Double Atwood Machine and acceleration

In summary, the tension in the upper string is 2T due to the massless pulleys and cords. If the lower pulley had mass, the forces on the pulley would not necessarily balance.
  • #1
y3nx3ng3a
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  • #2
Start by drawing free-body diagrams for each box and pulley and write out their Newton's second law equations. You'll also need a few constraints based on the (constant) lengths of the cords that connect each pair of objects.
 
  • #3
so far i have FTA-mAg = mAa1 for mA, mBg-FTA=mBa1 for mB, FTC=2FTA for the pulley, and mCg-FTC=mCa2 for the mC block... am i on the right track? I don't quite understand what you mean by the constraints...
 
  • #4
Yeah, that's the right start. I can't exactly remember how to solve this problem though, so I won't be of too much help :p
 
  • #6
Start with the bottom section, forget the top. Determine the acceleration of the two masses, and determine the tension in the rope.

Then the tension above the bottom section is 2T. This is the same as the tension pulling up mass "c". Use this to calculate the acceleration for mass "c".

edit: good god, didn't notice the OP date!
 
  • #7
I think I asked this question 3 semesters ago :P was surprised to find an email in my inbox about the response! thanks for the help though :P
 
  • #8
philnow said:
Start with the bottom section, forget the top. Determine the acceleration of the two masses, and determine the tension in the rope.

Then the tension above the bottom section is 2T. This is the same as the tension pulling up mass "c". Use this to calculate the acceleration for mass "c".

edit: good god, didn't notice the OP date!

Philnow, I'm glad you didn't notice the date, because I'm wrestling with this question, and I appreciate your help!

I'm stuck on this point: why is the tension above the bottom section 2T? If the bottom pulley were in equilibrium, this would make sense, because we would have Ftc - 2Fta = 0 (where Ftc is the tension in the rope going around the upper pulley, and Fta is the tension in the rope going around the bottom pulley - I'm using the labels from the diagram given by the OP at http://session.masteringphysics.com/problemAsset/1057076/4/GIANCOLI.ch04.p56.jpg)

However, the bottom pulley is accelerating, or might be at least, depending on the masses.

I'm guessing that the answer is that because the pulleys are massless, we have Ftc - 2Fta = 0*a = 0. That is, Ftc - 2Fta = 0, even though the lower pulley isn't in equilibrium.

Is that right? Intuitively, this is hard for me to accept, but it makes sense mathematically.
 
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  • #9
fconst said:
However, the bottom pulley is accelerating, or might be at least, depending on the masses.

I'm guessing that the answer is that because the pulleys are massless, we have Ftc - 2Fta = 0*a = 0. That is, Ftc - 2Fta = 0, even though the lower pulley isn't in equilibrium.

Is that right? Intuitively, this is hard for me to accept, but it makes sense mathematically.

Hi! Even with the lower pulley accelerating, by balancing the forces on the bottom pulley, we see that the tension in the upper string must be 2T. However... this tension will be dependent on the acceleration of the masses.

The three F=ma equations:

2T-mCg=mCaC
T-mAg=mAaA
T-mBg=mBaB

and the fact that (just by looking at the string set-up) the average position of mA and mB moves the same distance as the bottom pulley, which in turn moves the same distance (but in the opposite direction) as mC:

aC = -(aB + aA)/2

Should get you started!
 
  • #10
philnow said:
Hi! Even with the lower pulley accelerating, by balancing the forces on the bottom pulley, we see that the tension in the upper string must be 2T.

Thanks for the response. I'm still not sure that I understand why TUpperString = 2T, although I do understand how the rest of the solution you've outlined follows from that.

My confusion has to do with what is meant here by "balancing the forces." When I do an FBD for the lower pulley itself, I get:

TUpperString - 2T = mLowerPulley * aLowerPulley

If the lower pulley is in equilibrium, then TUpperString balances 2T. But since aLowerPulley is nonzero, it would seem that TUpperString is not equal to 2T - that the forces on the lower pulley do not balance.

However, since mLowerPulley = 0, I can see that TUpperString does indeed equal 2T. But this seems like a mathematical "trick" to me. I'm having trouble understanding the physical meaning of this.

Is it correct to say that TUpperString = 2T is only true if mLowerPulley = 0? That if the lower pulley had mass, the forces on the pulley would not necessarily balance? I realize that if the lower pulley had mass, the problem would change in a number of ways (I'm just now learning about rotational dynamics in my physics class), but I just want to be sure I understand the reasoning behind your solution.

Thanks again for your help.
 
  • #11
That's a great question. As an undergrad in physics myself, my hunch is that it's because the pulley has no mass, so ma=0. In fact, most Atwood machine problems that I have seen have been very careful to include that the pulley indeed has no mass... as for the physical meaning, I too would be interested in the answer.
 
  • #12
I would like to thank you guys. Even a significant time later this has helped me a lot. Especially considering I am only a high school AP student.
 
  • #13
Sorry to revive this post - could someone explain to me why we can't treat the free pulley as a system with mass mA+mB, solve for mC's acceleration as if it were a single Atwood machine? That would then give the acceleration of the free pulley-system, which we could use to compute the effective g of the free pulley-system due to its acceleration and solve that system as a single Atwood machine.
 
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Related to Double Atwood Machine and acceleration

1. What is a Double Atwood Machine?

A Double Atwood Machine is a mechanical system composed of two Atwood Machines connected in series, with one weight hanging from the other's string. It is used to demonstrate the principles of pulley systems and acceleration.

2. How does a Double Atwood Machine work?

The weight of the hanging object on one side creates tension in the string, which is transmitted to the other side and causes the weight on that side to accelerate. The acceleration of the two weights is inversely proportional to their masses, and the total acceleration of the system is the difference between the two individual accelerations.

3. What factors affect the acceleration in a Double Atwood Machine?

The acceleration in a Double Atwood Machine is affected by the masses of the two weights, the tension in the string, and the force of gravity. The acceleration can also be affected by external factors such as friction and air resistance.

4. How is acceleration calculated in a Double Atwood Machine?

The acceleration of a Double Atwood Machine can be calculated using the formula a = (m1-m2)g / (m1+m2), where m1 and m2 are the masses of the two weights and g is the acceleration due to gravity. This formula assumes that the string and pulley have negligible mass and there is no friction or air resistance.

5. How is a Double Atwood Machine used in scientific experiments?

A Double Atwood Machine is commonly used in physics experiments to study the principles of pulley systems and acceleration. It can also be used to demonstrate the concept of conservation of energy and to calculate the acceleration due to gravity. In addition, it can be used to study the effects of different variables, such as mass and tension, on the acceleration of the system.

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