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
Best Answer
The answer you have found is in fact the correct one, including the range of values; the question does not give enough information to be any more exact.
Consider that you are pulling this block up the slope, and decide that you need to take a break. You know that gravity is exerting $170\text{ N}$ down the slope at all times, so you lower the tension you're exerting on the rope to that value (there's a convenient spring scale incorporated into the rope). Nothing happens, so you lower the force a bit more. Again, nothing happens, so you know that static friction is taking up some of the load. You keep on reducing the up-slope force you exert on the rope, until, at about $110\text{ N}$ applied force, the block begins to slide down-slope. Friction is doing all it can to prevent down-slope slide, and you've reached the most reduced force rest you're going to get.
After a while, you continue up the hill. You pull up harder and harder on the rope, but for a while nothing happens. You are assuming more of the up-slope force, as static friction contributes less and less. You pass $170\text{ N}$, and still nothing happens. Static friction is now acting down the slope, adding to the down-slope gravity force. Finally, at about $230\text{ N}$ applied up-force, friction reaches its limit in helping gravity, and you continue your up-hill slog...
At any time during your stopping, resting, and resuming your journey, the conditions would meet those of Part (3), and any of the forces of tension between $110\text{ N}$ and $230\text{ N}$ could be observed...