Solving Slope Problems - MCAT

Solving slope problems.

Let’s start simply with a frictionless inclined plane and build from there, since the steps are largely the same for any block on an inclined plane like the one pictured here:

Getting the forces, probably the hardest part
The first part of solving an inclined plane problem is figuring out all of the forces acting on the block. We’ll start by looking at gravity, which has 2 components. The first is the component of gravity that tries to slide the block down the plane, called the sliding force (F=mg sin x). The second is the component of gravity forcing the block against the plane, called the y component of the block’s weight (F=mg cos x). Since I never could remember which component was sin and which was cos, I’ll give you an easy way to just figure it out. Think about if the angle was 0 (i.e., the plane was flat). Then there’s no force pushing the block down the plane. Now, which trig function gives us 0 with an angle of 0? It’s sine. Likewise, the force pushing down is the full force of gravity; which trig function gives us 1 with an angle of 0? That would be cosine.

Friction, it justs rubs you the wrong way.
Ok, so now you’ve got the force sliding the block down the plane and the force pushing it against the plane; what next? Well, you need to consider, is there friction? If not, you’re done; lucky you. The net force is just the sliding force; you can use F = ma to get the acceleration, mass, or whatever they’re asking for.

So if you’re not lucky, they put in friction. The next question is whether the force is enough to overcome static friction. Maybe they’ll tell you that the block is in fact moving or has just started to move. Cool, in that case, static friction has been breached; you get to figure out dynamic friction, and you can skip to the next section. Of course, these problems are rarely that simple, and you’ll probably have to work through static friction. The good news is that it’s not that much more work. Static friction seems complex given that the formula is F < or = uN, where u is the coefficient of static friction and N is the normal force. The normal force is the force that the incline plane pushes up against the block. (It is very convenient to keep the block from sinking into the inclined plane that the normal force does this.  ) Anyway, N is equal in magnitude to the force of gravity pushing the block down into the plane, i.e., F = mg cos x, the y-component to the weight of the block. However, N points in the opposite direction of the y component of the weight.

So what’s the deal with the inequality anyway? What static friction tries to do is generate exactly enough force to resist motion. So if we had a 10N force sliding the block down the plane and static friction could generate anything up to 100N to resist this motion, it would actually generate a 10N force in the opposite direction and your block will just sit there. However, you just need to calculate the maximum force of static friction by using the formula F = uN. (Oh, and if you used signs and got a negative number, make the result positive, or you’ll probably confuse yourself.) So ask yourself at this point, "Is the force sliding this block down the plane greater than the maximum force of static friction?" If you answered no, then the prof threw you a break and the block isn’t going anywhere. Otherwise, we can now forget that we ever figured out static friction and go to the next section.

Dynamic (Kinetic) Friction
We have good news: no inequalities are needed for dynamic friction, and the formula is exactly the same:

Fdynamic friction = uN

Ok, if your block is moving, and friction is present, you need to subtract the friction force opposing the motion from the x component of the gravitational force. This gives you the net force. If we call the x component of the gravitational force the sliding force, we would get the following equation for the net force:

Fnet = Fsliding - Fdynamic friction

Oh, if somehow you got a negative number at this point, you definitely screwed something up, since negative forces in this context mean that the block is sliding up the plane for no real reason. (Of course, if you had a couple of rocket engines strapped to the block, maybe it really IS sliding up the plane, but somehow I doubt that.)