Transcript Free Body Diagrams

Day 51 Friction
Aim: What are the different types of Friction?
LO: Relate friction to the normal force
LO: Calculate friction for different surface combinations
LO:
AGENDA
Do Now - Worksheet
Notes
Worksheet
HW# Due
Friction
Friction
 Friction is a special force that is caused by
the surface roughness of an object.
 It always acts in the opposite direction of
the motion of the object.
 There are two types of friction
– Static, and kinetic
Coefficient of Friction
 All surfaces exhibit friction, some more
than others.
 It depends on the roughness of the surface
of the object.
 It is represented by the symbol m.
– For static friction: ms
– For kinetic friction: mk
Sliding Friction – Microscopic
model
 Depends on microscopic (electrostatic) bonding
forces
 Depends on roughness of the surface
Kinetic Friction
 Kinetic friction is the force of friction on an
object when it is moving
 The formula is:
Ff = mkFN
Static Friction
 Static Friction is the force of friciton on an
object when it stands still.
 We find that it is harder to start an object
moving than it is to keep it moving.
 The formula is:
Fs  msFN
Graph of the behavior of sliding
friction
f s  ms N
f k  mk N
A Table of coefficients of sliding
friction
Example
 A boy exerts a 36N horizontal force as he
pulls a 52N sled across a cement sidewalk
at a constant speed. What is the coefficient
of friction between the sidewalk and the
sled (ignoring air resistance)?
36N
52N
Solution
 Known:
FN = Fg = 52 N
Fpull = Ffriction = 36N because the sled is
moving at constant velocity
Ffriction = FNmk Therefore mk = Ff/FN
mk= 36N/52N = ?
Example 2
 Suppose the sled runs on packed snow. The
coefficient of friction is now only 0.12. If a
person weighing 650N sits on the sled what
is the force needed to pull the sled across
the snow at a constant speed?
m= 0.12
Fw = mg= 650N
What force to pull sled?
Inclined Plane
 A common free body
diagram used is often
the inclined plane.
 Another name for an
inclined plane is a
ramp.
 Look at the diagram to
the right showing the
usual forces on an
inclined plane
FN
Ff
W
Vector
Diagram
FN
Ff
W
 If we look at just
the vector diagram
we see some
interesting things
 We usually know
the weight of the
object, so we can
find the normal
force.
 The normal force is
perpendicular to the
friction force and
the force of the
inclined plane
Example 3
 A skier (Ki) has just begun to descend a 30o
slope. Assuming the coefficient of kinetic
friction is 0.10 calculate:
(i) his acceleration and (ii) his speed after 4 s
Example 3

(i)
A skier (Ki) m = 7 kg has just begun to descend
a 30o slope. Assuming the coefficient of kinetic
friction is 0.10 calculate:
his acceleration and (ii) his speed after 4 s
Approach:
(i) Resolve forces | | and
to slope
(ii) Calculate frictional force
(iii) Find net force down the slope => acceleration
(iv) Use vf = vi + at => vf
Solution
 Force of gravity down the slope is:
Fgpara = FgSin()
Fgpara = 7kg*10 m/s/s*0.5 = 35 N
 Calculate Normal = Fgperp
Fgperp = FgCos()
Fgperp = 7kg * 10m/s/s *0.866 = 60.62 N
Solution Continued
 Calculate Frictional Force:
Ff = mkFnormal
Ff = 0.1 * 60.62 N = 0.6062 N
 Caluculate Net force down the slope
Fnet = Fgperp – Ff
Fnet = 35 N – 0.6062 N = 34.4 N
Solution last page
 Calculate Acceleration down the slope:
Fnet = ma
a = 34.4 N/7kg = 4.9 m/s/s
 Calculate velocity at t = 4 seconds
Vf = Vi + at
Vf = 0 m/s + 4.9 m/s/s * 4 s = 19.7 m/s
Inclined Plane
 A common free body
diagram used is often
the inclined plane.
 Another name for an
inclined plane is a
ramp.
 Look at the diagram to
the right showing the
usual forces on an
inclined plane
FN
Ff
Fp
W
Fp = the
force
caused by
the ramp
Vector
Diagram
FN
Ff
Fp
W
 If we look at just
the vector diagram
we see some
interesting things
 We usually know
the weight of the
object, so we can
find the normal
force.
 The normal force is
perpendicular to the
friction force and
the force of the
inclined plane