for any non-zero acceleration, the string is irrelevant. It might prevent acceleration, but it cannot otherwise affect acceleration.

    We might be in a situation where the breaking strength of the string by itself is insufficient to prevent slippage and where static friction is also insufficient to prevent slippage but the two taken together are more than enough to prevent slippage. In this case, the situation is still statically indeterminate. We know that the string must be under some tension. We know that there is some upwards force from static friction. But we do not have enough information to know how much force from each is present.

    The situation could be modelled by a set of equations:

    $$F_{breaking} >= F_t >= 0$$
    $$+N\mu _s >= F_s >= -N\mu _s$$
    $$F_t + F_s = F_{needed}$$

    Subject to the givens of this particular sub-case:

    $$F_{breaking} < F_{needed}$$
    $$N\mu _s < F_{needed}$$
    $$F_{breaking} + N\mu _s > F_{needed}$$

    Two unknowns, one equality, five inequalities. Not enough information to solve for both tension and static friction. But enough information to see that a range of solutions exists.
     
    Last edited: Sep 26, 2018
  17. Sep 26, 2018 #16

    kuruman

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    I don't see what's confused about this, but perhaps I was not clear when I said what I said. I am considering one of the possibilities as raised by the OP, namely when static friction is insufficient to keep the block from accelerating if there were no string. Then the acceleration would be ##a=g\sin \theta -\mu_k g \cos \theta##, and an uphill force ##F_{uphill} = m(g\sin \theta -\mu_k g \cos \theta)## would be needed to bring this acceleration to zero. My question then is, in this case and this case only, whether it is fair to say that if a string were attached to the mass and the mass were at rest, the tension would be the same as ##F_{uphill}##. In other words, can one draw the parallel between the dynamic case when the acceleration is zero and the static case? If not, why not? Oh, it is also assumed that the breaking strength of the string is greater than ##mg\sin \theta.##
     
  18. Sep 26, 2018 #17

    jbriggs444

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    No. It is not fair to say this. The tension in the string is indeterminate. It could fall anywhere within a range of values without changing the physical situation or the givens of the problem.

    If the string were just a little bit stretchable and if you had slowly increased the tension (possibly by unwinding it from a spool at a controlled rate) until it was just enough to bring the block to a stop and if you had then immediately stopped the spool, then the tension in the string would be and remain at ##F_{uphill}##.

    Which is perhaps to say that for all realistic purposes, your conjecture is completely correct but that for the idealized problem, it breaks down.
     
  19. Sep 26, 2018 #18

    kuruman

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    I think you helped me answer my own question an OP's initial question. Here is how I would state it.
    The sum of forces in the incline direction is zero, ##f+T=mg \sin \theta## where ##f## is the contact force due the incline. My previous argument was that, by analogy to the sliding case, we can identify ##f## as the kinetic friction, ##\mu_k mg\cos\theta##, in which case what is left over is ##T##. However, I could also have argued that in the static case ##f## is the maximum force of static friction, ##\mu_s mg\cos\theta##, in which case ##T## would be something else.

    So yes, there is a range of values for ##T##. I have one more question, then I'll shut up. Is it fair to say that this range is, ## m(g\sin \theta -\mu_s g \cos \theta)<T< m(g\sin \theta -\mu_k g \cos \theta)##? The assumption, of course, is that the we are still in the situation where angle ##\theta## is greater than the angle of repose.
     
  20. Sep 26, 2018 #19

    Dale

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    No, one cannot draw that parallel because the static case is indeterminate and the dynamic case is not. This is because the static force of friction can take on a range of values whereas the dynamic force of friction has a unique value.
     
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