The Inclined Plane
Resolving the
gravitational forces
on an inclined plane
First draw a box on the
inclined plane
Now draw the weight vector.
In what direction should the weight
vector be?
Don’t forget
to label the
vector!
Fw
The weight vector has two
components.
• 1) the perpendicular force
• 2) the parallel force (this force pulls the object
down the incline)
Fp
Fw
Fⱶ
Envision the parallel force in
the triangle.
The weight vector is the resultant vector
which is the diagonal of a parallelogram.
Fp
Fw
Fⱶ
Label theta in the force
triangle.
The force triangle is SIMILAR to the hill
triangle. What does this mean about the angle
of the incline and theta?
Fp
θ
θ
Fw
Fⱶ
This angle!
Yes!
They are
the
same
angle!
Important things to remember about
inclined plane problems:
• 1) The weight of the object is always found
by multiplying the mass by 9.8 m/s2 and its
vector is ALWAYS straight down.
• 2) The parallel force and the normal force are
ALWAYS equal!
• 3) The angle of the incline is ALWAYS the
same as the force triangle’s angle.
Let’s work a problem!
• A cart weighing 420 N rests on a 23o incline.
Calculate the component of its weight that
presses the cart to the hill.
• Which force are we looking for?
• Perpendicular force
• Step 1: Draw the free body diagram.
• Step 2: Label the magnitude of the vectors you
know and the angles you know.
• Step 3: Use SOHCAHTOA to find the missing
force’s magnitude. In this case, the
perpendicular force.
This is how you do it!
Another Example:
• Calculate the parallel force component of the
weight of a 3600 kg car that is resting on a hill
which is 35o above the horizontal.