Why Does the Atwood Machine Calculation Yield the Same Tension on Both Sides?

[Jaimie]

Hello,
I have a question about a sample problem in the McGraw Hill Physics 12 book. (p. 38-39). "An Atwood machine is made of two objects connected by a rope that runs over a pulley. The objects on the left (m1) has a mass of 8.5 and the object on the right (m2) has a mass of 17kg. a) What is the acceleration of the masses? b) What is the tension of the rope."

Okay. So my question is for b). I understand that to find the tension I can calculate it using either the left of right sides of the diagram (if I were to sketch this out). The acceleration was calculated at 3.27 m/s^2. Therefore w/ left side:
-Fg1 + FT = ma
FT= m1g + m1a
FT= (8.5)(9.81) + (8.5)(3.27) = 111.18N

If I want to check with the right side of the diagram
-FT + Fg2 = m2a
Fg2 - m2a= FT
(17)(9.81) - (17)(3.27) = FT
111.18N

I know that I should be getting a negative value for the latter calculation. Can someone guide me as to what I am doing wrong. Thank you so much for your help!​

 

[Simon Bridge]

What makes you think you are doing anything wrong?
Did you draw free-body diagrams for part (a)?

The tension force points upwards for both masses, and gravity points down for both of them. You seem to have defined the +ve direction for forces in terms of the direction the rope slides over/turns the pulley.

Since you have used the sign of the force to represent direction, explicitly, in your equations, then the variables themselves represent magnitudes, so they should always be positive. i.e. you write "-Fg1" to show a positive magnitude Fg1 force in the negative direction. Thus, when you write "-FT" in the second relation, that is a positive value of FT in the negative direction... and you solved for the value.

 

[Doc Al]

I know that I should be getting a negative value for the latter calculation. Can someone guide me as to what I am doing wrong. Thank you so much for your help!

You did nothing wrong. For the right side you chose down to be positive. And you used FT to represent the magnitude of the tension force on the mass (thus the minus sign in front of it). So of course its positive. (And the tension in the rope should be a positive number anyway.)

Looks like Simon beat me to it!​

 

[Jaimie]

The example given in the book and was calculated as follows and different to mine:
Fg2+ Ft= m2a
m2g + Ft= m2a ** shouldn't at this point the ft be '-' and m2g be '+' ?*
FT=m2a -m2g
FT= (17)(3.27) - (17)(9.81)
= -111.18N

I don't quite understand how you can get this result as '-', if all forces are assigned their respective force directions. Please advise?

 

[Doc Al]

The example given in the book and was calculated as follows and different to mine:
Fg2+ Ft= m2a
m2g + Ft= m2a ** shouldn't at this point the ft be '-' and m2g be '+' ?*

Personally, since I know the tension force acts upward, I would have written the tension force as - Ft, which would make Ft positive. Just as you did above.

But here they just called the force Ft and let the equations tell you which way it acts.

FT=m2a -m2g
FT= (17)(3.27) - (17)(9.81)
= -111.18N

So the unknown tension force acts upward, that's why it's negative.

I don't quite understand how you can get this result as '-', if all forces are assigned their respective force directions. Please advise?

But the tension force was not assigned its correct direction--that's why it turned out negative.

 

[Jaimie]

Thank you all for your help!