friction - mr.neave's...
TRANSCRIPT
![Page 1: FRICTION - Mr.Neave's Websitemrneave.weebly.com/.../sph3u_day_32_october_21_friction.pdfFRICTION ´a force between two surfaces that opposes motion «acts opposite the direction of](https://reader035.vdocument.in/reader035/viewer/2022071501/61212c15ba3d2949384833bd/html5/thumbnails/1.jpg)
FRICTION
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FRICTION
´ a force between two surfaces that opposes motion « acts opposite the direction of motion
´ static – object is stationary (trying to move)
´ kinetic – object is moving
´ Ff does not depend on velocity or surface area
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COEFFICIENT OF FRICTION
´ Coefficient of Friction – the ratio of the magnitude of the force of friction between 2 surfaces to the magnitude of the normal force between the surfaces f
N
FF
P •mu
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NORMAL FORCE
´ Equal to the weight of the object ´ FN=mg ay
= 9.8%2
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FRICTION
Ff=μFN Ff=force of friction (N)
μ=coefficient of friction FN=normal force (N)
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COEFFICIENT OF STATIC FRICTION
´ static – stationary (not moving ) ´ Coefficient of Static Friction:
« the ratio of the magnitude of the maximum force of static friction to the magnitude of the normal force
ss
N
FF
P
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COEFFICIENT OF KINETIC FRICTION
´ kinetic – moving ´ Coefficient of Kinetic Friction:
« the ratio of the magnitude of the force of kinetic friction to the magnitude of the normal force
kk
N
FF
P
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WHERE DO YOU WANT FRICTION... ´ as low as possible?
´ as high as possible?
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EXAMPLE #1
´In the horizontal starting area for a bobsled race, 4 athletes, with a combined mass of 295 kg, need a horizontal force of 41 N [forward] to get the 315 kg sled moving. Calculate the coefficient of static friction. Fs - 41N Fw
⇒•→FA ? 41N
b Fg
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Fs = 41NFn =
? - mgMs = ?
Gives Fs .
. 41Nm .
. 315kgRequired Ms
Analyst
FN=mgMs=¥
n
Steps: Fn.. (3151*9.845)Fw=3087N
Us : 41¥#Ms = 0.013
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EXAMPLE #2
´A truck’s brakes are applied so hard that the truck goes into a skid on a dry asphalt road. The truck and its contents have a mass of 4.2x103 kg, calculate the force of kinetic friction on the truck.
Use µ,
-0.65
pgµFnGiron .
.
4.2×103
••
µ , ,=FLTable pg .
1709 btg t.ua#FN
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tn.mg2×103 kg) ( 9.8 ;DFn = 41 160 N
Esq,,F÷
5 ( 41160 D
Fk = 26754N
;.Fk=270€
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EXAMPLE #3
A loaded 4-man bobsled with a mass of 615kg experiences a frictional force of 66N as it slides down the track. Calculate the coefficient of friction.
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HOMEWORK
Pg 171 #t3Pg 172 #
:1