There are two types of friction: static and kinetic. The force of friction depends on the normal force and the coefficient of friction between the two surfaces. The coefficient of friction does not depend on the surface area. When objects are on an incline, their weight can be broken into components parallel and perpendicular to the surface. These components along with equations of motion or equilibrium can be used to solve problems involving friction and inclined planes.

1. Friction & Inclined Planes
Honors Physics

2. TWO types of Friction
 Static – Friction that keeps an object at rest
and prevents it from moving
 Kinetic – Friction that acts during motion

3. Force of Friction
 The Force of Friction is
directly related to the
Force Normal.
 Mostly due to the fact
that BOTH are surface
forces
Nkkf
Nssf
Nf
FF
FF
FF
µ
µ
µ
µ
α
=
=
=
=
frictionoftcoefficien
alityproportionofconstant
Note: Friction ONLY depends on the MATERIALS sliding against
each other, NOT on surface area.
The coefficient of
friction is a unitless
constant that is
specific to the
material type and
usually less than
one.

4. Friction & N.F.L
If the coefficient of kinetic friction between a 35-kg crate and the floor is
0.30, what horizontal force is required to move the crate to the right at
a constant speed across the floor?
mg
Fn
Fa
Ff
(0.30)(35)(9.8)
a f f k N
a k N
N
a k
a
a
F F F F
F F
F mg
F mg
F
F
µ
µ
µ
= =
=
=
=
=
= 102.9 N

5. Friction & N.S.L.
Suppose the same 35 kg crate was not moving at a constant speed, but
rather accelerating at 0.70 m/s/s. Calculate the applied force. The
coefficient of kinetic friction is still 0.30.
mg
Fn
Fa
Ff
(35)(0.70) (0.30)(35)(9.8)
NET
a f
a k N
a k
a k
a
a
F ma
F F ma
F F ma
F mg ma
F ma mg
F
F
µ
µ
µ
=
− =
− =
− =
= +
= +
= 127.4 N

6. Inclines
θcosmg
θsinmg
mg θ
FNFf
θ
θ
θ
θ
θ
Tips
•Rotate Axis
•Break weight into components
•Write equations of motion or
equilibrium
•Solve

7. Friction & Inclines
A person pushes a 30-kg shopping cart up a 10 degree incline with a force of
85 N. Calculate the coefficient of friction if the cart is pushed at a constant
speed.
cosmg θ
mg
Fn
Fa
Ff
sinmg θ
sin
sin cos
cos sin
sin cos
sin
cos
85 (30)(9.8)(sin10)
(30)(9.8)(cos10)
a f f k N
a k N N
a k
a k
a
k
k
F F mg F F
F F mg F mg
F mg mg
F mg mg
F mg
mg
θ µ
µ θ θ
µ θ θ
θ µ θ
θ
µ
θ
µ
= + =
= + =
= +
− =
−
=
−
= = 0.117
θ
θ

8. Example
A 5-kg block sits on a 30 degree incline. It is attached to string that is thread
over a pulley mounted at the top of the incline. A 7.5-kg block hangs
from the string.
 a) Calculate the tension in the string if the acceleration of the system is 1.2
m/s/s
 b) Calculate the coefficient of kinetic friction.
m
2
m1
m1g
m2g
FNT
T
Ff
30
30
m2gcos30
m2gsin30
1 1
2 2
2
( sin )
cos
NET
f
N
F ma
m g T m a
T F m g m a
F m g
θ
θ
=
− =
− + =
=

9. Example
1 1
1 1
(7.5)(9.8) (7.5)(1.2)
NETF ma
m g T m a
m g m a T
T
T
=
− =
− =
− =
= 64.5 N
2 2
2 2
2 2
2 2
2 2
2
2 2
2
( sin )
sin
sin
sin
sin
cos
sin
cos
64.5 (5)(1.2) (5)(9.8)(sin30)
(5)(9.8)(cos30)
f
f
k N
k N
k N
N
k
k
k
T F m g m a
T F m g m a
T F m g m a
T m a m g F
T m a m g
F m g
F
T m a m g
m g
θ
θ
µ θ
θ µ
θ
µ θ
θ
µ
θ
µ
µ
− + =
− − =
− − =
− − =
− −
= =
− −
=
− −
=
= 0.80 N

10. Example
1 1
1 1
(7.5)(9.8) (7.5)(1.2)
NETF ma
m g T m a
m g m a T
T
T
=
− =
− =
− =
= 64.5 N
2 2
2 2
2 2
2 2
2 2
2
2 2
2
( sin )
sin
sin
sin
sin
cos
sin
cos
64.5 (5)(1.2) (5)(9.8)(sin30)
(5)(9.8)(cos30)
f
f
k N
k N
k N
N
k
k
k
T F m g m a
T F m g m a
T F m g m a
T m a m g F
T m a m g
F m g
F
T m a m g
m g
θ
θ
µ θ
θ µ
θ
µ θ
θ
µ
θ
µ
µ
− + =
− − =
− − =
− − =
− −
= =
− −
=
− −
=
= 0.80 N