Inclined Planes – No friction
If there is NO friction, no matter the angle or the mass the object WILL MOVE because there is
a NET FORCE in the direction of motion (down the ramp).
If the Mass is 10kg, and the angle
of the incline plane to the table is
30°, what is the object’s
acceleration?
Problem Solving Steps:
1. Draw your Force Vector Diagram as above. Be sure to
clearly label your angles. You may like to draw the Force of
Gravity and its components separately.
2. The Force that pulls the object down the ramp is the
Force of Gravity x-component or Fgx.
3. Determine your coordinate system for you problem. In
general, I use down to be negative, up to be positive; down
the ramp to be negative and up the ramp to be positive.
4. List your knowns and unknowns. Solve for the Force of Gravity and its xcomponent.
Fg = mg
m = 10kg
Fg = ?
θ = 30°
g = −9.8 m
Fg x = ?
(
Fg = (10kg ) −9.8 m
Fg = −98 N
s2
s2
)
Fg x = Fg sin θ
Fg x
opposite
sin θ =
=
Fg x = (−98 N ) sin(30°)
hypotenuse Fg
Fg x = −49 N
5. This becomes a simple Forces problem now. The applied force is the Fgx, the
mass is given and you solve for acceleration!
m = 10kg
Fg x = −49 N F = ma Fg x = ma −49 N = (10kg )a a = −4.9 m
a=?
s2
Inclined Planes – WITH friction
If there is friction, the object MAY NOT move down the ramp. You will have to sum the forces
in the direction of motion (down the ramp) to see if it accelerates down the ramp!
If the Mass is 10kg, and the angle
of the incline plane to the table is
30°, and the coefficient of friction
is 0.1, what is the object’s
acceleration?
Problem Solving Steps:
1. Draw your Force Vector Diagram as above. Be sure to clearly
label your angles. You may like to draw the Force of Gravity
and its components separately.
2. The Force that pulls the object down the ramp is the Force
of Gravity x-component or Fgx. BUT the Force of Friction
OPPOSES that motion!
3. Determine your coordinate system for you problem. In
general, I use down to be negative, up to be positive; down the
ramp to be negative and up the ramp to be positive.
4. List your knowns and unknowns. Solve for the Force of Gravity and its xcomponent.
Fg = mg
m = 10kg
Fg = ?
θ = 30°
g = −9.8 m
Fg x = ?
s2
(
Fg = (10kg ) −9.8 m
Fg = −98 N
s2
)
Fg x = Fg sin θ
Fg x
opposite
sin θ =
=
Fg x = (−98 N ) sin(30°)
hypotenuse Fg
Fg x = −49 N
5. The Force of Friction depends on the NORMAL FORCE. You must solve for
the Force of Gravity y-component to find the Normal Force, to find the Force
of Friction!
Fg = −98 N
µ = 0.1
Fg y = ?
cos θ =
Fg y = − FN
Fg Fg y = Fg cos θ
adjacent
Fg y = −84.9 N
= y
hypotenuse Fg Fg y = (−98 N ) cos(30°)
FN = 84.9 N
6. This becomes a Sum of Forces problem now. The applied force is the Fgx,
opposed by the Force of friction, the mass is given and you solve for
acceleration!
FN = 84.9 N
Ff = + µ FN
m = 10kg
−40.51N = (10kg )a
∑ Fx = Ff + Fg x = ma
Fg x = −49 N Ff = +0.1( 84.9 N )
∑ Fx = 8.49 N − 49 N = (10kg )a a = −4 m s 2
Ff = +8.49 N
µ = 0.1
a=?
NOTE: Just like in the Pulley Problems With Friction, Force of Friction
changes sign. This is done because NO MATTER WHAT Force of Friction has
to oppose motion. So if going down the ramp is negative, than moving up the ramp,
the direction that the Force of Friction is going, is positive.
REMEMEBER
Normal Forces are always Perpendicular to the Surface!
FRICTION ALWAYS OPPOSES MOTION!
Define your coordinate
system like you tilted your
head to look at it!