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Midterm 1: July 9. Will cover material from Chapters 1-6 Go to the room where you usually have recitation July 6 Recitation will be a review session Practice exam available Friday Note: HW 3 is still due on Wed, July 7. Review. Newton’s 1 st Law: Inertia Newton’s 2 nd Law: F=ma - PowerPoint PPT Presentation
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17/1/04
Midterm 1: July 9
Will cover material from Chapters 1-6 Go to the room where you usually have
recitation July 6 Recitation will be a review session Practice exam available Friday
Note: HW 3 is still due on Wed, July 7
27/1/04
Review
Newton’s 1st Law: Inertia Newton’s 2nd Law: F=ma Newton’s 3rd Law: “Action and Reaction”
Forces are vectors Units: Newtons [kg m/s2]
37/1/04
Review: Free Body DiagramsThe first step in solving any force problem:
1. Sketch the object in question2. Draw an arrow for each force acting on the
object3. Label each force4. Indicate the direction of acceleration off to the
side (acceleration is NOT a force)
aF1
F3
F2
47/1/04
Review: How to Solve a Force Problem
1. Draw a free body diagram
Guess at forces of unknown magnitude or direction
2. Break forces into components
3. Sum of all the forces in one direction = mass * acceleration in that direction
i.e. Fnet,x=m ax
4. Repeat step 3 for each direction
5. Solve for unknown quantities
57/1/04
Mass on a string
If we pull steadily on the bottom string, which will break first?
A) Top
B) BottomC) It’s a matter of luck
67/1/04
Mass on a string
F
Tension in a string transmits a force along the direction of the string.
77/1/04
Mass on a string
Free body diagram for bottom string (at rest):
F
Tbottom
F
Tbottom = F
87/1/04
Mass on a stringFree body diagram for mass m (at rest):
F
Ttop
mg
F
Ttop = F+mg
97/1/04
What happens if we pull fast?
Which breaks first?
A) TopB) BottomC) It’s a matter of luck
107/1/04
Why did the bottom string break?
If we attempt to accelerate too fast tension on bottom string becomes too large and string snaps.
Newton’s First Law: mass is at rest and cannot accelerate instantaneously to speed of the hand yanking down…
7/1/04 11
Chapter 6
Forces (Part the Second)
127/1/04
Dissipative Forces
So far - Forces don’t depend on history
Direction of motion
Velocity of particle
Dissipative Forces DO
Friction
Viscosity (Air Resistance)
137/1/04
Frictional Forces
Frictional forces are from object's surface interacting with another material
Frictional forces always oppose (intended) motion
FpushFfriction
147/1/04
Kinetic Friction
For an object that is moving, the magnitude of the frictional force is proportional to the magnitude of the normal force on an object:
NfF kkfrictionkinetic
k is the “coefficient of kinetic friction”
157/1/04
Static FrictionThe maximum magnitude of the static frictional force is proportional to the magnitude of the normal force on an object:
NfF ssfrictionstatic
Once an object begins to move, kinetic friction takes over.
k is the “coefficient of kinetic friction”
167/1/04
Kinetic vs. Static Friction
Material Kinetic Static
Glass on Glass 0.40 0.94
Copper on Glass 0.53 0.68
Rubber on Concrete (dry) 0.8 1.0
Rubber on Concrete (wet) 0.25 0.30
Teflon on Teflon 0.04 0.04
Objects are harder to start moving than to keep moving
s > k
Some typical values:
No need to memorize, this will be given when needed
177/1/04
Static and Kinetic Friction
If you push with steadily increasing force:
Ffriction
Fpush
μsN
μkN
Fpush
Ffriction
187/1/04
A Classic Demo…
Pull the “table cloth” slowly, then quickly
197/1/04
Pig on a Frictional Plane
What angle should we tilt the plane so that the pig slides?
μs=0.5
N
Fg
Ffr
y
x
207/1/04
Pig on a Frictional Plane
Fg,x = mg sin θ
N
Fg
Ffy
x
Breaking Fg into components:
Fg,y = mg cos θ
x direction:
NfmgF
fF
maF
ssxg
sxg
xxnet
max,,
max,,
,
sin
0
0
Summing forces in the y direction:
cos
0
0
,
,
,
mgFN
FN
maF
yg
yg
yynet
217/1/04
Pig on a Frictional Plane
N
Fg
Ffy
x
We have:
Nmg s sin
cosmgN
)cos(sin mgmg s
)(cossin s
s tan
So for the pig to slip:
6.26)5.0(tan 1
227/1/04
Pig on a Frictional Plane
N
Fg
Ffy
x
x direction:
xkxg
xkxg
xxnet
maNF
mafF
maF
,
max,,
,
Summing forces in the y direction:
cos
0
0
,
,
,
mgFN
FN
maF
yg
yg
yynet
Say we tilt the ramp just pass = 26.6° anda = 2m/s2. What is μk?
237/1/04
Pig on a Frictional Plane
N
Fg
Ffy
x
xkxg maNF ,
27.0
)6.26cos()/8.9(
)6.26sin()/8.9()/2(2
22
sm
smsmk
cos
sin
g
ag xk
xk mamgmg cossin
247/1/04
Heavy Box
Say you can bench 500 N (about 110 lbs), can you push a 1000 N box across the floor?
