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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

5

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)