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Static friction of masses connected by a rod

  1. Dec 17, 2013 #1
    This isn't homework...I'm reviewing physics after many years of neglect.

    Given 2 masses, [itex]m_1, m_2[/itex], connected by a rigid, massless rod, stationary with respect to a ramp which makes an angle of [itex]\theta[/itex] with the horizontal, with coefficients of static friction between the masses and the ramp = [itex]\mu_{s1}, \mu_{s2}[/itex] respectively, what is the magnitude of the tension or compression in the rod, and what are the magnitudes of the static friction, [itex]f_{s1}, f_{s2}[/itex], acting on each mass?

    This assumes [itex]\theta[/itex] is small enough that the masses do not lose traction, i.e.,

    [tex]\theta \leq \arctan \frac{\mu_{s1} m_1 + \mu_{s2} m_2}{m_1 + m_2}[/tex]

    Note that if the masses are assumed to already be moving, then the problem is straightforward:

    [tex]T = (\mu_{k1} - \mu_{k2})\frac{m_1 m_2}{m_1 + m_2}g\cos\theta[/tex]
    [tex]f_{k1} = \mu_{k1} m_1 g\cos\theta[/tex]
    [tex]f_{k2} = \mu_{k2} m_2 g\cos\theta[/tex]

    Where T is the tension in the rod (T<0 implies compression). Such problems are common in basic physics, e.g., Halliday, Resnick, & Krane, 4th Ed., chap.6, problem 29.

    So I assumed it would be straightforward to do the same problem, but with the masses stationary. But I get 2 equations (sum of the forces for each mass) and 3 unknowns (T, [itex]f_{s1}, f_{s2}[/itex]).

    I find it surprising that such a simple situation is undetermined, and assume I've missed something simple.

    Note: I've looked through several physics texts and can find only problems in which the masses are already moving.

    Also, it's driving me crazy!
     
  2. jcsd
  3. Dec 17, 2013 #2

    tiny-tim

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    hi inkliing! :smile:
    yes, it is undetermined: you need more information, including the compression coefficient of the rod

    (eg if the slope is such that both masses would remain stationary on their own, then, without more information, clearly the compression could be anything, and the static friction forces could be anything ≤ µmg)

    (it's not like the kinetic friction case, where the friction forces are fixed)

    it's like asking the reaction forces for a rod supported in 3 places, or a table supported in 4 places … that's indeterminate unless you know the flexibility of the rod or table :wink:
     
  4. Dec 17, 2013 #3

    AlephZero

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    It is undetermined if the rod is rigid.

    But if the rod is flexible, and nothing moves, you don't need to know the flexibility of the rod. (Think about the words in bold text.)

    You could have a situation where without the rod, one mass would slide and the other would not, but when joined by the rod, they don't slide. In that case, the system is determinate whether or not the rod is flexible, because the static friction force on the mass that wants to slide will be at the maximum value.

    I don't have the book in front of me so I can't read that problem, but the questions in textbooks are (or should be!) carefully invented so there is a unique solution. If you invented this problem yourself, you have gone beyond those restrictions, unless you add some more conditions to make the solution unique
     
  5. Dec 17, 2013 #4

    jbriggs444

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    Agreed. All that would matter would be the rod's compression or tension, not the amount by which that compression or tension would change if the bolded hypothetical were to fail.

    I cannot agree with this. The mass that would otherwise slide could be subject to any amount of static friction less than or equal to the maximum as long as the net force (including the contribution from the rod) is zero.
     
  6. Dec 17, 2013 #5

    tiny-tim

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    i agree :smile:
     
  7. Dec 17, 2013 #6

    AlephZero

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    That would not be true for a flexible rod. One end of the rod would "slide" until the friction force plus the force in the rod was big enough to stop it. At that position, the friction force is equal to the dynamic friction. Once the rod is stationary, there is nothing to make it move again so the forces stay constant.

    That assumes there was no tension in the rod when the masses were put onto the plane, but there is a similar unique solution for any initial tension in the rod. Either one end slides and the above type of argument applies, or nothing moves and the tension in the rod remains at its initial value. The tension in a flexible rod can't change unless its length changes.

    But I take your point, for a rigid rod if the coeff of static friction was greater than the coeff of dynamic friction. In theory there could be a range of forces in the rod that would give equilibrium. But this is hypothetical, since rigid rods don't exist.
     
    Last edited: Dec 17, 2013
  8. Dec 18, 2013 #7
    Thx for the helpful responses.

    It now seems clear to me that, since rigid rods do not exist, the rod's strain determines its tension or compression, leaving 2 equations (sum of forces on the masses) and the 2 unknown static frictions.

    In other words, if the angle in the ramp is small enough such that both masses, when standing alone with no rod connecting them, experience static frictional forces strictly less than their respective maximums, then, in general, at the same angle, when the masses are then connected by a flexible rod, the rod will experience a nonzero tension or compression resulting in a nonzero tensile or compressive strain of the rod. If this strain is known then the system is determined.
     
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