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Incline and Friction direction

  1. May 29, 2017 #1
    1. The problem statement, all variables and given/known data
    It's Problem 3 of this page:
    http://www.problemsphysics.com/forces/inclined_planes_problems.html

    "A box of mass M = 10 Kg rests on a 35° inclined plane with the horizontal. A string is used to keep the box in equilibrium. The string makes an angle of 25 ° with the inclined plane. The coefficient of friction between the box and the inclined plane is 0.3."

    2. Relevant equations
    See link.

    3. The attempt at a solution
    I don't have a problem understanding most of it, but there is one thing that left me thinking: how do we know, in advance (if we can know it), where the direction of the Friction force points?

    The thing is, the solved problem has Friction force (Fs) pointing downwards (left), but as far as I can tell we could have just as rightfully assumed that it pointed upwards (right). In other words, on the X axis we could have two possible situations:

    1. Like they did: Wx and Fs pointing left; Tx pointing right. So at the equilibrium it becomes (chosen +left, -right):
    Wx + Fs - Tx = 0

    2. Like I tried doing before checking their solution: Wx pointing left, Tx and Fs pointing right. Then we have (chosen +left, -right):
    Wx - Fs - Tx = 0

    I tried to think about it, but only managed to come up with two things. First, if we knew how much Wx and Tx are, then we can certainly know that, to keep the equilibrium, Fs must "assist" the "weaker" one of them, and therefore must be on the "weaker one's side". In other words, if Wx was +3, and Tx was -5, then Fs must be +2 (pointing left, like Wx); viceversa, if Wx was +5, and Tx was -3, then Fs must be -2 (pointing right).

    But then I realized that we don't know Tx when this decision should be made (when setting the equations). So how do we make the decision?

    The best I could come up with was this: do both methods, and then double-check if there are any contradictions in either one. If there are, discard that one.

    So I went on and did my thing. Here's what I got by assuming that Fs pointed up.

    Equilibrium on X (+left, -right):
    +Wx - Fs - Tx = 0

    Equilibrium on Y (+down, -up):
    +Wy - N - Ty = 0
    => N = Wy - Ty
    => N = mg cos (35°) - T*sin (25°)
    => N = 80.36 - T*0.42

    Back to the first equation:
    +Wx - Fs - Tx = 0
    => Wx = Fs + Tx
    => mg sin (35°) = N*μ + T*cos (25°)
    => 56.27 = (80.36 - T*0.42) * 0.3 + T*0.91
    => (...) => T = 41 N

    So Tx = 37.2 N

    Now Fs:
    +Wx - Fs - Tx = 0
    => Fs = Wx - Tx => Fs = +19 N

    Alternatively:
    N = Wy - Ty => N = 63 Newtons
    So Fs = N*μ => Fs = +19 N again

    So to recap the set of data I got:
    W = 98.1
    Wx = 56.3
    Wy = 80.4

    T = 41
    Tx = 37.2
    Ty = 17.3

    Fs = 19
    N = 63

    X axis: Wx - Fs - Tx =? 0 => 56.3-19-37.2 = 0.1 => seems good (aside from approximations)
    Y axis: Wy - N - Ty =? 0 => 80.4-63-17.3 = 0.1 => close enough again

    Is there a contradiction somewhere that I'm missing, or are there actually two possible ways to solve this problem?
     
  2. jcsd
  3. May 29, 2017 #2

    BvU

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    Gold Member

    I think you are doing just fine: the book solution assumes Fs is pointing to the right, without any motivation.
    You worked out an alternative assumption (I didn't check the numbers), which is fine too.

    The weakness in the exercise statement is in the wording 'to keep the box in equilibrium' . I think any tension in the range from your value to the book value achieves that result. My point is that the friction coefficient helps find the maximum friction force, but the actual friction force can be anything between + and - ##\mu F_N## (in this configuration with different ##F_N## for each string tension ##T## ).
     
  4. May 29, 2017 #3

    ehild

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    @BvU is right, the tension can be anything between those both values. The problem does not specify if it asks the minimum or maximum tension. The minimum tension prevents the block sliding down the slope, and the box would not start to move upward along the slope till the tension is less than the maximum value.
     
  5. May 29, 2017 #4
    Alright, thanks guys.
     
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