# Forces and Newton's Laws of tension

1. Sep 18, 2008

### lmbiango

A 204-kg log is pulled up a ramp by means of a rope that is parallel to the surface of the ramp. The ramp is inclined at 29.6 ° with respect to the horizontal. The coefficient of kinetic friction between the log and the ramp is 0.825, and the log has an acceleration of 0.734 m/s2. Find the tension in the rope.

I drew a beautiful free body diagram like the one in my book that is kind of like this problem, but I can't show you that. Anyway;

The sum of the forces in the x direction = ma in the x direction
and The sum of the forces in the y direction = 0

So, sigmaFx = (204kg)(0.734m/s^2) = mg(sin(29.6)) - 0.825 + T

T = 149.736/(204 x 9.8 x sin(29.6) - 0.825)
T = 0.1518 N

Keep in mind when reading this, my teacher is a total douche and won't explain what "sum of the forces" even means so I am getting what I can out of the book but obviously I am not doing it right because the homework thing is telling me the answer is wrong.

2. Sep 18, 2008

### hage567

Well I'll ignore the childish "total douche" comment about your teacher to tell you that 0.825 is the coefficient of kinetic friction, it is NOT a force on its own. So having the 0.825 term in your equation is meaningless. What is the term for frictional force? Also, you must sum up the forces in the y direction as well. You will need that information to figure out the frictional force.

By the way, I think "sum of the forces" is pretty self-explanatory.

3. Sep 19, 2008

### lmbiango

I think it is fair to call him whatever I'd like to after he literally said I was "stupid and needed to go back to high school algebra" when I asked him a question about kinetics. It is known at my school that you avoid this guy, but I am stuck with him. I'm not stupid, I have a 3.9 actually- but I am a bio major, not math. I just don't get this stuff that is why I spend many hours on this board asking for help to preserve my GPA instead of just giving up because I can't get help anywhere else. We don't even have tutors....

OK anyway back to physics.

The point of my comment was to say that I don't know the first thing about forces. I don't see how you can just add up all the forces you are given in the problem and end up with the right answer. There has to be certain forces you are supposed to add up, right? I am not sure what they are.

The thing I want to find is tension which is in the x direction the way I drew my axis, and although I am sure you are right, I don't understand why I need to solve for the y direction also. The only force I even see in the y direction is the weight of the log- so that would make the equation: (204)(9.8) = 0 . I'm lost.

4. Sep 19, 2008

### hage567

You need to consider ALL the forces on the log. Do you know how to draw a free body diagram? Start with that and put everything on it. I will tell you there are four forces present. Do you know what they are?
The reason you need to consider the y direction (so the one perpendicular to the surface of the incline in this case) is because the frictional force (look this up) depends on the normal force, which is in the y direction. There is also a component of the weight in this direction. So sum them up and solve for the normal force. Once you have that, you can solve your equation for the x direction that you got by summing up the forces in that direction.

Maybe you should try to find a tutor on your own. An upper-year student or something? I'm sure there's someone there who can help.

Anyway, this site might help illustrate some of the stuff I've said: http://hyperphysics.phy-astr.gsu.edu/hbase/N2st.html#c2

5. Sep 22, 2008

### lmbiango

Ok, no luck so far on the tutor- I tried to understand and execute what you said, and it made pretty good sense, but then I got an answer that seemed way off, and sure enough, it was as the homework providor tells me it is incorrect, but here is what I did:

$$\Sigma$$Fy = MAy = N - Mg cos (29.6)
Since there is no acceleration in the y direction I got:

0 = N - 204(9.8) cos(29.6)
N = 1738.29 N

(This already seems really wrong to me)

So then the coefficient of friction...

fk = $$\mu$$kN
fk = (.825)(1738.29)
fk = 1434.09 N

And the X direction to find tension, because that is what I need:

$$\Sigma$$Fx = max = T- fk - Mg sin (29.6)

204(.734) = T - 1434.09 - 204(9.8) sin(29.6)

So, T = 596.338 N

But of course, that is incorrect. I still don't really understand what I am doing here. I am just taking every force that exists and summing it or subtracting it depending on direction and setting it equal to either "ma" or zero. I feel like units matter here and what forces are used matter. I don't think it's ok to just grab whatever forces they give you in that particular problem and use them in the equation. It just doesn't seem like it is going to work like that, and it isn't. Any help in explaining this and helping me figure out the problem would be greatly appreciated.

6. Sep 22, 2008

### hage567

You did everything right up until the last line!
Try solving for T again being really careful of the signs on each term.
Yes, that's the whole point! It's how you mathematically describe what is going on.
You must consider every force that is present because they all influence how the object behaves under those conditions. You are not just grabbing at forces, the trick is to realize which ones are present in the situation of the problem and interpret them correctly. It takes practice.