Transcript Physics 130 - University of North Dakota

9/30 Friction
Text: Chapter 4 section 9
HW 9/30 “Skier” due Thursday 10/3
Suggested Problems:
Ch 4: 56, 58, 60, 74, 75, 76, 79, 102
Talk about friction
Lab “Atwood’s Machine”
This is about Energy conservation
Lab
vf2 = vi2 + 2ax
Fnet = ma
I never let you use
this and here is why.
combining we can get:
PE =
mgy
y
1/
2 = 1/ mv 2 + F x
mv
2
f
2
i
net
This is where the idea of
“kinetic energy” comes from!
“Potential Energy” of gravity is
mgh, or mgy. (must pick y = 0
location)
Lab
vf2 = vi2 + 2ax
Fnet = ma
I never let you use
this and here is why.
combining we can get:
PE =
mgy
y
v
v
1/
2 = 1/ mv 2 + F x
mv
2
f
2
i
net
The System of both blocks has a net
loss of PE which is equal to the net
gain of KE. Note that both blocks
have the same velocity.
Static Friction
The maximum static friction force depends on the type of surface
and the contact force between the surfaces. The actual static
friction force depends on Newton’s second law!
fT,BMAX = sNT,B note subscripts!
s, called the “coefficient of static friction”, is a number that that
is small for slippery surfaces.
Note this is not a vector equation as N is 90° to f and only relates
their magnitudes.
Example
A 12 kg block is placed on a ramp angled at 20° with respect to
the horizontal. The coefficient of static friction is s = 0.5 . Does
the block slip? If not, how big is the frictional force?
y
NR,B
fR,B
x
20°
s = 0.5
WE,B
Friction forces always parallel
to the contact surface
First pick a coordinate system
and draw a free body
diagram.
Normal forces always
perpendicular to the contact
surface
Example
The next step is to divide the weight force into components along
the x and y axes.
y
WE,B = mg = (12)(9.8) = 118 N
NR,B
fR,B
20°
s = 0.5
Wy
Wx
WE,B
x
Wx = 118 sin 20° = 40 N
Wy = 118 cos 20° = 111 N
Example
The next step is to divide the weight force into components along
the x and y axes.
y
WE,B = mg = (12)(9.8) = 118 N
NR,B
fR,B
20°
s = 0.5
Wy
Wx
WE,B
x
Wx = 118 sin 20° = 40 N
Wy = 118 cos 20° = 111 N
Example
Now we consider the motion in the x and y directions separately.
y
NR,B
fR,B
Wx
Wy
WE,B
Wy = 111 N
Wx = 40 N
s = 0.5
x
y direction
ay = 0
Fnet,y = NR,B - Wy
Fnet,y = may
NR,B - Wy = 0
NR,B = Wy = 111 N
x direction
ax = ? does it slip?
Fnet,x = Wx - fR,B
Fnet,x = max but we don’t know ax
Wx - fR,B = max
If the ramp is rough enough the
block will not slip, ax = 0, and Wx = f.
fR,BMAX = sNR,B = (0.5)(111) = 55.5 N
The maximum static friction force exceeds Wx so
the block remains at rest and fR,B = Wx = 40 N.
Kinetic Friction:
Always acts parallel to the surface of
contact.
Relative motion of surfaces (sliding)
fa,b = kNa,b always
(Remember, static by second law only)
k is the coefficient of kinetic friction.
Jack pulls the box at constant
speed
Draw a FBD of the box.
a=0
v=
Since a = 0, Fnet = 0 and
TS,B = fF,B = kNF,B
NF,B
fF,B
TS,B
Box
WE,B
Jack pulls harder, what
happens?
Kinetic Friction
Jack pulls harder
Draw a FBD of the box.
a=
v=
Fnet = TS,B - fF,B = ma
y
fF,B
NF,B
TS,B
x
Box
WE,B
The tension force increases
but the kinetic friction force
stays the same!
Kinetic friction does not
depend on velocity or
acceleration.
fF,B = kNF,B
Jack pulls the box at constant
speed
Draw a FBD of the box.
a=0
v=
30°
NF,B
fF,B
Box
WE,B
How will the string angle
change the FBD?
Tension force now has a
vertical component which
means NF,B is smaller.
fF,B is proportional to the
normal force so it is smaller
also.
Jack pulls the box at constant
speed
Draw a FBD of the box.
a=0
v=
30°
NF,B
fF,B
TS,B
Box
WE,B
The new FBD looks
something like this.
Fnet = 0 so
Tx = fF,B
Ty + NF,B = WE,B
Homework Problem
A 70 kg skier skis down a slope that
is angled 36.8° with respect to the
horizontal. The coefficient of
kinetic friction between the skis and
the snow is 0.10. What is the skiers
acceleration?
36.8°
If the instantaneous velocity is 16m/s, how far up the slope
did he start from rest?