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No friction needed a phenomena I can't understand

  1. Jul 4, 2011 #1


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    http://img820.imageshack.us/img820/8610/triangler.jpg [Broken]

    I know that if I could apply force F to meet in a single point with the normal force and the tension, there would be no need for friction for equilibrium. Why? What's the physical explanation to this phenomena?
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jul 4, 2011 #2
    is it not called equilibrium?

    I mean, when you stand straight up on your own feet, your weight is counter by a normal reaction from the ground and there are no horizontal (friction) components, either.

    If you now stand inclined, leaning against the wall, there will not be friction on the wall, simply a force normal to it, keeping you from falling over...BUT, there will be a horizontal (friction) component at the bottom of your shoes, that will keep you from sliding over, and keeps you in equilibrium....in your diagram, the stick could represent you leaning against the wall and the point where the stick and the rope meet, is the bottom of your shoes and it is the rope that is providing both components to balance you..the one the carries you vertically (and hence, there is no need for frictional component along the wall) and the one pushing you against the wall and just countering the normal force from the wall...again, you are in equilibrium.
  4. Jul 4, 2011 #3
    I am not sure what is meant by 'no need for friction' either. I presume the rope is attached at C and that the stick is merely resting against the wall at B. In that case there is a coefficient of static friction required at B. The coefficient entails that as you increase the downward force at F it increases the total friction at B, even though the coefficient remains the same.
  5. Jul 4, 2011 #4
    I see.

    I think it would help a lot if you drew all the forces involved and their components.

    While it seems that we need a frictional component along the wall...I think a system like this could be possible but you really need to include the weight of the stick and place it right there in the middle, if you will, this will help in balancing the system...otherwise, if the stick is weightless and all you have is a weight F hanging from the end, the horizontal component of the countering force from the rope will simply push the stick flat against the wall as the top tip slides upwards against a friction-less surface....or, if the weight of the stick is toooo large compare to all other forces, it will work its way down (temporarily pushing point A away from the wall) to rest, again, flat against the wall.
  6. Jul 5, 2011 #5


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    Well, here's the issue. In my problem, they tell me that F = 20 [N] and that there's friction and I must calculate it.

    I found out that friction coefficient = 0.7

    In the next question they ask where must the force F act so the beam stays in equilibrium, even if the wall is completely smooth. I must calculate Nb and T in this case.

    It's kind of a weird question. I don't know how am I supposed to know it.

    I saw this is what they did in the solution:

    http://img199.imageshack.us/img199/6651/forcestriangle.jpg [Broken]

    i.e. if they build a force triangle, there's no friction! But, in actuality, I can draw F wherever I want and not just draw Fs, and therefor there will be equilibrium if I use the statics equations. So I completely don't understand why this solution is necessarily the correct one. It just depends on the concept of not writing friction force, not on where I place F!

    Don't you find it too a bit odd?
    Last edited by a moderator: May 5, 2017
  7. Jul 5, 2011 #6
    I think this has to do with what I was saying before...if you now have a friction-less wall, you had to include the weight of the stick...

    ...in your problem, the stick seems to remain mass-less, but they are allowing you to move F to any point along the stick.

    Needless to say, you can always find the reactions at B and C so that the system is static, i.e., it does not move in the horizontal or vertical directions...

    ...what you need to do now is find the position of F along the stick so that when there is no friction at B, the forces that tend to make the stick rotate also balance out...

    The forces that make the stick rotate clockwise are the normal at B (horizontally to the right) and the reaction from the rope at A (horizontally to the left).

    The forces that make the stick rotate counterclockwise are F (down) and the reaction from the rope at A (up).

    I think that should do it.
  8. Jul 5, 2011 #7

    Philip Wood

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    This calls for you to use a very simple and beautiful theorem about an object acted upon by just three coplanar, non-parallel forces. The theorem says that for the body to be in equilibrium all three forces (or their lines of action) must pass through a single point. But where does this theorem come from?

    Any 2 of the three lines of action must pass through some point. [Just try drawing non-parallel lines on a big piece of paper!] Call this point 'P'. Now the Principle of Moments says that (for coplanar forces) the sum of the clockwise moments about any point must equal the sum of the anticlockwise moments about that point. So we can choose the point to be P. We know that neither of the first two forces has any moment about P, because they both pass through that point. So if the object is in equilibrium, the third force must also pass through P, or it would give rise to an unbalance moment about P.
  9. Jul 5, 2011 #8


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    This makes a lot more sense than the other replies! Thank you :smile: Though, I appreciate the other replies.
  10. Jul 5, 2011 #9
    You should note that this is an unstable equilibrium configuration. If the end of the stick touching the wall is bumped slightly up or down, or the downward force F is jostled slightly left or right, the whole contraption collapses (without friction on the wall).
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