# Atwood Machine – Inertia Example Problem

The Atwood Machine is a common classroom experiment showing the laws of motion of two coupled systems undergoing constant acceleration. An Atwood Machine consists of two masses mA and mB, coupled together by a inextensible massless string over a massless pulley. When the two masses are equal, the system is in equilibrium and no motion occurs. The two masses will remain stationary. When the two weights are not equal, the system will move where the heavier mass is pulled down while the lighter mass is pulled up. This example problem shows how to derive the acceleration of the system and the tension in the string.

Problem:
a) Find the acceleration of an Atwood Machine if mA = 3 kg and mB = 5 kg.
b) Find the tension in the string connecting the two masses.

Here is an illustration of the setup.

We will ignore the values of mA and mB at this point to show the derivation of the answer. Block B is heavier than Block A, so the overall direction of motion will be down on Block B’s side of the pulley and up on Block A’s side. Choose your coordinate system so the acceleration is always positive.

This system is coupled together by the massless string. The heavier Block B pulls down on the string a distance Δd in some time t. In the same time, Block A moves up Δd. That means each block’s velocities are the same.

vAΔdt = vB

The velocity directions can be adjusted by the coordinate system chosen for each system. Since the velocities are always the same, their accelerations are the same.

a = aA = aB

Since the string is massless and inextensible, the tension is uniform throughout the system. The tension pulling Block A up is the same as the tension pulling Block B up.

This means the acceleration and tension is the same for both blocks. Now find the forces acting on each block.

For Block A, the weight mAg is pulling down while the tension force T is pulling up.

ΣF = T – mAg

Since these forces are in motion, these forces are equal to mAa.

ΣF = mAa = T – mAg

For Block B, the forces are nearly the same. The difference is the overall acceleration is in the opposite direction of Block A.

ΣF = mBg – T = mBa

We now have two equations and two unknowns, T and a. Add these two equations together and T will drop out leaving only acceleration.

mAa = T – mAg
mBa = -T + mBg
mAa + mBa = mBg – mAg

Factor out the acceleration and gravity variables on each side.

(mA + mB)a = (mB – mA)g

Divide both sides by (mA + mB) to solve for a.

Since we have the acceleration, we can use this to find the tension using either of the two force equations. Let’s use Block A.

mAa = T – mAg

Solve this for T and get:

T = mAa + mAg

Substitute the formula for acceleration (a) into the equation.

Rewrite the equation as

Factor out mAg

Change the 1 so you have a common denominator inside the parenthesis.

Add the two fractions together to get

Simplify to get

which becomes

Use these formulas for acceleration and tension to find the answer to the initial problem.

Part a) Find acceleration when mA = 3 kg and mB = 5 kg.

Earlier, we found the formula for acceleration to be

Enter the values for mA and mB.

a = 2/8 g
a = 0.25 (9.8 m/s2)
a = 2.54 m/s2

The blocks are accelerating at 2.54 m/s2.

Part b) Find the tension in the string.

T = 36.75 kg·m/s2

kg·m/s2 is the same as the Newton unit of force.

T = 36.75 N

The tension in the string is 36.75 Newtons.

Tip:

The key to this type of problem is to choose your coordinate system so the direction of the acceleration is always in the positive direction. As long as your accelerations are all acting together, sign errors will not be a problem. In the case of an Atwood machine, acceleration will always run towards the larger mass.