Why is one Tension Negative and the Other Positive - Atwood Machines

In summary: Both methods have their advantages and disadvantages. The method used in the picture uses a principle that we already know what's going on, so it's easier to understand. However, this method can lead to mistakes if we don't take into account the effects of gravity. On the other hand, the method that uses measurements may be more accurate, but it can be harder to understand.
  • #1
Blakeasd
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Why is one Tension Negative and the Other Positive -- Atwood Machines

Hello,

Please consider the following:

atwd.gif


I am trying to understand how to solve for acceleration when dealing with Atwood machines. I keep getting hung up on one part: the direction of tension. In the image above, why is tension in the sum of the forces for m2 negative when if I were to draw a force diagram it is pointing up. Furthermore, why is gravity positive? Tension is pulling the masses up and gravity is pulling them down. A force diagram's directions are not different between m1 and m2, so why are they different when summing the equations.

Many Thanks.
 
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  • #2
Obviously the author is using coordinates where the positive axis point down for the mass 2. for the mass 1 they are using a different coordinate axis where the positive side points upward as usual. This is common choice designed to keep the acceleration positive for both masses. If the axis had been chosen with the positive end pointing up (as you expected), the equation would be [tex]T - m_2 g = - m_2 a.[/tex] That equation is identical to the one found by the author except that the whole equation was multiplied by -1. That, of course, makes no difference at all for the solution.
 
  • #3
I see, thanks. But why is T - m2g = to a negative m2a ?

What is the reasoning behind m2a being negative. If when dealing with the first mass ma is positive; what makes the difference here?
 
  • #4
Blakeasd said:
I see, thanks. But why is T - m2g = to a negative m2a ?

What is the reasoning behind m2a being negative. If when dealing with the first mass ma is positive; what makes the difference here?

The difference is that the mass 2 is accelerating downwards which has a negative component in the vertical direction (If we chose upwards as positive). The acceleration, just as the forces, may have negative components. That's easy to forget. That's why it is common to chose axis such that the acceleration is positive avoiding having to deal with negative accelerations.
 
  • #5
dauto said:
The difference is that the mass 2 is accelerating downwards which has a negative component in the vertical direction (If we chose upwards as positive).

Yes, but isn't mass 1 accelerating downwards as well. It also is being affected by gravity. Why then is it equal to positive ma and not negative ma. I expect it to be negative, but that is not the case. Am I missing something obvious?

Thanks!
 
  • #6
Blakeasd said:
Yes, but isn't mass 1 accelerating downwards as well.
They can't both be accelerating downward. Mass 2 accelerates down while mass 1 accelerates up.
 
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  • #7
Doc Al said:
They can't both be accelerating downward. Mass 2 accelerates down while mass 1 accelerates up.



I see!

Thank you.
 
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  • #8
Blakeasd said:
Yes, but isn't mass 1 accelerating downwards as well. It also is being affected by gravity. Why then is it equal to positive ma and not negative ma. I expect it to be negative, but that is not the case. Am I missing something obvious?

Thanks!

Yes, you're missing the obvious fact that if one mass goes down than the other mass must go up
 
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  • #9
There are two different ways to attack this sort of problem.

The picture in the OP uses the principle that "we know which way everything moves, so let's draw a picture where all the quantities are positive."

The other way says, "We don't know what happens, because we haven't solved the problem yet. So let's measure everything positive in the same direction (e.g. upwards), and if some things turn out negative, that's not a big deal." Using this method, because the string is a fixed length, if m1 has acceleration a, m2 has acceleration -a.

You can use either method, but the mixing them up usually leads to mistakes.
 
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1. What is an Atwood machine?

An Atwood machine is a simple mechanical device used to demonstrate the principles of classical mechanics. It consists of two masses, connected by a string, that are suspended over a pulley. The masses are free to move vertically, and the tension in the string changes as the masses move.

2. Why is one tension negative and the other positive in an Atwood machine?

The tension in a string is always directed away from the object it is attached to. In an Atwood machine, one mass is heavier than the other, causing it to accelerate downwards. This creates a positive tension in the string connected to the heavier mass, as the string is being pulled downwards. On the other hand, the lighter mass is accelerating upwards, creating a negative tension in the string connected to it, as the string is being pulled upwards.

3. How does the difference in mass affect the tension in an Atwood machine?

The tension in an Atwood machine is directly proportional to the difference in mass between the two masses. This means that the greater the difference in mass, the greater the tension in the string. If the masses are equal, the tension will be zero as there is no difference in mass.

4. Why is the tension in an Atwood machine constant?

In an Atwood machine, the masses are connected by a string that is assumed to be massless and frictionless. This means that the tension in the string is the only force acting on the masses. According to Newton's second law, the acceleration of an object is directly proportional to the net force acting on it. Since the tension remains constant, the acceleration of the masses also remains constant, resulting in a constant tension throughout the motion.

5. Can the tension in an Atwood machine ever be zero?

Yes, the tension in an Atwood machine can be zero if the masses are equal and there is no external force acting on them. In this case, the acceleration would also be zero and the masses would remain stationary. However, in most cases, the masses will have a slight difference in mass and the tension will not be zero.

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