Force of Friction: Use lower case f to represent it.
Two types:
Static friction (when object is at rest on a surface)
Direction: Against “desired”, “possible” motion
Kinetic friction (when object is in motion on a surface)
Direction: Against actual motion
At any given time, only has one of these two.
Surface roughness: µs , µk coefficient of friction
Static friction: A variable amount
fs can be from 0 to µs N to prevent slipping.
fs always matches pulling force,
but only up to the limit µsN,
fsmax = µs N is the point of setting object in to motion
Kinetic friction: constant size fk =µk .N (always)
fs=
Fpull
(always)
Opposite direction vs. pull (always)
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f without vector hat, only mean the size of the force
2. Up to some limit. Fpull is too big for fs (static friction)
to balance,
When F pull > fs (max) = µs N , object starts to move.
The upper limit of fs(max) is determined by
------N (support force=press force)
------µs (surface roughness)
3. When it is moving, there is only kinetic friction:
fk has constant size
fk =µk .N (always) (no matter a > = <0)
fk opposite direction vs. relative velocity always
fsmax= µs .N,
in order to make objects more steady against slipping
(to increase fsmax) you should
fk= µk .N
to make objects move more easily on surface
(to reduce fk kinematic friction) you should
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For both fk and fs contact area Does not matter
Example questions: On a rough incline,
V=0 no motion
Given m,
asking for
θ,
f
µs,
maximum θ for
object to stay
µk,
1- object
2- forces? mg, N, f
3- coordinate system
4- break down forces component
Red θ + α = 90
Blue θ + α = 90
Blue θ always = Red θ
5- Fnetx = m ax = mg sin θ –fs=0
Fnety = m ay = N- mg cos θ=0
6- solve the equation
fs = mg sin θ
larger θ, larger sinθ, need more fs
N = mg cos θ
larger θ, less cosθ, less support.
You can increase θ till the maximum limit, before it
starts to glide.
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fs = mg sin θ
N = mg cos θ
You can increase θ
till the maximum limit,
before it starts to glide.
at Maximum angle, θmax ,
N= mg cosθmax
Demo:
On rough incline, We measure that
θmax = 35 degree
µs = tan(35)= 0.70 between the two surfaces
Glass on glass µs = 0.94 θmax= 43
Ski on snow, steel on ice µs = 0.1
θmax= 5.7
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Steel on steel µs = 0.74 θmax = 36.5
steel on steel µs = 0.74
θmax = 36.5
Example: when θ is 45o steel on steel slope, what‘s a?
fk or fs ?
Only fk ; f = µk N,
no fs now
m a y = F net y = N – mg cos θ = m ay = 0
m ax = F net x = mg sinθ –µk N = m g ( sin θ – µk cos θ)
ax = g ( sin θ – µk cos θ )
= 9.81 (0.707 – 0.57 x 0.707)
= 2.98 (m/s2)
If v0 = 0 , x0 =0,
x = ½ ax t2 = …. v= ax t = ….
You can then calculate v , ∆x, etc.
Sizes of static friction and kinetic friction
To start the motion is harder than to keep the motion on surfaces
To start, we need F (drag, pull …) ≥ fs max = µs . N
Once it started, we know that existing F( drag, pull…) is more
than fk , object will accelerate, if the same F is kept.
That’s why to maintain the motion is easier. (inertia is working)
If F (drag, pull …) > fs max, it must > fk. Once it moves, the same
drag force is always enough to accelerate it against kinetic f.
Friction on rolling is less than friction during slipping.
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Translational equilibrium
If ΣF=0,
net force ( in all directions) = 0 ⇒ Zero a
Object is in equilibrium
rest or constant velocity motion, along a straight line
Translational equilibrium
F net x = 0;
F net y = 0
Both x and y direction total force = 0
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Connected objects :
inextensible massless string (rope)
|a1| = |a2|
|∆x1| = |∆x2|,
|T1| = |T2|
Tension inside a massless rope is equal everywhere.
:
Magnitude of a1 and a2 are the same!
Direction of a1 and a2 can vary….
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