Max static friction = μsN = (0.75)(1000 N) = 750 N
Ffriction
μs = 0.75 and μk= 0.45
Kinetic friction = μkN = (0.45)(1000 N) = 450 N
Not a chance…
No problem!
257/1/04
Pulling a Box
μk=0.25
Frope
θ
x: Fropecosθ - Ffr = ma Fropecosθ - μk(mg - Fropesinθ) = ma
y: Fropesinθ + N - Fg = 0 N = mg - Fropesinθ
Frope
Fg
N
Ffr
a
Summing forces:
267/1/04
Pulling a Box Frope
μk=0.25
θ
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0 5 10 15 20 25
Acce
lera
tion
Angle
m
mgFa kkrope
)sin(cos
0)cossin(
)sin(cos
krope
kkrope
m
F
m
mgF
d
d
d
da
25.0for 0.14)(tan 1 kk
To find the maximum:
sincos k
277/1/04
Accelerating a Car
The coefficients of friction for rubber on concrete are:
Dry: s=1.0 k=0.80
Wet: s=0.30 k=0.25
A car of mass 1000 kg tries to accelerate from a stop sign. What is the minimum time to accelerate to 30 m/s on dry pavement? On wet pavement?
287/1/04
Accelerating a Car
Dry: s=1.0 t = (30 m/s)/(1.0 × 9.8 m/s2) = 3.06 s
Wet: s=0.30 t = (30 m/s)/(0.30 × 9.8 m/s2) = 10.2 s
If a tire does not slip, friction is static
Ff=μsN
ga s
atvv f 0 g
v
a
vt
s
ff
mamgf ss max,
297/1/04
Stopping a Car
What are the minimum stopping distances for a 1000 kg car at 30 m/s in the following circumstances?
Dry pavement, wheels rolling? (ABS)
Dry pavement, wheels locked?
Wet pavement, wheels rolling? (ABS)
Wet pavement, wheels locked?
307/1/04
Stopping a Car
Wheels locked:
Ff=μkN
v0=30 m/s
Dry: μk = 0.80 xf = 57.4 mWet: μk = 0.25 xf = 183.7 m
)(2 02
02 xxavv ff
a
vx f 2
20
ga
mamgf
k
ks
max,
g
vx
kf 2
20so…
317/1/04
Stopping a car
Wheels rolling:
Ff=μsN
vi=30 m/s
Dry: μs = 1.0 xf = 45.9 mWet: μs = 0.30 xf = 153.1 m
Same as before, but we replace k with s.
g
vx
sf 2
20
Distance is shorter with ABS
327/1/04
Drag Force
Opposes Motion
Depends on velocity
A correct treatment is complicated,but an approximation of the “drag force” in the case of high velocity is given by:
is the fluid density, C is the “drag coefficient”
221 AvCFD
337/1/04
Terminal Velocity
An object stops accelerating when Fg=FD
gFAvC 221
AC
Fv g
22
Fg
FD
This is the “terminal velocity”
347/1/04
Force and Uniform Circular Motion
Object travels around a circle at constant speed
a
v
R
Recall: centripetal acceleration
center thetowards2
R
va
center thetowards2
R
vmF
amF
c
c
Centripetal Force
357/1/04
Demo: Water in a bucket
367/1/04
Why Didn’t I Get Wet (hopefully...)?
a
v
Accelerates faster than g downward, water cannot fall out and is pushed in a circle.
377/1/04
Ferris Wheel
Radius: R = 9mPeriod: T = 20 sec (fast)
What is the force on an 80 kg rider from the seat when he is at the top and at the bottom?
387/1/04
Ferris Wheel
The centripetal force is given by:
center thetowards2
R
vmamF cc
T
Rv
2Recall:
Ns
mkg
T
RmFc 71
)20(
)9)(80(442
2
2
2
Then,
397/1/04
Ferris Wheel
At the top:
NNsmkgN
FFN
FFN
FmaF
cg
cg
ccynet
71371)/8.9)(80( 2
,
At the bottom:
NNsmkgN
FFN
FFN
FmaF
cg
cg
ccynet
85571)/8.9)(80( 2
,
N
Fg
ac
N
Fg
ac
407/1/04
Rotation with FrictionHow fast can a car round an unbanked curve (dry pavement) with a radius of 50 m without slipping sideways?
R=50 m
vFf
vmax = μsgR
Dry: μs = 1.0 vmax = 1.0×9.8×50 = 22.1 m/s
R
vac
2
Nf ss max,
fs ac
RmvN
mafF
s
csnet
/2
417/1/04
Rotation with FrictionHow much does the curve need to be banked to provide the centripetal acceleration without using sideways frictional force from the tires?
θ
N
mg
y
x
y: N cosθ = mg N = mg / cosθ
x: N sinθ = mac but ac = v2/R N sinθ = mv2/R
Thus:R
mvmg 2
cos
sin
45)/8.9)(50(
)/1.22(tantan
2
21
2
smm
sm
Rg
v
427/1/04
Example: Problem 6.41
A puck of mass m slides on a frictionless table while attached to a hanging cylinder of mass M by a cord through a hole in the table. What speed must the puck travel at to keep the cylinder at rest?