I have the following question (and the beginning of the solution)

enter image description here

I am very confused about signs in this question. When I started solving the question without looking at the answer, I took

$F_{net}=-mgsin\theta + bl\dot \theta $

because the particle is released at an angle $\theta _0$ and therefore its velocity is towards the equilibrium position for the whole motion, and so the drag force is always positive (in the direction of x and theta increasing).

However I look at it, I get that the drag force and the restoring force act in opposite directions in this case when the particle is simply released at rest, so should have opposite signs. I do not understand why in the answer to the question they are given the same signs so that

$F_{net}=-mgsin\theta - bl\dot \theta $

It is causing me some trouble as my choice of sign of the drag force is giving me an $\alpha$ of $-\frac{mg}{bl}$ which would lead to a solution of exponential increase of the angle theta which is clearly not the case!

EDIT: I now understand the signs in the equation here. The velocity and $\dot \theta$ are negative and so using $F_{drag}=-bv$ gives a positive resistive force (as required). However now my confusion just lies in the diagram. I understand that here the arrows are meant to dictate the sign of the force acting (towards equilibrium means you should stick a minus sign in front when using this force in calculations); how would you draw this diagram? My 'free body diagrams' are quite messy I think- I put arrows in and signs in, for example I would draw '$-mgsin\theta $ pointing towards equilibrium and $-bv$ pointing away.

I realise that this is probably seems quite a silly question, but I think I often get the signs wrong because I am not quite sure how to put signs on free body diagrams? How do you use the free body diagrams and is there a resource that goes through some examples?


The drag force does not necessarily act in the direction opposite to that of the returning force, it acts in the direction opposite to that of the velocity of the mass. When the mass is rising, the directions of the returning force and the velocity of the mass do not coincide.

EDIT (12/9/2016): the velocity in the picture (when the mass is going from the right to the center) is directed to the left (that is, it is negative), the drag force is in the opposite direction, so it is positive, which is why they use minus sign in the answer.

  • $\begingroup$ Thank you for your reply- sorry I was not clear. I meant that drag force always acts in the opposite direction to the velocity and in this case because the mass is released from its highest displacement at t=0, the velocity will be in the same direction as the restoring force and so the drag force is in the oppsotite direction to the restoring force, hence the oppostie signs in the expression for the net force. $\endgroup$ – Meep Dec 9 '16 at 18:44
  • $\begingroup$ @21joanna12: Please see the EDIT. $\endgroup$ – akhmeteli Dec 9 '16 at 18:55

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