1. Motion on Inclined Planes

2. Inclined Plane Problems
A tilted coordinate system is convenient, but not necessary! a
Its important to understand ∑F = ma &
How to resolve it into x,y components
in the tilted coordinate system!!

3. FGx= FGsin(θ) = mgsin(θ) FGy= -FGcos(θ) = -mgcos(θ)
You MUST understand
this case to understand
the case with friction!!
By geometry, the 2
angles marked θ are
the same! a
FG = mg
Both Angles = !
By Trigonometry:
FGx= FGsin(θ) = mgsin(θ)
FGy= -FGcos(θ) = -mgcos(θ)

4. Example: Sliding Down Incline
A box of mass m is placed on a smooth (frictionless!)
incline that makes an angle θ with the horizontal.
Calculate: a) The normal force on the box.
b) The box’s acceleration.
c) Evaluate both for m = 10 kg & θ = 30º
Free Body Diagram
Figure Caption: Example 4–16.
(a) Box sliding on inclined plane.
(b) Free-body diagram of box.
Answer: On an incline (or any surface), the normal force is perpendicular to the surface and any frictional forces are parallel to the surface.
(a) The normal force is equal to the component of the weight perpendicular to the incline, or mg cos θ.
(b) The force causing the acceleration is the component of the weight parallel to the incline; therefore the acceleration is g sin θ.
(c) The normal force is 85 N and the acceleration is 4.9 m/s2.

5. Is the normal force FN equal & opposite to the weight??? NO!!!!!!!
Example: The Skier
A skier descends a 30° slope, at constant
speed. What can you say about the
coefficient of kinetic friction?
Figure 5-8. Caption: Example 5–6. A skier descending a slope; FG = mg is the force of gravity (weight) on the skier.
Solution: Since the speed is constant, there is no net force in any direction. This allows us to find the normal force and then the frictional force; the coefficient of kinetic friction is 0.58.
Is the normal force FN equal & opposite to the weight??? NO!!!!!!!

6. Summary of Inclines: An object sliding down an
incline has 3 forces acting on it: the normal force FN,
gravity FG = mg, & friction Ffr. FN is always
perpendicular to the surface & is NOT equal &
opposite to the weight mg. The friction force Ffr is
parallel to the surface. Gravity FG = mg points down.
If the object is at rest, the
forces are the same except
that we use the static
frictional force, & the
sum of the forces is zero.

7. THE NORMAL FORCE IS NOT EQUAL TO THE WEIGHT!!!
Problem
Newton’s 2nd Law
∑F = ma
x: mgsinθ – Ff = ma
y: FN - mgcosθ = 0
Friction: Ff = μkFN
NOTE!!!
 FN = mgcosθ
FN  mg FN Ff
mg sinθ
mg cosθ
FG = mg
THE NORMAL FORCE IS NOT EQUAL TO THE WEIGHT!!!

8. Example: A ramp, a pulley, & two boxes
Box A, mass mA = 10 kg, rests on a surface inclined at
θ = 37° to the horizontal. It’s connected by a light cord,
passing over a massless, frictionless pulley, to Box B, which
hangs freely. (a) If the coefficient of static friction is s = 0.4,
find the range of values for mass B which will keep the
system at rest. (b) If the coefficient of kinetic friction is k =
0.3, and mB = 10 kg, find the acceleration of the system.
Figure 5-9. Caption: Example 5–7. Note choice of x and y axes.
Solution: Pick axes along and perpendicular to the slope. Box A has no movement perpendicular to the slope; box B has no horizontal movement. The accelerations of both boxes, and the tension in the cord, are the same. Solve the resulting two equations for these two unknowns.
The mass of B has to be between 2.8 and 9.2 kg (so A doesn’t slide either up or down).
0.78 m/s2

9. Example: A ramp, a pulley, & two boxes Static Case (i): mB << mA
(a) Coefficient of static friction s = 0.4, find the range of values
for mass B which will keep the system at rest.
Static Case (i): Small mB << mA: mA slides down the incline, so
friction acts up the incline.
Static Case (ii): Larger mB > mA: mA slides up the incline, so
friction acts down the incline.
Figure 5-9. Caption: Example 5–7. Note choice of x and y axes.
Solution: Pick axes along and perpendicular to the slope. Box A has no movement perpendicular to the slope; box B has no horizontal movement. The accelerations of both boxes, and the tension in the cord, are the same. Solve the resulting two equations for these two unknowns.
The mass of B has to be between 2.8 and 9.2 kg (so A doesn’t slide either up or down).
0.78 m/s2
Static Case (ii): Larger mB > mA
mA slides up incline
Ffr acts down incline
Static Case (i): mB << mA
mA slides down incline
Ffr acts up incline

10. Example: A ramp, a pulley, & two boxes
(b) The coefficient of kinetic friction is k = 0.3, & mB = 10 kg.
find the acceleration of the system & the tension in the cord.
Motion: mB = 10 kg
mA slides up incline
Ffr acts down incline
Figure 5-9. Caption: Example 5–7. Note choice of x and y axes.
Solution: Pick axes along and perpendicular to the slope. Box A has no movement perpendicular to the slope; box B has no horizontal movement. The accelerations of both boxes, and the tension in the cord, are the same. Solve the resulting two equations for these two unknowns.
The mass of B has to be between 2.8 and 9.2 kg (so A doesn’t slide either up or down).
0.78 m/s2