Post on 19-Mar-2018
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Ch06: Force and Motion - IIStatic FrictionKinetic Friction Friction propertiesUniform circular motionCentripetal accelerationNewton’s law in Uniform circular motion
6.2: FrictionFrictional forces are common and important in our daily lives:
due to friction we can walk on earthcar under danger can stop due to friction when breakingyou can’t move heavy crate when pushing it due to friction
Frictional force is due to the interaction of two adjacent surfaces at the atomic or molecular level force is needed to break the welding between surfaces frictional force due to action reaction
Friction is always parallel to the surface of interaction and opposite the direction of motion.
6.2: FrictionFrictional Force: motion of a crate with applied forces
With no applied force F , There is no attempt at sliding no friction
Force F attempts sliding but is balanced by the frictional force No motion. Static frictional force fs
Force F is now stronger but is still balanced by the frictional force. No motion. Larger static frictional force fs
6.2: FrictionFrictional Force: motion of a crate with applied forces
Finally, the applied force F has overcome the static frictional force Block slides and accelerates Kinematics friction fk
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6.3: Properties of FrictionProperty 1. If the body does not move, then the static frictional force fs and
the component of F that is parallel to the surface balance each other. They are equal in magnitude, and is fs directed opposite that component of F.
Property 2. The magnitude of fs has a maximum value fs,max that is given by
Property 3. If the body begins to slide along the surface, the magnitude of the frictional force rapidly decreases to a value fk given by
µk is coefficient of kinetic friction
µs is coefficient of static friction
Friction is proportional to the normal force.Friction is independent of area.
The coefficients do not depend on weight or speed and generally range between 0.003 and 1.µs is generally larger than µk .
6.3: Properties of Friction: ExampleA car driving with a speed v0. an emergency breaking made It to stop in a distance of 290 m as shown. If µkbetween tiers and street is 0.6, find it’s initial speed v0
2/88.50.6)(9.8)( sm
gamamgF
mafmaF
k
kNk
k
xx
−=−=
=−=⇒=−=−⇒
=−⇒
=∑
µµµ
smv
xav
xav
xavv
/58)029)(88.5(2
2
20
2
0
0
20
20
2
=−−=⇒
∆−=⇒
∆+=
∆+=
fk
FN
mg
)........(225sin
0)(a 025sin y
°−=⇒
==−°+
=∑
Fmgn
mgFnmaF yy
12 kg Box on surface with friction. If µs=0.5 and µk=0.3 finda) how large should F be to get box start moving?b) once it’s moving, find the acceleration
)........(1 025cos0)(a 0cos
=−°==−
=∑
nFfFmaF
s
s
xx
µθ
Solution: analyze F to its components
a) Start moving not moving (a=0), but, fs is at its maximum
= 0
fs Fcos25°
Fsin25°
We can find n from y-direction
6.3: Properties of Friction: Example - box on rough surface
NF
mgF
mgFFmgF
s
s
ss
ss
5325sin)5.0(25cos
)8.9)(12(5.025sin25cos
)25sin25(cos025sin25cos
=+
=
°+°=⇒
=°+°=°−−°⇒
µµ
µµµµ direction)-y (from 25sin
,25cos25cos
cos
°−=
−°=⇒
=−°=−
=∑
Fmgnm
nFa
manFmafF
maF
k
k
k
xx
µµ
θfk Fcos25°
Fsin25°
b) When moving we have acceleration a and kinetic friction fk as shown
²/6.112
)25sin53)8.9)(12)((3.0(25cos53
)25sin(25cos
sma
mFmgFa k
=−−
=
−−=⇒
µ
6.3: Properties of Friction: Example - box on rough surface
3
(1)..........0sin0)(a 0sin
=−⇒==−
=∑
nmgfmg
maF
s
s
xx
µθθ
)1()2.........(cos
0)(a 0cos y
inFforsubmgn
mgn
maF
N
yy
θ
θ
=⇒
==−
=∑
12 kg Box on incline surface with friction. If µs=0.5 and µk=0.3 finda) At what angle the object start movingb) once it’s moving, find the acceleration
Solution: a)
°==
=⇒
=⇒=⇒
=−⇒
−
−
6.260.5tan
tan
tancossin
0cossin
1
1s
s
s
smgmg
µθ
µθθµθ
θµθ
Object not movng fs
6.3: Properties of Friction: Example - box on rough incline surface
12 kg Box on incline surface with friction. If µs=0.5 and µk=0.3 find
b) once it’s moving, find the acceleration
2/7.1)6.26cos0.3)(9.8)(()6.26)(sin8.9(
cossincossin
sin
smaa
ggamamgmg
mafmgmaF
k
k
k
xx
=
°−°=⇒−=⇒
=−⇒=−
=∑
θµθθµθ
θ
Static friction becomes kinetic fk<fs object accelerates at same angle in part (a)
a
6.3: Properties of Friction: Example - box on rough incline surface
Two objects of masses m1 and m2 are connected by a lightweight cord over alightweight, frictionless pulley as shown. force of magnitude F at an angle θ with the horizontal is applied to the block m1 which ison a rough surface.The coefficient of kineticfriction between the block and surface is µk. Determine the magnitude of the acceleration of the two objects.
