- Good morning. Billy could you please tell me the names
of all the forces and their directions that are acting
currently on this book which is on this incline?
♫ Flipping Physics ♫
- The force of gravity is, as always, straight down.
The force of static friction; static because the book
isn't moving, is parallel to and up the incline
to prevent the book from sliding down the incline.
And the force normal is perpendicular to the incline
and is a push, so the force normal is up
and normal to the incline.
- Correct. Now the next step when dealing with
a problem with forces in it, is to break
forces into components. Bobby, which forces do you think
we need to break into components?
- Well, the force normal and the force of static friction
are neither in the x or y direction, so we need to break
those two forces into components.
- When an object is on level ground,
that is typically how we would approach this
free body diagram. However, on an incline,
we usually don't use the x and y directions.
We instead tilt our axes to match the incline angle,
and rather than having x and y direction axes,
we have axes which are parallel and perpendicular
to the incline. In other words,
on an incline, we usually use a direction which is
perpendicular to the incline and a direction
which is parallel to the incline.
Now that our directions are perpendicular to the incline
and parallel to the incline, Bobby which forces do we
need to break into components?
- Well, the force normal is in the perpendicular direction
and the force of static friction is in the parallel
direction, so neither of those
need to be broken into components.
However the force of gravity does need to be broken
into components because it's neither in the parallel
nor perpendicular direction.
- Yes. Notice how we only have one force
to break into components instead of two.
And when the book moves, the motion of the book
is typically parallel to the incline,
which means the acceleration of the book in the
perpendicular direction is 0, which
makes summing the forces in the perpendicular
and parallel directions much easier than summing
the forces in the x and y directions.
Now let's draw the components of the force of gravity.
The force of gravity perpendicular is always
perpendicular to the incline and down.
And the force of gravity parallel is always
parallel to the incline and down the incline.
Every once in a while a student will draw
the force of gravity parallel up the incline,
which is incorrect; the force of gravity is always
down towards the center on the planet,
so the force of gravity parallel is always down the incline.
Now, let's identify some angles.
Theta 1 is called the incline angle.
Theta 2 is the angle between the force of gravity
perpendicular and the force of gravity.
Theta 3 is the angle between the force of gravity
and the incline, and because the incline and the force of
gravity parallel are, by definition, parallel to one another
theta 3 is also the angle between the force of gravity
and the force of gravity parallel.
Oh, and lastly, because the force of gravity is,
by definition, vertical, it is at a 90 degree angle
than to the horizontal. In other words,
this is a right angle right there.
The interior angles of a triangle in a two dimensional
Euclidean plane always add up to 180 degrees.
Therefore, 180 degrees equals theta 1 plus 90 degrees
plus theta 3. Subtracting 90 degrees from both sides
gives us 90 degrees equals theta 1 plus theta 3.
Subtracting theta 3 from both sides gives us
theta 1 equals 90 degrees minus theta 3.
We can also see that theta 2 and
theta 3 add up to 90 degrees.
This is because the incline and the force of gravity
perpendicular are perpendicular to one another.
Again, subtracting theta 3 from both sides gives us
theta 2 equals 90 degrees minus theta 3.
Therefore, theta 1 is equal to theta 2.
They both equal 90 degrees minus theta 3,
therefore, theta 1 and theta 2 are equal to one another.
This means the angle we are going to use to
resolve the force of gravity into its parallel and
perpendicular components is
the same as the incline angle.
Bo, could you please resolve the force
of gravity into its components?
- Okay, sine theta equals opposite over hypotenuse
and opposite theta is the force of gravity parallel,
on the hypotenuse is the force of gravity.
[Mr. P] And now, please solve for
the force of gravity parallel.
- We can multiply by the force of gravity
to get the force of gravity parallel equals
the force of gravity times the sine of the incline angle.
- And the force of gravity equals the mass of the object
times the acceleration due to gravity,
so the force of gravity parallel equals mass
times the accerlation due to gravity times
the sine of the incline angle.
- Billy, please do the same thing with the cosine.
- Cosine theta equals adjacent over hypotenuse
which then equals the force of gravity perpendicular
divided by the force of gravity,
multiplied by the force of gravity to get
the force of gravity perpendicular equals
the force of gravity times the cosine of theta,
substituting in for the force of gravity,
we get the force of gravity perpendicular
equals mass times acceleration due to gravity,
times the cosine of the incline angle.
- We now have equations for the force of gravity parallel
and the force of gravity perpendicular,
in terms of the incline angle.
Oh and we should probably redraw
our free body diagram. Bo could you please do that?
- The force normal and force of static friction
have not changed, however now rather than having
the force of gravity straight down, we have its components.
We have the force of gravity parallel,
which is parallel to and down the incline,
and the force of gravity perpendicular,
which is perpendicular to the incline and down.
- There you go. Thank you very much for learning
with me today, I enjoyed learning with you.