In this lesson, we will be doing basically the same thing as last lesson, except that now
the object will be on a slanted surface instead of horizontal. Here's a typical picture of an
object on a slant.
Notice how gravity points straight down and the other two forces are at an angle.
Remembering the earlier lesson on 
To do this though, because of the various properties of sine and cosine, we have to switch
which is the x and y component. Now instead of:
x component = R cos 
y component = R sin
The equations are:
x component = R sin
y component = R cos
Remember this only happens when an object is resting on a slanted surface. Any other time,
slanted vectors are handled the normal way.
Let's do a problem. If a 5kg object is sitting on a hill with an angle of 30°, with a force
of static friction of 40N, what is the coefficient of static friction of the hill on the object?
First, let's find the force of gravity and break it down.
FG = g * (5.98e24) * (5kg) / (6.37e6m)2
FG = 49.168N
x component = 49.168N sin 30° = 24.584N
y component = 49.168N cos 30° = 42.581N
And, since we know that the normal force is equal to the vertical or y component of gravity,
we can draw a force diagram.
Since we know the force of static friction and the normal force, we can easily find the
coefficient of static friction.
40N = 
S * 42.581N
S = .939
You've probably noticed that all we have worked with is static friction, that is because to
deal with kinetic friction, we must first deal with Newton's Laws of Motion, which happens to be
the next set of lessons.
Forces & Friction Sample Problems
