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I want to solve a simple mechanical problem whereby three forces act upon an object.

  • $F_1 = (10 - x)$ N
  • $F_2 = -3 N$ (friction force)
  • $F_3 = -10x$ N

So at $x$ = 0 we see that $F_1$ + $F_2$ + $F_3$ = 7 N and so the object will accelerate. I want to know at what distance $d$ it will stop. To do so, I think I can write the static conservation of energy, that is, the sum of the works done by all my forces equal 0.

  • $W_1 = 10d - \frac{d^2}{2}$ ($\int_0^d F_1 dx$)
  • $W_2 = -3d$
  • $W_3 = \frac{(-10d^2)}{2}$

So: $W1 + W2 + W3 = 0$ leads to $d(\frac{-11d}{2} + 7) = 0$ and I can find two solutions:

  • $d = 0$
  • $d = \frac{14}{11}$

$d = 0$ makes sense because no work at all is done by any forces initially, and $d = 14/11$ means, I guess, that the object will stop at $x = 14/11$. But when I calculate the sum of the forces at $x = 14/11$ I find that F equals -7 N!

Which clearly implies that the system is not stopped and that the object will now move backward. But by thinking about it I realized that then my friction force should change direction (so it should change sign, from -3 N to +3 N), and I guess that's why my reasoning is flawed. But how can I account for the friction forces in a static conservation of energy equation then?

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  • $\begingroup$ The friction force won't change sign, but it will change magnitude. If your object is moving to the right, friction will always act toward the left until the object stops. Friction would only change sign if the object starts moving the other way (after it has come to rest). Note that the friction force is actually the maximum of -3 and -(F1 + F3). Otherwise, you could have a net external force of only 2N, and the object would start to accelerate due to friction, which is impossible. $\endgroup$ – Nuclear Wang Jul 30 '15 at 9:18
  • $\begingroup$ I have been thinking about your answer and I think the real question is: can we use a static approach to solve a problem involving friction forces that should equal 0 when the system is put to rest (and as you said, oppose the direction of any movement with an absolute value of 3N)? $\endgroup$ – Loic Jul 30 '15 at 9:26
  • $\begingroup$ Hi Loic and welcome to Physics.SE! Please see this help post to learn how to write your equations in a way nicer way i.e. in $\LaTeX$, in order to improve legibility. Thanks! $\endgroup$ – Gonenc Jul 30 '15 at 9:49
  • $\begingroup$ If there is friction, how is there conservation of energy? $\endgroup$ – ja72 Jul 30 '15 at 13:47
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Typically, the Friction force will be proportional to the velocity of the object it affects (or at least it is usually assumed to be proportional). The Force you describe is constant and pointing in negative x-direction.

The situation you are describing resembles an object on a Hookean spring in a gravity field. At first, F1 pulls it upward (positive x) far stronger than the 'gravity' force F2, so it accelerates. Once it passes the equilibrium point of F1 (x=10 in your case), both forces point to negative x ('downwards'). The mass is decelerated and comes to a halt at some certain height. But this does not mean it will stop there indefinitely, the force acting on the object there is still present and will accelerate the object further 'downwards'.

You could introduce a friction force like this: $F2' = -A \cdot \frac{dx}{dt} = -A \cdot v$ So the direction of the 'friction' always counteracts the momentary movement of the object.

Using your force F1 and this new F2', the equation of motion will describe a damped harmonic oscillator - you can find the complete derivation on Wikipedia

--> So, even with friction in place - in this form at least - the object will never come to a final halt. Either it oscillates with decreasing amplitude, or it approaches its final position - but only reaches it after an infinite time.

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  • $\begingroup$ Thank you very much! So my basic definition of the friction force was flawed. I guess my system should really behave like a damped oscillator. I will look in this direction to solve my problem :). $\endgroup$ – Loic Jul 30 '15 at 9:43
  • $\begingroup$ Hi Patrick Kraus and welcome to Physics.SE! Please see this help post to learn how to write your equations in a way nicer way i.e. in $\LaTeX$, in order to improve legibility. Thanks! $\endgroup$ – Gonenc Jul 30 '15 at 9:49

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