Q. The Atwood Machine consists of a single pulley with a rope wrapped around it and two masses, one attached to each end of the rope. In this example, m1 = 4 kg and m2 = 2.5 kg. The system is released from rest with the 4-kg mass 1.48 m above the floor and the 2.5-kg mass on the floor. Assume that the mass of the pulley is negligible.
How long does it take the 4-kg mass to hit the floor?
When you have object's that will be accelerating, you should feel the urge to create free body diagrams and sum up those forces. But wait - there's is one catch in this type of problem (one with a pulley)!

We need to be very careful about the sign convention we use. Normally, you're probably comfortable with defining anything that points upwards as positive and the downward direction as negative. That will not work in this case! With pulleys, we need to let the rotation of the pulley dictate our sign convention. You see, if one block moves up, the other block moves down - both motions need to have the same sign!


Let's agree for the sake of this problem, that any motion which causes the pulley to rotate counterclockwise is a positive motion. With that in mind, let's look at the free-body-diagrams (FBD's) for each block.


For m1, it's downward motion causes the pulley to rotate counterclockwise, so that's the direction we'll label as positive. This equation contains two unknown quantities, the tension of the rope T and the acceleration of the mass ay.

Performing a similar procedure for m2 we can see that an upward movement on it's part will cause the pulley to rotate counterclockwise, so that's the direction we'll label as positive. The unknown quantities in this equation are the same as before, T and ay.

The tension of the rope in both situations is the same, since it's the same rope.

We now have two algebraic equations, both of which contain the same two unknown variables. Whenever you have two equations with two unknowns, you can solve it. But it makes sense to plan ahead a bit. Since we're interested in the motion of the system, we should strive to eliminate T from these equations and solve for ay.


We have already two expressions for the tension in the rope. We can set these equal to each other.

Now that we know the acceleration of the object, we can easily find how long it takes for the 4-kg mass to drop.

Notice in the figure to the right, we have the value of y increasing downward. This is totally consistent with keeping the downward direction positive as we talked about above. Given the fact that the block falls from rest we can use the kinematic equation of motion:

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