# Free Body Diagrams - Finding Acceleration, and tension

1. Jul 8, 2008

### pitaaa

1. The problem statement, all variables and given/known data
Three blocks, of masses m1 =26 kg, m2 = 38 kg, and m3 = 41 kg, are connected by two strings over two pulleys, as in the given attachment. Friction is negligible. Determine (a) the magnitude of the acceleration of the blocks and (b) the magnitude of the tension in each of the two strings.

2. Relevant equations
Fnet = ma
F = mg
Mtotal = M1+ M2+ M3

3. The attempt at a solution

a) Hmm, well what have I not tried... probably everything but the actual right method. I started out by drawing the free body diagram with all the forces acting on all of the weights - gravity, normal forces, kinetic frictions, and applied forces. That's where I'm stuck... Since I have no other values other than weights, I don't even know where to begin. I tried finding the net x forces, and net y forces with the hopes that through combining them, I could eliminate some sort of unknown - that failed :P. And I'm also looking at my given formulas thinking maybe I could rearrange them somehow, but nothing is working for me. Any hints as to how to get on the right track would be much appreciated!

b) I think b would be relatively easy to solve once the acceleration could be found. Then you'd substitute your known values into the F=ma formula (that's what I'm thinking anyway) ... but the difficulty lies in the actual starting of the question.

https://www.physicsforums.com/attachment.php?attachmentid=14663&d=1215571969

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• ###### PHYS1.jpg
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Last edited: Jul 8, 2008
2. Jul 8, 2008

### alphysicist

Hi pitaaa,

Since the attachements are unavailable right now, would you be able to explain the diagram in words?

What equations did you get from your free body diagrams?

3. Jul 8, 2008

### pitaaa

m2 (38 kg) is on a table, with m1 suspended from the left side from a pully, and m3 hanging from the right side from a pully ... the pully's are attached to the corners of the table. Here's a link to the photobucket file: http://i347.photobucket.com/albums/p475/piiita/PHYS1.jpg

The equations I got are:

For m1:

$$\Sigma$$F = FT - FG

For m2:

$$\Sigma$$Fx = Fapp - FK

$$\Sigma$$Fy = FN - FG

For m3:

$$\Sigma$$F = FT - FG

4. Jul 8, 2008

Those are the right idea for m1 and m3. But go ahead and pick variables for the forces. For example, from question b you can tell that the ropes have different tensions, so the $F_T$ in the m1 equation is different from the $F_T$ in the m3 equation (because they are different ropes!). I would suggest calling them something like $T_1\mbox{ and }T_2$ and going ahead and plugging them into your equation. The same for F_G[/itex]; go ahead and keep track of the three separate $F_G$ forces by calling them $m_1 g, m_2 g, \mbox{ and } m_3 g$. Also, the reason we sum the forces is so we can set them equal to ma; so go ahead and write out the right side of each equation by putting in the ma term. For m2, you have an applied force and a frictional force, but those are not in the problem. There are two horizontal forces acting on m2 from the two ropes, so only T1 and T2 appear. What do you get for you equations? 5. Jul 9, 2008 ### pitaaa Okkkay, so: m1: ma = FT1 - FG1 m2: mya = FT1 - FT2 mxa = FN - m2g mxa = FN - FG2 m3: ma = FT2 - FG3 Last edited: Jul 9, 2008 6. Jul 9, 2008 ### alphysicist Except for a couple of sign errors, these look okay. If we focus on your m1 equation: ma = FT1 - FG1 this is correct, but go ahead and label the mass as m1 since you will have to combine the equations together: m1 a = FT1 - FG1 So this is perfect; in particular, you have the (m1 a) and tension terms positive because they are upwards, and the gravity force negative because it is downwards. Now look at the m3 equation: m3 a = FT2 - FG3 This is okay, except for a sign error. Based on my previous paragraph, do you see what it is? For the m2 equation: m2 a = FT1 - FT2 There is a problem with the signs here, too. From your force diagram, which way is a? what about FT1? What about FT2? Once you have those three directions, most people call forces/accelerations to the right positive, and to the left negative. (You can do the opposite of course, but you have to be consistent for all three terms.) Once you correct those sign errors you're almost done. You'll have three equations with three unknowns (the two tensions and the accleration). You can then solve the three simultaneously and find all three. 7. Jul 9, 2008 ### pitaaa Okay, in that case, m1a = FT1 - FG1 m3a = Fg3 - FT2 So, would m2a = FT1 + FT2? 8. Jul 9, 2008 ### alphysicist These are fine. (In the m3 equation, you have downwards as positive and upwards as negative, but that's fine as long as your consistent with every term in your equation like you are here.) FT1 and FT2 are pulling on m2 in opposite directions--and so they'll need to have opposite signs in the equation. Is the acceleration of m2 in the same direction that FT1 is pulling, or is it in the direction that FT2 is pulling? Things in the same direction need the same sign. 9. Jul 9, 2008 ### pitaaa Acceleration is in the direction FT2 is pulling, so, m2a = FT2 - FT1? So for m3a, couldn't it = FT2 - Fg3? That's what I had initially, but you said I had a sign error? 10. Jul 9, 2008 ### alphysicist Yes, perfect! You have to deal with the whole equation. For m3, the acceleration is in the direction of the FG3, so they must have the same sign. So for m3, you can use either of the following two: m3 a = FG3 - FT2 or - m3 a = FT2 - FG3 The first of these calls downwards positive, and the second calls upwards positive. But the important things is that the acceleration term and the gravitational force must have the same sign in the m3 equation since they are in the same direction (and FT2 must have the opposite sign since it is in the opposite direction). With three unknowns and three equations, you can solve for a. What do you get? 11. Jul 9, 2008 ### pitaaa Er, and how do you go about doing that... heh. I just finished grade eleven, and I'm taking grade twelve physics in summer school, so as of yet, I've never been faced with a three unknown situation. 12. Jul 9, 2008 ### alphysicist Here are your three equations: m1a = FT1 - FG1 m2a = FT2 - FT1 m3a = Fg3 - FT2 If we rewrite them a bit to help keep track: [tex] \begin{align} m_1 a &= T_1 - m_1 g\nonumber\\ m_2 a &= T_2 - T_1\nonumber\\ m_3 a &= m_3 g - T_2\nonumber \end{align}

I would solve the first one for $T_1$, and the third one for $T_2$, and then plug them both into the second equation.

Then both $T_1\mbox{ and }T_2$ will be eliminated from the second equation, which will only have one unknown (the acceleration a) that you can solve for.

13. Jul 9, 2008

### pitaaa

Andddddd I finalllllllllllly got it haha - acceleration is 1.4 m/s2. The thing with a question like this, is that I would have never thought to do what was supposed to be done - which troubles me big time! :/ I get to a certain spot in the procedure and then I just get stuck because I don't know what else to do with the question. Thanks SO MUCH for taking the time to answer my questions again! :) I probably learn more on these forums than I do in my actual class... -.-

14. Jul 9, 2008

### alphysicist

Sure, glad to help! When you get stuck on a problem like this, the difficulty often is just that there are so many parts that have to be solved together. It helps (me at least) to think of the different parts as being as separate as possible; like in this problem, the way to approach it was to treat each mass as almost a separate force diagram problem. Then, it's easier to put them together to get the final answer.