# 3-Pulley / 3-Weight system; Find Tension

1. Sep 19, 2013

### oddjobmj

1. The problem statement, all variables and given/known data

http://imgur.com/vgLv6NY

The figure shows a system of masses and pulleys. Assume the masses slide without friction and the pulleys have negligible mass. Calculate the tension in the string.

Data: M1 = 2.1 kg; M2 = 1.4 kg; M3 = 1.6 kg;

2. Relevant equations

f=ma

3. The attempt at a solution

I don't know what it is about this pulley system that is throwing a wrench in the cogs but I just can't find the correct answer.

I know that the acceleration of M1 will be lower than that of M2. I also believe that the acceleration of M3 must equal that of M1 + M2.

a1=(M3g-M2a2) / (M1+M2+M3)

Similar for a2.

a3=M3g / (M1+M2+M3)

Are my assumptions correct? I tried to use these equations to solve for a2. Since a2 is the limiting factor on the tension due to it being the lightest of the two resisting forces the total tension should just be equal to M2*a2.

What I got:
a2=0.76541 m/s
Tension=m2*a2=1.072 N

Does that sound right?

Thank you!

Last edited: Sep 19, 2013
2. Sep 19, 2013

### vela

Staff Emeritus
A reasonable guess, but it's not correct. It's good to look at limiting cases to check your intuition. Suppose, for instance, that mass 1 was infinite so it doesn't move. According to your guess, the accelerations of mass 2 and mass 3 would therefore be equal, but I'd say that mass 3's acceleration would be half of mass 2's.

I'm not sure how you came up with these equations.

Just apply $\vec{F}=m\vec{a}$ to each block separately, and find the correct relationship between the accelerations. You'll end up with four equations and four unknowns.

3. Sep 19, 2013

### oddjobmj

The only unknown in F=m*a for each individual part is a. When you say you don't know where I came up with those equations I have to say that is what I came up with when I applied F=ma. So I am back where I started. I am guessing then that my equations are wrong. For example:

I was thinking that the force acting on m1 is m3*g - m2a2. Is that not correct?

4. Sep 19, 2013

### vela

Staff Emeritus
How can those equations come from F=ma? No forces appear in your expressions. Isn't the tension an unknown?

5. Sep 19, 2013

### Staff: Mentor

The key to doing this problem is to properly analyze the kinematics of the motion. The total length of string between M1 and M2 must be constant. This is a kinematic constraint on the problem. Another constraint is that the length of string between pulley 1 and pulley 3 must equal the length of string between pulley 2 and pulley 3. In addition to being able to move downward, pulley 3 can also rotate to compensate for the fact that M1 and M2 are not accelerating at the same rate. You need to do a sort of "length balance" on the string to derive the proper relationship between the acceleration of M3 and the accelerations of M1 and M2. Once you do this, you can do a free body diagram on each of the three masses and write down the force balances on the three masses. These three mass balances, together with the relationship between the acceleration of M3 and the accelerations of M1 and M2 will give you enough to calculate the tension in the string (which is independent of location in the system).

So lets see your analysis of the kinetmatics to get a3>

6. Sep 19, 2013

### vela

Staff Emeritus
No, that's not correct. There are three forces on mass 1: its weight, the normal force from the surface, and the tension T.

7. Sep 19, 2013

### oddjobmj

Because there is no vertical motion and no friction then all we care about is T. That is exactly what I am trying to solve for. I am just representing T differently... The force acting on m1 is T. Taking a step forward we can represent T as other parts of the system. The tension is the result of m3 being acted on by gravity but because there is a mass on the other side of the system also pulling back on m3 the tension does not simply equal m3*g.

I understand that the only forces that are acting on mass 1 are the three you listed. I am trying to represent T by its origin.

I'll try to step back and start fresh with your suggestion of applying F=ma.

F1=m1a1
F2=m2a2
F3=m3a3

I can't simplifying F1, F2, and F3 to T because T is dynamic with respect to the other objects. Obviously T does not equal m1a1 while simultaneously being equivalent to m2a2. The accelerations are also unknown. I know the masses. The only way to proceed is to try to represent F1, F2, and F3 as T by considering their relationship with the other two components.

EDIT:

If I then simplify F1, F2, and F3 to T what do I get on the other side? That's what I was trying to get to earlier.

8. Sep 19, 2013

### oddjobmj

I tried earlier and got it wrong. Is this correct? a3=(a1+a2)/2

EDIT:

I just don't know the relationship between the lengths of string. a3 obviously can not be greater than the sum of the other two accelerations. It has to be some fraction of them. If the masses were equal then a3=a1=a2. The only way I get that using a fraction is effectively the same as an average which is the above suggestion.

Last edited: Sep 19, 2013
9. Sep 19, 2013

### vela

Staff Emeritus
Actually, T does equal $m_1 a_1$ and $m_2 a_2$. That's what Newton's second law applied to masses 1 and 2 tells you. Why do you think that's not the case?

10. Sep 19, 2013

### oddjobmj

Because that would be too simple? Actually, it's probably just because I am an idiot and am probably over thinking this whole problem. Thank you for your patience :)

T=m1a1
T=m2a2
T=m3a3 or 1.6g - something?

(a1+a2)/2=a3 ??

11. Sep 19, 2013

### Staff: Mentor

Well, that's a good guess. But you'd like to be confident about the result, wouldn't you. To do that, you need to study the kinematics. Let mass 1 move to the right d1 from its initial position, and let mass 2 move to the left d2 from its initial position. Consider the geometry, and determine from these distances and the kinemtic constraints on the system, the amount that pulley 3 moves down (and the amount that pulley 3 rotates) in terms of d1 and d2.

Chet

12. Sep 19, 2013

### vela

Staff Emeritus
Try again for mass 3. The first thing you need to do is identify all the forces on mass 3.

13. Sep 19, 2013

### oddjobmj

Last edited: Sep 19, 2013
14. Sep 19, 2013

### vela

Staff Emeritus
Where did this come from?

One piece of advice: don't plug numbers in until the very end. Leave the mass of block 3 as $m_3$.

15. Sep 19, 2013

### oddjobmj

What forces are acting on m3?

Gravity which the resulting force equals m3*g

The other two masses are also pulling up on m3. Up being opposite of the pull of gravity so negative.

Last edited: Sep 19, 2013
16. Sep 19, 2013

### vela

Staff Emeritus
You should abandon the notion that the other masses are pulling on mass 3. It's the tension in the rope pulling up on the mass.

17. Sep 19, 2013

### oddjobmj

I see, thank you. Any suggestions then on what the third Tension equation is? I'm pulling my hair out.

18. Sep 19, 2013

### vela

Staff Emeritus
Well, in terms of tension T and the weight $m_3g$, what's the net force on mass 3? Set that expression equal to $m_3 a_3$.

19. Sep 19, 2013

### oddjobmj

m3a3=m3g-2T

Thank you, solving that system did it.

20. Sep 19, 2013

### vela

Staff Emeritus
Good work!