newton’s laws -contidinuedsuperlab.roma2.infn.it/07_newton_s_laws_cont.pdf · newton’s third...
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i dNewton’s Laws - continued
Friction Inclined Planes N T LFriction, Inclined Planes, N.T.L.
TWO types of Frictionyp
Static – Friction that keeps an object at rest and prevents it from movingKinetic – Friction that acts during motion
Force of Friction
FF αThe Force of Friction is directly related to the F N l
Nf FF
µ
α
= alityproportion ofconstant Force Normal.
Mostly due to the fact that BOTH are surface Nssf FF µ
µ=
= frictionoft coefficienThe coefficient of f i i i i lthat BOTH are surface
forcesNkkf FF µ=
friction is a unitless constant that is specific to the
t i l t dmaterial type and usually less than one.
Note: Friction ONLY depends on the MATERIALS sliding against each other, NOT on surface area.
ExamplepA 1500 N crate is being pushed
across a level floor at aa) What is the coefficient of kinetic f i ti b t th t d thacross a level floor at a
constant speed by a force F of 600 N at an angle of 20° below the horizontal as shown in the
friction between the crate and the floor?
the horizontal as shown in the figure.
FNFa
82563)20(600
= Nkf
NFFF
FF
θ
µN
20Fay
2117051500)20( i600
1500sin
82.563)20(cos600cos
+
+=+=
====
aayN
aaxf
NF
FmgFF
NFFF
θ
θ
Fax
331021.170582.563
21.17051500)20(sin600
==
=+=
k
N NF
µµ
mgFf
331.0=kµ
ExamplepIf the 600 N force is instead pulling the
block at an angle of 20° above the horizontal as shown in the figure
FN
20
Fa
Fayhorizontal as shown in the figure, what will be the acceleration of the crate. Assume that the coefficient of friction is the same as found in (a)
20
Fax
friction is the same as found in (a)
mgFf
maFFmaFNet =
)i(cos maFF
maFF
Na
fax
=−
=−
θθµθ
1.153)20sin6001500(331.020cos600)sin(cos
amaFmgF aa
=−−=−− θµθ
2/883.01.15357.4288.563
smaa
=
=−
Inclines
FNFf
θθ
θθ
θcosmg θ
Ti
θsinmg
mg θTips•Rotate Axis•Break weight into components•Write equations of motion orθsinmg •Write equations of motion or equilibrium•Solve
ExamplepMasses m1 = 4.00 kg and m2 = 9.00 kg are connected by a light string that passes over a frictionless pulley. As shown in the diagram, m1 is held at rest on the floor and m2 rests on a fixed incline of angle 40 degrees. The masses are released from rest, and m2 slides
F
1.00 m down the incline in 4 seconds. Determine (a) The acceleration of each mass (b) The coefficient of kinetic friction and (c) the tension in the string.
gmamTamgmTmaFNET
1111 +=→=−=FNT
Fm gcos40
m g
T
Ff
40 amTFgm f 22 )(sin =+−θm2gcos40
m1
m2g40
m2gsin40
m1g
Example maFNET =pgmamTamgmT 1111 +=→=−
amTFgm f 22 )(sin =+−θ
2
2
)4(2101
21
a
attvx ox
+=
+= NT 7.39)8.9(4)125(.4 =+=
θsin amTFgm =
2/125.0
)4(201
sma
a
=
+
µθ
θ
θ
sin
)(sin
sin
2112
2112
22
amgmamFgm
amgmamFgm
amTFgm
Nk
f
f
=−−−
=+−−
=−−235.0
57.67125.12.395.07.56
=−−−
=kµ
θµθθµθ
µθ
cossincossin
sin
22112
21122
2112
gmamgmamgmamgmamgmgm
amgmamFgm
k
k
Nk
=−−−=−−−
θθµ
cossin
2
2112
gmamgmamgm
k−−−
=
Newton’s Third Law“For every action there is an EQUAL and
OPPOSITE reactionOPPOSITE reaction.This law focuses on action/reaction pairs (forces)They NEVER cancel outThey NEVER cancel out
All you do is SWITCH the wording!you do s S C t e o d g•PERSON on WALL•WALL on PERSON
N.T.LThis figure shows the force during a collision between a truck and a train. You can clearly see the forces are EQUAL and OPPOSITE. To help you understand the law better, look at this situation from the point of view of Newton’s Second Law.
