Theoretical Acceleration of an Atwood machine

[andrewdavis23]

Homework Statement

The descending mass of the pulley system (atwood machine) is 60g and the ascending mass is 55g. Using DataStudio to measured the slope of the velocity over time (read by the spinning spokes of the pulley) the acceleration (constant) is experimentally given as 0.316m/s^2. Now they want to know the theoretical acceleration so you can find the % error.

Homework Equations

They ask you to solve for (m2-m1)g in the same data table in the lab book. (m2-m1)g = (65-55)9.8 = 49N

The Attempt at a Solution

I used f=ma and other equations and I keep on getting 9.8m/s^2 as the acceleration. This makes sense to me because the objects are acted on by gravity, but the main confusion is: WHY IS THE ACCELERATION MEASURED AS 0.316m/s^2 and not 9.8m/s^2?

Please help me

 

[Doc Al]

I used f=ma and other equations and I keep on getting 9.8m/s^2 as the acceleration. This makes sense to me because the objects are acted on by gravity, but the main confusion is: WHY IS THE ACCELERATION MEASURED AS 0.316m/s^2 and not 9.8m/s^2?

9.8 m/s^2 is the acceleration of a body in freefall--where gravity is the only force acting. But the masses in an Atwood's machine are not freely falling--they constrained by strings.

Show how you derived the acceleration.

 

[andrewdavis23]

I actually tried again, figuring m2a2=m2g-m1g, plugged in the numbers 60g*a2=60g * 9.8m/s^2 -55g * 9.8m/s^2. I solved for a2 and got 0.8m/s^2, does that seem right?

If its not I originally used the equation g = a (m1+m2) / (m2-m1) and solved for acceleration and got 9.8m/s^2 and I'm sure that, that equation is meant for the atwood machine.

 

[Doc Al]

I actually tried again, figuring m2a2=m2g-m1g, plugged in the numbers 60g*a2=60g * 9.8m/s^2 -55g * 9.8m/s^2. I solved for a2 and got 0.8m/s^2, does that seem right?

No.

If its not I originally used the equation g = a (m1+m2) / (m2-m1) and solved for acceleration and got 9.8m/s^2 and I'm sure that, that equation is meant for the atwood machine.

That's the correct equation, but the acceleration is a, not g. g is a constant! (Which is equal to 9.8 m/s^2, of course.)

 

[andrewdavis23]

I tried the correct equation and got 0.8m/s^2 for a (inputing g=9.8m/s^2). This has to be right then...so thank you very much for the help!

 

[Doc Al]

I tried the correct equation and got 0.8m/s^2 for a (inputing g=9.8m/s^2). This has to be right then...so thank you very much for the help!

No, not right.

In post #3 you had two equations. The first one, which I think you used to get your answer, was not correct. The second equation is the correct one. Use it to solve for the acceleration.

 

[andrewdavis23]

I think I see what was throwing me off. We wrote down m2=15g and m1=10g, but we also had a base weight on each piece of 55g (to "slow down the system"). So when I go to put it in the equation for acceleration [ a=((m2-m1)g)/(m2+m1) ] I need to make m2=(15g+55g) and m1=(10g+55g) because while the difference of masses (m2-m1) is still 5g, the sum of the masses (m2+m1) changes from 25g to, the correct; 135g which is what was making my theoretical acceleration so different from the experimentally record acceleration.