Friction - cgrahamphysics.files.wordpress.com · Friction ©cgrahamphysics.com 2015 Friction is the force that occurs when two surfaces are in contact Friction opposes motion Dynamic
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Friction
©cgrahamphysics.com 2015
Friction
©cgrahamphysics.com 2015
Friction is the force that occurs when two surfaces are in contact
Friction opposes motion
Dynamic friction: an object is moving
Static friction: friction prevents object to move
Both, dynamic and static friction depends on the surfaces in contact
Recall that friction acts opposite to the intended direction of motion, and parallel to the contact surface.
Suppose we begin to pull a crate to the right, with gradually increasing force.
We plot the applied force, and the friction force, as functions of time:
T f
T f
T f
T f
T f
Fo
rce
Time
tension
friction
static
friction
dynamic
friction
static dynamic
Solid Friction compared
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During the static phase, the static friction force Fs exactly matches the applied (tension) force.
Fs increases linearly until it reaches a maximum value Fs,max.
The friction force then almost instantaneously decreases to a constant value Fd, called the dynamic friction force.
Take note of the following general properties of the friction force:
Forc
e
Time
tension
friction
static dynamic
Fs,max
0 ≤ Fs ≤ Fs,max Fd < Fs,max Fd = a constant
Fd
Static friction changes to dynamic friction
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Static and dynamic friction
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Static friction
𝐹𝑆 ≤ 𝜇𝑆𝑁
𝐹𝑆 = frictional force
𝜇𝑆 = coefficient of static friction
N = the normal reaction force
≤ indicates that static friction can vary from zero to a maximum value.
When 𝐹𝑆 = pulling force object just about to start moving
Dynamic friction
𝐹𝐷 = 𝜇𝐷𝑁
𝜇𝐷= coefficient of dynamic friction
If the object is moving at constant velocity, then 𝐹𝐷= pulling force
The values 𝜇𝑆 and 𝜇𝐷 vary greatly depending on the two surfaces in contact
What causes friction
If two metal surfaces are smoothened and polished, the more strongly they stick together if brought in contact.
In fact, if suitably polished in a vacuum, they will stick so hard that they cannot be separated.
We say that the two pieces of metal have been cold-welded.
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At the atomic level small peaks on one surface cold weld with small peaks on the other surface.
Applying the initial sideways force, all of the cold welds oppose the motion.
If the force is sufficiently large, the cold welds
break, and new peaks contact each other and cold weld.
If the surfaces remain in relative sliding motion,
fewer welds have a chance to form.
We define the unitless constant, called
the coefficient of friction μ, which
depends on the composition of the two surfaces, as the ratio of Ff
and R.
surface 1
surface 2
surface 1
surface 2
cold welds
surface 1
surface 2
Atomic level of friction
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Example
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A box is pushed across a level floor at a constant speed with a force of 280N at 450 to the floor. The mass of the box is 50kg.
Calculate
a) the vertical component of the force
𝐹𝑦 = 𝐹 sin 45 = 280 sin 45 = 198𝑁
b) the weight of the box
W = mg = 50 x 10 = 500N
c) the horizontal component of the force
𝐹𝑥 = 𝐹 cos 45 = 280 cos 45 = 198𝑁
F cos 45
F sin 45
…continued
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d) the coefficient of dynamic friction between the box and the floor
𝜇𝐷 =𝐹𝑓
𝑁
Since it is moving at constant speed, 𝐹𝑓 = 𝐹𝑥 = 198N
𝜇𝐷 =𝐹𝑓
𝑁=
198
698= 0.28
N
W
𝐹𝑦 N
𝐹𝑦
W
N = 𝐹𝑦 + W = 198 + 500 = 698N
Different alignment for coordinates
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If we have to solve problems that involve a slope, we can choose to change the direction of the x and y axis.
A skier places a pair of skies on a snow slope that is at an angle of 1.70 to the horizontal.
y
x
The coefficient of static friction between The skies and the snow is 0.025 Determine whether the skies will Slide by themselves
Continued
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FBD:
N = W cos 1.7
𝐹𝑓 = 𝑊 sin 1.7
𝜇𝑠 =𝐹𝑓
𝑁=
𝑊 sin 1.7
W cos 1.7 = tan 1.7
1.7
W
N
friction
W cos 1.7
Wsin 1.7
If 𝜇𝑠 > 0.025, the skier will slide away 𝜇𝑠= tan 1.7 = 0.029 it will slide
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You are pulling your friends, who are sitting in a wagon,
up a hill in a steady speed. Your friend and the wagon have a combined mass of 82 kg and the street has a slope
of 𝟔. 𝟓𝟎. If the coefficient of dynamic friction between the wagon and the street is 0.10, what is the tension in the
string?
FBD
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Y – direction: N = W cos 6.5
X – direction: 𝐹𝑓 + 𝑊𝑠𝑖𝑛 6.5 = 𝑇
𝐹𝑓=𝜇 𝑁 = 𝜇𝑊 cos 6.5
At steady speed, a = 0
𝑇 = 𝜇𝑊𝑐𝑜𝑠 6.5 + 𝑊 sin 6.5 = 𝑊(𝜇 cos 6.5 + sin 6.5) = 82 × 10(0.10 cos 6.5 + sin 6.5) = 174.3 ~1.7 × 102𝑁
6.5
W
N
Wcos 6.5
𝐹𝑓
T
W sin 6.5
Example
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A trunk on top of an inclined ramp, which makes an angle of 350 with the ground, starts to slide down the ramp. If the coefficient of friction between the ramp and the trunk is 0.42, what is the acceleration?
350
W
N 𝐹𝑓
W cos 35
W sin 35
Y – direction: N = W cos 35
X – direction:
W sin 35 𝐹𝑓
ma
W sin 35
𝐹𝑓
ma + 𝐹𝑓 = W sin 35
…continued
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N = W cos 35
ma + 𝐹𝑓 = W sin 35
𝐹𝑓= 𝜇 𝑁 = 𝜇𝑊 cos 35
ma + 𝜇𝑊 cos 35 = W sin 35
a = 𝑊 sin 35 −𝜇𝑊 cos 35
𝑚=
𝑚𝑔(sin 35 −𝜇 cos 35)
𝑚
a = 10 (sin 35 – 0.42 cos 35) = 2.3𝑚𝑠−2
b) if the trunk slides 1.3m before reaching the bottom of the ramp, for what time interval did it slide?
𝑠 = 𝑢𝑡 +1
2𝑎𝑡2
𝑡 =2𝑠
𝑎=
2×1.3
2.3= 1.063~1.1𝑠
0
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