Hello folks, I'm being tricked by the pesky friction force again. I know that static friction acts just to put everything in equilibrium as long as its magnitude doesn't exceed a certain value. That is ok. But when there are other adjustable force such as TENSION and NORMAL, (which in this case DOESN'T have to be related to the friction by f=\mu \times normal) , how do these three reach an agreement? To put an example. I modified a figure (attached here in this post) to illustrate this: We have a rope attached to a wall on one end and attached to a box on the other. The box is resting on an incline with angle theta that has a certain coefficient of static friction. Let's not bother about the value, I'll just assume the friction never exceeds its allowed maximum. There are three unknowns: normal, tension and friction. Granted, there is an inequality for the friction (shouldn't exceed mu times normal) but that isn't helpful to solve the system. However, we can intuitively see that this system HAS to be determinate. the equations I get, assuming that friction goes down the slope (direction could be fixed if this were solvable and we looked at the sign) are: [itex]-T sin(\theta) - mg cos (\theta) + n= 0[/itex] and [itex]-f+Tcos(\theta) -mg sin(\theta) = 0[/itex] \theta, and m are given, so we have three unknowns and two equations. This is not an extended body so torques can't possibly help. What am I missing? I know some textbook problems give an additional (all-important) piece of data: the body is just about to slide down or go uphill, but in this case, I think the tension cannot be adjusted to have two cases, this seems intuitively to have a unique case. BTW, this is no homework problem. I already graduated some years ago... just not practicing in a while. Thanks for any help.