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Finding friction on a onedimensional tube

  1. Jul 6, 2013 #1
    Find the friction coefficient between the bar and the surfaces. Both surfaces are equal (and thus have equal coefficients).

    Ok so, my problem is that, generally in physics when I assume an wrong direction for a vector, I will be able to discover that in my answer (generally speaking the vector will be negative), however, in this question assuming a wrong direction will give me a completely wrong answer and I don't understand why this happens or how to define the right orientation right from the start.

    My original trial:
    [itex]\\\sum F_x = 0 \rightarrow N_a\mu_e + Nb\left(-\mu_e \cos(\theta)-\sin(\theta) \right ) = 0
    \\\sum F_y = 0 \rightarrow N_a + N_b(\cos(\theta)-\mu_e \sin(\theta)) = W
    \\\sum \tau_A = 0 \rightarrow W = \frac{N_b}{\sin(\theta)}
    \mu_e = 0.569345 \text{ or } \mu_e = -1.75641[/itex]

    Book (I suppose it's also the correct one):
    [itex]\\\sum F_x = 0 \rightarrow N_a\mu_e + Nb\left(+^{[1]}\mu_e \cos(\theta)-\sin(\theta) \right ) = 0
    \\\sum F_y = 0 \rightarrow N_a + N_b(\cos(\theta)+^{[2]}\mu_e \sin(\theta)) = W
    \\\sum \tau_A = 0 \rightarrow W = \frac{N_b}{\sin(\theta)}
    \\\text{Only differences in [1] and [2].}
    \mu_e = 0.225979 \text{ or } \mu_e = 4.42518[/itex]
  2. jcsd
  3. Jul 6, 2013 #2


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    The problem arises when you select whether FB is + or - NBμecos(θ). As soon as you choose which, you have specified which way that force acts in relation to FA.
    You can determine the direction easily from first principles. Friction always acts to oppose relative motion of the surfaces in contact. If it acts downwards at the top of the rod that must be because, in the absence of friction, the rod would slide upwards. But since the rod is of fixed length, that would mean the base of the rod would slide towards the vertical face. To get a valid alternative solution you must reverse both friction directions. I would guess that will then give you the correct solution but with the coefficient negative.
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