Download Closer look at friction notes

A closer look at friction
The factors that effect the magnitude of frictional
forces are:
• the nature of the two surfaces in contact.
Rough surfaces will cause a larger frictional force
than smooth surfaces.
•
the weight of the object
The force due to friction along a horizontal
surface is directly proportional to the normal force
(FN) between the object and the surface.
Normal force: The force a body exerts on another,
perpendicular to the surface of contact.
FN
Fg
Note that on a horizontal surface:
F N = Fg
If the surface is not horizontal, such as a ramp, then
F N ≠ Fg .
Coefficient of friction
The coefficient of friction (designated µ) is a number
that describes the tendency of two surfaces to
produce friction. The value of µ depends on the
nature of the two surfaces.
For example, a rough wood dragged across a
carpeted floor will have a large µ value, and hard
plastic dragged across smooth ice will have a small µ
value.
FFR = µFN
There are two types of coefficient of friction, static
and sliding (also known as kinetic).
• static coefficient of friction (µs) describes the FFR
that must be overcome to start an object sliding
• kinetic coefficient of friction (µk) describes the FFR
acting on a moving object.
The static FFR required to start an object moving is
greater than the kinetic FFR acting on a moving
object.
Generally:
µs > µk
The value of µ depends on the specific surfaces, but
some approximate examples are:
µs
µk
waxed wood on dry snow
0.22
0.18
rubber on dry concrete
1 to 4
1.02
EX: A horizontal force of 50 N is required to pull an
8.0 kg block of aluminium at a uniform velocity
across a horizontal wooden desk. Find µk.
Since a = zero, Fnet = zero, as Fnet = ma.
∴ FApp = FFR (in opposite directions) = 50 N
Note:
F N = Fg
FN = Fg = mg
= 8.0 kg x 9.81 m/s2
= 78.48 N
FFR = µkFN
µk = FFR / FN
= 50 N / 78.48 N
= 0.64 (µ has no units as the N units cancel)
EX: A 800 kg car, moving at 72 km/hr, slams on
the brakes, locking the wheels. The rubber tires skid
on the concrete road, producing a frictional force.
Find the distance required by the car to stop. Use
the coefficient of friction from the sample coefficients
given earlier.
given:
vi = 20 m/s
vf = zero (car stops)
m = 800 kg
and
µk = 1.02
To find the d, we first need to find a, and then use:
v 2f = vi2 + 2ad
To find a, we need to find Fnet.
The friction between the rubber tire and the concrete
supplies the only force acting on the car.
∴ FFR = Fnet.
To find FFR we use:
FFR = µkFN
Recall that FN = Fg = mg.
FN = Fg = mg
= 800 kg x 9.81 m/s2
= 7848 N
FFR =
=
=
µkFN
1.02 x 7848 N
8004.96 N
Fnet = FFR
= -8004.96 N
(opposite direction to car’s motion∴ is -)
Fnet
= ma
a = Fnet / m
= -8004.96 N / 800 kg
= -10.0062 m/s2
v 2f = vi2 + 2ad
d=
v2f − vi2
2a
d
= (0 m/s)2 - (20 m/s)2 / 2 x -10.0062 m/s2
d
= 19.987607683236393 m
= 20 m
EX: Repeat the same calculation for the same
situation, except that the car is skidding on ice. The
coefficient of friction for rubber on ice can be as low
as 0.0050. We assume the worst possible case.
FN = 7848 N (as before)
FFR =
=
=
µkFN
0.0050 x 7848 N
39.24 N
Fnet = FFR
= -39.24 N
Fnet
= ma
a = Fnet / m
= -39.24 N / 800 kg
= -0.04905 m/s2
d=
v2f − vi2
2a
d
= (0 m/s)2 - (20 m/s)2 / 2 x -0.04905 m/s2
d
= 4077.4719673802243 m
= 4.1 km (a very long skid!!!)
EX: A child sits on a wooden sled, which is sitting
on dry snow. The combined mass of the child and
the sled is 28.5 kg. Find the minimum force needed
to start the sled moving.
recall µs = 0.22
FN = Fg = mg
= 28.5 kg x 9.81 m/s2
= 279.585 N
FFR =
=
=
=
µsFN
0.22 x 279.585 N
61.5087 N
61.5 N
This is the minimum force required to start moving
the sled.