Friction
Holding in Place

Objects on an incline will often stay put.
There must be a force that holds the object in place.

Static friction is from the contact of resting objects.

• Force holds up to a certain point
• Force is based on the type of contact (rough, smooth)
• Maximum force is proportional to the pressing force of the
object (normal force)
Inequality

The approximate formula for static friction is:
F fr  m s FN
ms is the coefficient of static friction

This is an inequality.
• The force of static friction is generally less than the
coefficient times the normal force
Coefficient of Friction

Use rope and measure the
force when movement
begins.
• Measure weight
• Measure force at slipping
point.
F fr (max)  m s mg
FT
m1
m s mg  FT
m s  FT / mg  FT / W
Slippery Slope

F fr (max)  mg sin q
FN
mg sin q

If Ffr < msFN = msmg cosq,
then the block will hold.
At equality the block just
begins to move.
F fr  m s FN
q
mg cos q
Fg  mg
mg sin q  m s mg cos q
m s  tan q
Sliding

Sliding objects also have a frictional force exerted on
them.

This frictional force is kinetic friction.

An approximate formula:
F fr  m k FN
mk is the coefficient of kinetic friction
Static vs. Kinetic

Static and kinetic friction are
similar.

Static and kinetic friction are
different.
• Force in opposite direction
to motion
• Static friction is an inequality
up to a maximum
• Proportional to normal force
• Coefficient of friction is
typically greater for static
friction at the same surface
• Coefficient of friction
depends on materials
• Wood on wood (0.4 vs. 0.2)
Downhill Skiing
q  3°

A TV station has invited you
to be the science
commentator for an
upcoming ski race. You
observe that a skier needs at
least 3° of slope to move
forward at a constant
velocity.

What is the coefficient of
friction on the skis?
Force Diagram
FN = mg cosq

At constant velocity the
forces must all balance.

Friction doesn’t act in the
direction of the normal force.

The normal force cancels the
component of gravity.
Ffr = -mkFN
Fg = -mg
q
Fgy = -mg cosq
FN  mg cosq
Minimum Sliding

When the skier is sliding
slowly we can neglect air
resistance.

Kinetic friction balances the
component of gravity pulling
forward.
FN = mg cosq
Ffr = -mkFN
Fgx = mg sinq
Fgx  F fr  0
q  3°
mg sin q  m k mg cos q
m k  tan q  0.05
Normal Force and Friction

Friction depends on both the normal force and on the
coefficient of friction.

To reduce friction requires reducing one of those
factors.
• Reduce normal force by lightening the load
• Reduce normal force by adding additional upward force
• Add a lubricant to reduce the coefficient of friction