θµθ sin,)(cos
)(
1212 FgmnbutammgmnF
amF
k
system
−=+=−−
=∑ ∑
6.3: Properties of Friction: Example – two object connected by a cord 6.5: Uniform Circular Motion
Since object is rotating in circular path Centripetal acceleration acdirected towards the center of the circle of motion (perpendicular to the velocity v).
Since we have acceleration towards the center we have Fnet produce ac
Fnet (ΣFr ) is directed towards the center
The string is applying the force to keep the ball rotating in circular path
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When the String Breaks
When the string breaks (meaning no radial force is applied anymore), the ball continues to move at constant velocity, in a straight line along the tangent to circle at the break point. v
RvmmaamF crr
2
===∑ rr
Newton’s second law in the radial direction (اتجاه نصف قطري) is
If Fr is towards the center +ve force
If Fr outwards the center -ve forceThe string is applying the force to keep the ball rotating in circular path
6.5: Uniform Circular Motion: Newton’s law
A 0.15 kg Object on a string moves in horizontal circle with a speed of 2 rev./s. If the radius of the circle is 0.6m, find the tension in the string
RvmT
RvmmaF cr
2
2
=⇒
==∑
smrsrevv /54.7)6.0(4)2(2/.2 ==== ππ
To find T we need v
NrvmT 2.14
6.0)²54.7(15.0
2
===⇒
6.5: Uniform Circular Motion: Example
RvmmaF cr
2
==∑
At the bottom of the loop
T towards the center (up)
mg out of center (down)
RvmmgT
RvmmgT
2
2
+=⇒
=−
T
mg
6.5: Uniform Circular Motion: Example –moving on a vertical circle
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motion circular maintain to speed minimum is
22
gRv
gRvmg
Rvm
=⇒
>⇒>
RvmmaF cr
2
==∑
mgRvmT
RvmmgT
−=⇒
=+
2
2
At the top of the loop
Both T towards the center (down)
Tmg
to maintain circular motion T must be +ve
6.5: Uniform Circular Motion: Example –moving on a vertical circle
Find minimum speed a bicycle can go without falling. R= 2.7 m.
Rvmmgn
maF cr
2
=+
=∑
At minimum speed bicycle s about to fall n = 0
smgRvRvmmg /1.5
2
==⇒=⇒
At the top, there is a normal force n from track on the bicycle, directed downward, in addition to the weight
6.5: Uniform Circular Motion: Example –moving on a vertical circle
Ex: car rounding a curve without skidding?
What is the maximumspeed of the car
m = 1500 kgrcurve = 120 mµs = 0.5
Car rounding curveac force to the center
It is static friction force fs
fs
n
mg
Ex: car rounding a curve without skidding?
RvmmaF cnet
2
==
gRvgRvRvmmgn
Rvmf
ss
sss
µµ
µµ
=⇒=⇒
==⇒=
max2
22
Maximum speed is when the car is about to skid (at fs,max)
smv /2.24)8.9)(120)(5.0(max ==⇒
fs
n
mg
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Banked curveWe can reduce the possibility of skidding by designing a banked curvesAssume a car driving at speed v on banked curve
with no friction. Find the optimum banking angle without skidding out of the track
RvmmaFF crx
2
===∑∑
Rvmn
2
sin =θ
grv
Rvmmgmgn
mgn22
tansincoscos
cos direction,-y
=⇒=⇒=⇒
=
θθθθ
θ
Banked curve
°==
=⇒=
−
−
42)125)(8.9(
)33(tan
tantan
21
21
2
θ
θθgRv
gRv
If v=33m/s and curve radius r=125m, what is the banking angle so that the car will not slide
Review
Maximum static frictional force (object about to move)
Kinetic frictional force (object is moving)
RvmmaF cr
2
==∑Newton’s law in uniform circular motion
Static frictional force equals applied force (object is not moving)