Atwood Machines:
Some slick shortcuts
My name is Chris Mee, and I have made this website for APPC 02-03 to help future APPCers to understand Atwood Machines. When we first were introduced, I had no clue what was going on, but after I had an internalization, I think I can show everyone else how to work Atwood Machines in a very simplified manner.
The Device above is refered to as an Atwood Machine, invented by George Atwood. He created this device to “dilute” the effects of gravity to facilitate an accurate measurement of g. It is essentially two masses attached by an essentially massless rope slung over an essentially massless free-turning pulley.
The AP test writers for APPC love the Atwood machine, and you will commonly see it on free-response sections of the AP test. The Atwood machine will be first introduced to you during the fourth chapter about Newton’s 3 laws, and you will have some fairly difficult questions to answer about the acceleration of the system, and the tension on the ropes. Either DAY or The Reif will show you some pretty complicated ways to find the acceleration, but here is a way that is simplier even than the Truth that Eugene gives.
So…
F = MA
Best start with the baisics, right?
So lets suppose that on this Atwood Machine, the object on the left has a mass of m, and the object on the right has a mass of M, where M > m. And you are asked to find the acceleration of the system. Where do you begin?
Well, lets first make this “system” a lot simpler, ok?
A pulley serves to change the direction of the motion. So on the right object, there is a force down of Mg, and an upward force of tension from the rope called Ft. On the left object there is a force down of mg, and a force up of tension, Ft. This sounds complicated, huh? Here comes the simplification.
Using what I said earlier about pulleys, change the system to look like this:
Now the system just has one direction of motion, and you don’t have to worry about the Force of Tension.
So, lets start to plug in what we know. Think of this system as one object.
F = MA
The net force on the system is the big one minus the little one, or:
Mg – mg
And the mass of the system is the two masses added together, or:
M + m
So… (Mg – mg) = (m + M)A.
Now you just solve for A.
A = g(M – m)
( m + M)
This is the easy way to find the acceleration of an Atwood system.
You might also run across what is commonly refered to as a Half Atwood Machine.
It looks like the system below:
Lets work this one with numbers, ok?
Suppose that object A has a mass of 10 kg, and object B has a mass of 20kg (it’s a density thing, don’t worry), and that object A is resting on a frictionless table.
Find the acceleration on the system.
Once again, the pulley just changes direction, so rearange the system:
F = MA
So… The system only has one force on it, which is 20 * 9.81, or 196.2 N.
The mass of the system is the sum of the masses, or 10 kg + 20 kg = 30 kg.
196.2 N = (30 kg)A
A = 196.2 N / 30 kg =
6.54 m/s2
Well, there it is: Atwood Machines at a glance.
If you have any questions, you can E-mail DAY at dyoung@fayar.net
Or try these links:
Kenyon University shows how it’s done
Hyper Physics – This site has is all, just click on the Mechanics Link
Atwood at an incline: these are some tricky problems, remember F=MA
Good page, shows classic derivations and practice problems
This will help with that Tension thing