FF
TrainTrainTruckTruck
TrainTruck
aMAmFF
==
TrainTrainTruckTruck
There is a balance between the mass and acceleration. One object usually h LARGE MASS d SMALL ACCELERATION hil h h hhas a LARGE MASS and a SMALL ACCELERATION, while the other has a SMALL MASS (comparatively) and a LARGE ACCELERATION.
N.T.L ExamplespAction: HAMMER HITS NAILReaction: NAIL HITS HAMMERReaction: NAIL HITS HAMMER
Action: Earth pulls on YOUReaction: YOU pull on the earth
Newton’s Law of GravitationWhat causes YOU to be pulled down? THE EARTH….or
more specifically…the EARTH’S MASS. Anything thatmore specifically…the EARTH S MASS. Anything that has MASS has a gravitational pull towards it.
MmFgαgWhat the proportionality above is saying is that for there to be asaying is that for there to be a FORCE DUE TO GRAVITY on something there must be at least 2 masses involved where one ismasses involved, where one is larger than the other.
N.L.o.G.As you move AWAY from the earth, your DISTANCE increases and your FORCE DUE TO GRAVITY decrease. This is a special INVERSE relationship called an Inverse-Square.
2
1Fg α 2rg
The “r” stands for SEPARATION DISTANCE and is the distance between the CENTERS OF MASS of the 2 objects. We us the symbol “r” as it symbolizes the radius. Gravitation is closely related to circular motion as you will discover later.
N.L.o.G – Putting it all togetherg g
221
rmmFg α
Constant nalGravitatio UniversalGalityproportion ofconstant G
r
==
22271067.6
mmkg
NmxG = −
221
rmmGFg =
ththhthiUF
earth eLEAVING th areyou when thisUse
earthon theareyou when thisUse
221 →=
→=
rmmGF
mgF
g
g
N.L.o.GLet’s set the 2 equations equal to each other since they BOTH represent your weight or force due to gravity
thLEAVING thhthiU
earth on the areyou when thisUse
21 →
→=
mmGF
mgFg
p y g g y
eartheLEAVING thareyou when thisUse221 →=
rGFg
Mm SOLVE FOR g!
MGg
rMmGmg
=
= 2
SOLVE FOR g!
226
2427
/81.9)1097.5)(1067.6( smxxg ==−
kgxMr
Gg
−==
=
6
24
2
1097.5Earth theof Mass
26 /81.9)1037.6(
smx
g
mxr −== 61037.6Earth theof radius
An interesting friction/calc problem…!F
g p
Suppose you had a 30 kg box that
FN
Suppose you had a 30- kg box that is moving at a constant speed until it hits a patch of sticky snow where it experiences a frictional force of 12N
Ff
frictional force of 12N.
a) What is the acceleration of the box?
mg
the box?
b) What is the coefficient of kinetic friction between the b d h ? =→=
=amaF
maFNet
3012box and the snow?
=
=→=
a
amaFf 3012== kNkf mgFF µµ 0.4 m/s/s
==
k
k
µµ )8.9)(30(12
0.04kµ
Optional!!-The “not so much fun” begins….Now suppose your friend decides to help by pulling the box
across the snow using a rope that is at some angle from the horizontal She begins by experimenting with the angle of pullhorizontal. She begins by experimenting with the angle of pulland decides that 40 degrees is NOT optimal. At what angle, θ, will the minimum force be required to pull the sled with a constant velocity?constant velocity?
Let’s start by making a function for “F” in terms of “theta” using our equations of motion. θθ sinsin NN FmgFmgFF −=→=+
FN F θµθµθ
)sin(coscos
K
NkF
FmgFFFF−=
==
Fsinθ
Ff
θ
µθµθµθµθ
θµµθ
)sin(cossincos
sincos
kk
kk
mgFmgFF
FmgF
+=+
−=
Fcosθ
Fsinθ
mgθµθ
µθ
µθµθ
sincos)(
)sin(cos
k
k
kk
mgF
mgF
+=
=+
θµθ sincos k+
Optional-What does this graph look p g plike?
θθµθ
i)( kmgF =
Theta Force
1 11.7536
10 11.8579θµθ sincos k+ 10 11.8579
20 12.3351
30 13.2728
40 14.8531F Th t 50 17.4629
60 21.9961
70 30.9793
Force vs. Theta
30
35
15
20
25
orce
(N) Does this graph tell
you the angle needed f i i f ?
5
10
15Fo for a minimum force?
