Download notes about solving friction problems

All About Friction
When you did the friction lab, you graphed the frictional force between the block and table as a function of the
contact force between the block and table. Your graph came out to be a straight line, which tells us that the frictional
force between two objects is proportional to the contact force: if you double the contact force between the block and
table, the frictional force doubles. Mathematically, we write that like this:
F f   FN
(frictional force) = (coefficient of friction) x (contact force)
The symbol μ that we are using is a Greek letter “mu,” and it is going to stand for the “coefficient of friction,” which is
a unitless number that tells us how rough the two surfaces are. The coefficient of friction is different for different
surfaces; it can never be below zero and is usually less than 1. The coefficient of friction for rubber tires on pavement
is about 0.8; for skates on ice it is about 0.1.
Our Toolbox so far…
We can now include this friction equation in our toolbox,
since it is pretty useful to us. (There is one more equation
we’re going to include in the toolbox; you’ll figure that out
during the lab today.)
v 
a
Now, how can we use this new friction equation in problem
solving? Follow the steps just like we did last time:
1) Draw the free-body diagram(s).
2) Pick a positive direction.
3) Add up the forces in each direction to find Fnet.
4) Use Newton II: a 
v 
d
t
v f  vi
Fnet
m
Fg  mg
a
F f   FN
t
vi  v f
2
Fnet
m
5) Put in numbers and solve.
Example:
A 20-kg box of hockey equipment slides across the smooth ice until it hits a rough patch, where the coefficient of
friction is 0.15. The box is moving at 6 m/s when it hits the rough patch. (a) What is the acceleration of the box while
it is on the rough patch of ice? (b) How far will the box slide before stopping?
The free-body diagram for the box should
include a gravitational force downward, a
contact force upward, and a frictional force
opposite the direction of motion.
It is still a good idea to pick the positive
direction to be the direction of motion of
the box.
Add up the forces in each direction.
Vertical:
Fnet  Fg  FN
Vertical:
a
Use Newton II.
Put in numbers and solve. But we can’t
solve for the acceleration until we know the
frictional force. Use our new friction
equation to find it.
Vertical:
Fg  FN
m
200N  FN
20kg
FN  200N
0
m
s
s

Horizontal:
Fnet  F f
Horizontal:
a
F f
m
Horizontal:
a
m
  FN (0.15)(200N)

 1.5 ss
m
20kg
D = 12 meters (you should be able to do this)
The negative sign on the answer tells us that the box is slowing down, just as we knew it should! Now you can figure
out the answer to part (b) using the rest of the expressions in the toolbox! (Hint: the answer is 12 m.)