Q. Take a look at the diagram on the right. If m1 = 100 kg, and the coefficient of kinetic friction between m1 and the inclined plane is k = 0.3, then
(a) what must m2 be such that m1 moves up the inclined plane at a constant speed?
(b) Would your answer change if the block were to move down the inclined plane at a constant speed? Why or why not? (No calculations are necessary.)

The way to a physicist's heart is through Free-Body-Diagrams (FBD's). Whenever you're in a rut, and your brain freezes at the first problem of a physics test, just slap a quick FBD on every object that moves and you're guaranteed enough partial credit to at least pass the exam.

So that's exactly how we'll start with this problem. First, we'll deal with m1 and come up with its equations of motion. If you're not sure how to create an FBD for an object on an inclined plane I'll refer you to an inclined plane problem.


In this FBD,
T = tension
fk = force of kinetic friction (notice that it points in the direction opposite that of the block's motion)
N = the normal force (which always points perpendicularly away from the surface)
W1 = the weight of m1

When dealing with FBD's, it's crucially important that you define which directions you want to be positive directions, and then hold consistent to that definition throughout the problem. Here, I'm defining positive motion to be up the inclined plane.

Once we have the FBD properly drawn, the next step is to add the forces in their components form.


What simplified this problem considerably was the fact that the blocks moved at constant speed, which means that the accelerations are zero!

Okay, we've run through everything we know about m1. Let's do the same with m2.


Didja see how I assigned the downward direction as being positive, which goes against every convention that you're probably used to? Well, if m1 heads up the inclined plane, then m2 heads downward. So we need to label both of those motions as positive! See the problem concerning the Atwood machine which explores the sign conventions involved for pulleys.

On to the second part of the question : does anything significant change if m1 heads down at a constant speed? Well, you might initially think, "No, since we're still dealing with a constant speed." But something significant does indeed change! Recall that the force of friction is dependent on the direction of motion. Take a look at how the free-body diagram for m1 changes:


Friction now points in the same direction as the tension. This indeed changes things! The sum of the forces has changed, and the value we'll find for the tension of the rope will also change.

When looking at this question, some students thought that in the first part, when m1 moved up the inclined plane, it must be true that m2 > m1. This is not necessarily true. If we don't take into consideration friction, m2 needs to be comparable only to the x-component of m1's weight.

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