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I have difficulty knowing the direction of friction on static structures. The structure of the image is on equilibrium, and the exercise askes the maximum force P that the structure can take. Would friction force be to the right or left on point B ? why?

If the roll was free to rotate (without that tension), I think the friction force direction on B would depend of the application point of force P. Furthermore, on dynamics we can discover if our assumption is correct. Here in statics, to solve the problem correctly usually we have to take the correct directions of friction.

In this case, since force P is creating a positive moment, we would need something to cancel it, in order to be static. So, friction force would be to the left.

Is this the right thought? Or do I have to think what would be the direction of friction if that roll was free to rotate, and that would be the same direction to this problem?

  • $\begingroup$ First, what direction do you think it is, and why? That information helps us understand what your conceptual difficulty may be. Otherwise, it's just a homework problem and you won't get an answer. $\endgroup$
    – Bill N
    Jun 23, 2017 at 22:30
  • $\begingroup$ @BillN I understand, and I agree, but at this point I am so confused that I am mixing statics with dynamics, and I'm thinking if that roll would rotate freely the friction direction consideration would be different than here on statics... I don't know... but i'll try to edit my answer. $\endgroup$ Jun 23, 2017 at 22:46
  • $\begingroup$ @BillN I think I did expose my problem, because sometimes I got more than 1 "theory" to understand things, and then it becomes hard to see what's the correct one. I hope you help me. $\endgroup$ Jun 23, 2017 at 22:51

2 Answers 2


In no case, we need to guess the true direction of friction a priori. The laws of Physics - which at least in this domain can always be expressed mathematically - will always predict for you the correct direction of friction. I will illustrate this in the case of your question by assuming the friction to be in the $+x$ and the $-x$ direction respectively and getting that the friction should be in the $-x$ direction via both the methods.

Method 1: Let's assume $f$ is in the $+x$ direction:

We already know that the tension must be along the $+x$ direction. Therefore, the equilibrium in the horizontal will force us to write $$T+f=0$$ As you can see, $f$ is negative - implying that friction is actually in the $-x$ direction.

Method 2: Let's assume $f$ is in the $-x$ direction:

We already know that the tension must be along the $+x$ direction. Therefore, the equilibrium in the horizontal will force us to write $$T-f=0$$ As you can see, $f$ is positive - implying that friction is actually in the $-x$ direction.

From one of the equations above (as per your method), you can express $f$ in terms of $T$ (or vice versa) and solve for it in terms of $P$ using the rotational equilibrium condition.

  • $\begingroup$ ok, thank you very much. I thought we had to guarantee the equilibrium by cancelling the moments. $\endgroup$ Jun 24, 2017 at 10:43
  • $\begingroup$ "actually in the right direction." Do you mean "right" as in "correct" or "right" as opposed to "left?" I think it should be "correct" or "left". Right? :) $\endgroup$
    – Bill N
    Jun 24, 2017 at 14:57
  • $\begingroup$ @BillN I should have written "left" instead of "right" in the conclusion of Method 2. Anyway I am replacing left-right with -ve X and +ve X respectively. Also, I am replacing right with true when I mean so. Thanks for pointing out ambiguities and an error. $\endgroup$
    – user87745
    Jun 24, 2017 at 15:27
  • $\begingroup$ @VitorAguiar68 I had some ambiguities and errors in my previous version of answer. I have corrected them. I hope it makes things clearer. $\endgroup$
    – user87745
    Jun 24, 2017 at 15:32
  • $\begingroup$ @Dvij ok, thanks. By the way, I found an exercise where I can show you that not always we can demonstrate the right direction of friction, and we have to tell for the correct direction. physics.stackexchange.com/questions/341173/… $\endgroup$ Jun 24, 2017 at 16:21

If the object is in equilibrium, the net force has to be zero. Think about the net force in the x-direction. If the net force in the x-direction is zero, what direction should friction point?

  • $\begingroup$ to the left. But imagine a free cylinder where you have a force P to the right, above center of gravity, and you have friction, and you want it static. If you take the net force of x-direction, you have $F=P$, but the cylinder will move anyway because the moments aren't cancelled. $\endgroup$ Jun 23, 2017 at 23:24
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    $\begingroup$ @VitorAguiar68 I agree! It's not possible for the system to be in equilibrium in the situation you described. It will move, no matter how much friction there is. That shouldn't bother you. Obviously not EVERY combination of forces can result in an equilibrium, otherwise nothing would ever move... $\endgroup$ Jun 24, 2017 at 4:02

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