Physics(Tension)

Hi!

Im having trouble answering this tension problem.
(I know my little drawing is retarded but its the best I can do )

-x-
| |
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0A0B

x=pulley
0=mass
| & - =string


Ok, if mass A is 30 and mass B is 60, how would I find the tension and the acceleration.

I know Wa=MaG (G=9.8m/s/s) and Wb=MbG
Ahh, I just dont kno!! Im pretty good at inclined plane problems with tension and a mass, but am totally lost on this.

Well, I happened to stop in here, so let me tell ya:

It looks like your strings are pointed straight down. Okay, that's good. So, there is only one thing which could happen here. No matter what, the system is going to be "rightward", meaning that MassB is going to go down while MassA goes up.

What you have to do is set up a free-body diagram and label ALL of the forces acting on each mass. (I'm assuming that you are not worried about any of the pulley's characteristics, at this point. That comes later on in your learning).

First, let's analyze MassA's FBD :

^
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Ma
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V

So, as you can see, there are two forces acting on MassA. One is an upward force (label this T1) and the other is the gravitational force (label this Fg1). Fg1=(30kg)(9.8m/s^2). [Note that I have assumed the masses to be weighted in kilograms. If not, change the units around a bit]. So, where did this come from? It cames from Newton's Second Law--namely: Force=Mass*Acceleration. This is an amazing principle and no doubt helped you a lot in your inclined plane problems.

So, we should now analyze MassB's FreeBodyDiagram:

^
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Mb
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V

Again, we see two forces. The upward force will be labeled T2 (for good purposes, DO NOT ASSUME that T1=T2). Also, the downward force is, once again, a gravitational acceleration force. Meaning, Fg2=(60kg)*(9.8m/s^2).
=-=-=-=-=
Now, we are going to use Newton's 2nd extensively...although we do not have to break it up into y-axis and x-axis parts for this problem as you may have down before. [Note: If you did not do this for inclined plane problems, tell me and I will go over it for you, because your teacher is leading you down a hard path, then].

Anyway, since this is a MOVing system, it isn't quite as easy as it might be, but let's see what we can do:

First, restate Newton's Second Law: The NET Force acting on an object is equal to its Mass times the acceleration. (we will have to jump back and forth from one side to the other to complete this, since acceleration is not equal to zero).
[note: we have chosen down to be positive and up to be negative since our system is headed downward. It's just a simple difference between a negative sign].

Okay, So we can see for System A that:

F=m*a
Fg1-T1=m1*a1 (where a1 is the acceleration of m1 (mA))
294kg-T1=(30kg)*acceleration (obviously, this is not the value for gravity, or else it would have fallen 10 meters in the first second! The tension involved makes it go much more slowly).

So, a1=(294-T1)/30

Now, doing the same for MassB (m1) :

Fg2-T2=m2*a2
588-T2=60*a2
therefore, a2=(588-T2)/60

Now, here's where a very elementary (yet often overlooked) understanding of the world comes into play. How fast do you think MassA is moving? Hmm...what do you think is its acceleration? Heh, well, although you can't tell me yet, is it easy to see that the velocities and accelerations of both masses are equal? Good :0) Since the string is relatively inelastic, there is no difference in the rate the first edge of the string goes over the pulley than the rate of the second edge of the string.

So, now we can say: a1=a2=a

Taking our two equations above, what do we get?

(294-T1)/30=(588-T2)/60

I've got to go to church, now, but try this out and maybe I'll be able to get back on, later.


Buen Suerte!

Wow, thanks alot Davida. I understood everything you said until the last part, where you combined the two equations. Where do I go from there, how do I find T and A? My teacher doesn't explain things very well . Although she taught us vector components, which helps solves the inclined plane problems. We just finished dynamics and are just about done with statics (torque,centre of gravity).

Hehe, well, that's about everything ;0)

Okay, here's what I did with the acceleration...

We can look at the system as MassB goes down and MassA goes up. Since they are tied together by a string which will not stretch, then it is just like they are tied together by a (massless) metal rod!! Meaning, the string is so "stiff" that it falls down with MassB just as quickly as it flies up with MassA on the opposite side. This means that Masses A and B have equal properties of motion at ANY given time (meaning that their velocities are equal AND THEIR ACCELERATIONS are EQUAL!!) Think about this for a while, perhaps pretending that YOU are the string or are riding along the string, over the pulley, and back down. If we drew a black dot on a part of the string before it went up and over the pulley (on MassA's side), it would go up, around the pulley, and back down MassB's side at the same rate. This "rate" is its acceleration.

Once we know that the two accelerations are truly one equal acceleration and have no difference between each other (if you eventually still cannot see this, simply memorize it) then we may examine our equations which we made once again:

a1=(294-T1)/30
-and-
a2=(588-T2)/60

Seeing again that the first acceleration is equal to the second acceleration, we have:

a1=(294-T1)/30=a2=(588-T2)/60=mutual acceleration!! (a)

So, regrouping:

(294-T1)/30=(588-T2)/60

Ready for the easy part? ;0)

Actually, this part IS easier, b/c the fact that we are considering a massless pulley (again, I am assuming this) and a pefect pulley, at that, means that we don't have to worry about friction (which would add a complication to the Free Body Diagrams). Luckily, you're class doesn't seem to be worried about that, yet, so here's the magic:

You Ready?? Here she is: Tension1 IS EQUAL to Tension2 because Newton's 3rd Law of Motion tells us that the pull from the first mass is equal (and opposite...but we're talking about magnitude and not direction, here) to the pull from the second mass!! SO!! T1=T2=T!!!!

YES!!

Regrouping, again, we combine our Tensions and say:

(294-T1)/30=(588-T2)/60
(294-T)/30=(588-T)/60 (where T1=T2=T)
==> 30(294-T)=588-T
--> 8820-30T=588-T
-29T=588-8820=-8232

Therefore: T=283.86 Newtons!!

This is a resonably high force, since we are dealing with very large masses in the 10's of kilograms.

So, there you have it! Now, take this and memorize the processes. Apply them to all of your problems, and learn as you go.

In the future, I will try to work through these things with you instead of solving them outright, but this is a week chock-full of exams and headaches, so it was hard not to just type it all out ;0)

God Bless!
-Dave+
MrLaFazia.com

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EDIT: Oh yes, and to find the acceleration, you just use Newton's Second Law of Motion again because now you can use the Net Forces because NOW you know the Tensions! :0))