What we call wedges or slides in everyday language are called inclined planes in physics-speak. From our experience on slides during recess in elementary school, sledding down hills in the winter, and skiing, we know that when people are placed on slippery inclines, they slide down the slope. We also know that slides can sometimes be sticky, so that when you are at the top of the incline, you need to give yourself a push to overcome the force of static friction. As you descend a sticky slide, the force of kinetic friction opposes your motion. In this section, we will consider problems involving inclined planes both with and without friction. Since they’re simpler, we’ll begin with frictionless planes.
Frictionless Inclined Planes
Suppose you place a 10 kg box on a frictionless 30º inclined plane and release your hold, allowing the box to slide to the ground, a horizontal distance of d meters and a vertical distance of h meters.
Before we continue, let’s follow those three important preliminary steps for solving problems in mechanics:
- Ask yourself how the system will move: Because this is a frictionless plane, there is nothing to stop the box from sliding down to the bottom. Experience suggests that the steeper the incline, the faster an object will slide, so we can expect the acceleration and velocity of the box to be affected by the angle of the plane.
- Choose a coordinate system: Because we’re interested in how the box slides along the inclined plane, we would do better to orient our coordinate system to the slope of the plane. The x-axis runs parallel to the plane, where downhill is the positive x direction, and the y-axis runs perpendicular to the plane, where up is the positive y direction.
- Draw free-body diagrams: The two forces acting on the box are the force of gravity, acting straight downward, and the normal force, acting perpendicular to the inclined plane, along the y-axis. Because we’ve oriented our coordinate system to the slope of the plane, we’ll have to resolve the vector for the gravitational force, mg, into its x- and y-components. If you recall what we learned about vector decomposition in Chapter 1, you’ll know you can break mg down into a vector of magnitude cos 30º in the negative y direction and a vector of magnitude sin 30º in the positive x direction. The result is a free-body diagram that looks something like this:
Decomposing the mg vector gives a total of three force vectors at work in this diagram: the y-component of the gravitational force and the normal force, which cancel out; and the x-component of the gravitational force, which pulls the box down the slope. Note that the steeper the slope, the greater the force pulling the box down the slope.
Now let’s solve some problems. For the purposes of these problems, take the acceleration due to gravity to be g = 10 m/s2. Like SAT II Physics, we will give you the values of the relevant trigonometric functions: cos 30 = sin 60 = 0.866, cos 60 = sin 30 = 0.500.
