I'm a private Physics tutor, and I'm a bit stumped by one of my student's problems. It's #3 on this worksheet (http://www.hopewell.k12.pa.us/Downloads/Inclined%20Plane%20Worksheet.pdf -- not his school), but I can't find the answer key anywhere. The teacher didn't hand out an answer key either, so I don't know if he knows the answer either.
In this problem, we have a block with a mass m = 33.2 kg at rest on an inclined plane ($\theta = 31.5^{\circ}$). Static friction is present ($\mu = 0.214$), and the block is attached to a cable that is fixed at the other end. We are asked to find the tension ($F_T$) in the cable.
The question is, what direction is the friction force ($F_{fr}$) pointing? I know that it should be pointing against the direction of natural motion, but what is it here? In the free-body diagram, is the tension force or horizontal gravity component dominating? If you do Newton's second law (slanted coordinate system), you get
y-direction: $F_N - mg \cos \theta = 0$
x-direction: $mg \sin \theta - F_T \pm F_{fr} = 0$
$F_{fr}$ itself is positive, I put the $\pm$ sign in there to show my question. The gravity component is
$mg \sin \theta = (9.8)(33.2)(\sin[31.5^{\circ}])$ = 170.0 N
The magnitude of the friction is
$F_{fr} = \mu mg \cos \theta = (0.214)(9.8)(33.2)(\cos[31.5^{\circ}])$ = 59.4 N
So $F_T$ = 170.0 N $\pm$ 59.4 N = 229.4 N or 110.6 N. How do you know which one to pick? Is there some sort of law that says the tension is always minimized?
This isn't even accounting for the fact that the real definition of static friction is $F_{fr} \le \mu F_N$, so $F_{fr}$ could be less than 59.4 N. How would that change things?
EDIT: corrected one of the numbers