00 10 20 30 40 50 60 70 80
Theta (degrees)
Optional-What does this graph look like?theta Force
0.5 11.7563
1 11 7536 θµθµθ
sincos)( kmgF
+=
1 11.7536
1.5 11.7517
2 11.7508
2.5 11.7507
θµθ sincos k+
Force vs. Theta
3 11.7515
3.5 11.7532
4 11.7558
4 5 11 759311.762
11.764
11.766
4.5 11.7593
5 11.7638
11 756
11.758
11.76
Forc
e (N
)
Could this graph tell
11.752
11.754
11.756Could this graph tell you the angle needed for a minimum force?
11.750 1 2 3 4 5 6
Theta (degrees)
What do you notice about the SLOPE at this minimum force?
Optional-Taking the derivativep gHere is the point. If we find the
derivative of the function Force vs. Thetaderivative of the function and set the derivative equal to ZERO, we can find the ANGLE t thi i i
11.764
11.766
ANGLE at this minimum. Remember that the derivative is the SLOPE of 11.758
11.76
11.762
rce
(N)
the tangent line. The tangent line’s slope is zero at the minimum force and 11 752
11.754
11.756For
at the minimum force and thus can be used to determine the angle we
11.75
11.752
0 1 2 3 4 5 6
Theta (degrees)
need.Theta (degrees)
This tells us that our minimum force is somewhere between 2 & 3 degrees.
Optional-Taking the derivative using the Chain Rule
θµθµθ mgF
k
k
sincos)(
+=
θµθµθµθµµ
mgdmgd
dF kkk
k
k
))sin(cos()
sincos( 1−+
=+
=θθθ ddd
==
)16()3(cos)()3sin()(
2
2
xxxxfxxxf
++=′
+=
)()()(f
Derivative ofoutside function
Leave insidefunction alone
Derivative ofinside function
)3cos()16()( 2 xxxxf ++=′
Optional-Taking the derivative using the Chain Rule1))sin(cos(
θθµθµ
θkk
dmgd
ddF +
=−
2 functionoutsideofderivative)sincos1(
constants
θµθ
µθθ
kmgdd
+
=−
f ii idfd i iialonefunction inside theleaving as wellas
functionoutsideof derivative)sincos1(-
θθ
θµθ k =+
)cossin()(
functioninside ofderivativecossinθµθµθ
θµθ
kk
k
mgF +−−=′
=+−
2)sincos()(
θµθθ
k
F+
Now we set the derivative equal to ZERO and solve for theta!Now we set the derivative equal to ZERO and solve for theta!
Optional-Setting the derivative equal to zero
++−−
=′θµθ
θµθµθ)sincos(
)cossin()( 2kkmgF
+−−=
+θµθµθµθ
)cossin(0
)sincos(
kk
k
mg
+−=+
=
θµθθµθ
cossin0)sincos(
0 2
k
k
=+
θµθθµθ
cossincossin0
k
k
→=
=−− µθ
µθ
)04.0(tan)(tan
tan11
k
k
=→
θµθ )04.0(tan)(tan k
2.29°
NEWTON’S FIRST LAW ANDPHYSICAL THERAPYPHYSICAL THERAPY
PINCHED NERVESThe neck is a complex place when it The neck is a complex place when it comes to nerves, and neck pain can stem from all manner of things, including a pinched nerve in the g pshoulder. Nerves can become trapped in the shoulder itself, due to the cramped conditions in an area called the brachial plexus. The brachial plexus has a number of cervical nerves travelling through it to the upper limbs.
Note: Pinched shoulder nerves Note: Pinched shoulder nerves prevent people from lifting their arm.
TREATMENTS?Ice – This will reduce inflammation.
What if ice does not work?
Physical Therapy Routine - TRACTION
CERVICAL TRACTIONO t h i d b h i l th i t One technique used by physical therapist
and chiropractors to provide pain relief and improve motion is cervical traction. Traction gently extends the neck g yopening the spaces between the cervical vertebrae and temporarily alleviating pressure on the affected discs. Neck traction can either be done continuously traction can either be done continuously or intermittently, alternating between short periods of pulling and resting.
It’s also possible to do cervical traction at home such as the device shown here. There are pulley systems that you can There are pulley systems that you can hook up to a doorway, or devices that will enable you to perform cervical traction while lying down.
PHYSICS APPLICATION
Cervical traction works based on the idea of EQUILIBRIUM .
netF =∑ 0T
waterbagNeck mgT =∑
T
mg
mg