Download Apparent Weight – Downward Acceleration Alternative View

Apparent Weight – Downward Acceleration
Alternative View
The elevator is accelerated
downward.
!a
y
!w
!FN
Net force in the new y direction (downward)
acting on the person:
Fnet = w − FN = ma
So apparent weight is: FN = w − ma = 400 N
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Static Friction
The static friction force between dry, unlubricated surfaces is:
• independent of area of contact for hard surfaces – not true for
deformable surfaces, such as tire rubber.
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A cup of coffee sits on a table in a plane (µs = 0.30). The
plane accelerates. What is the maximum acceleration
before the cup starts to slide?
For the cup not to slide: fs = ma (second law)
fsmax = µsFN = µs mg and fsmax = mamax
FN
fs
mg
a
So, mamax = µs mg
→ amax = µs g = 0.3g
amax = 2.94 m/s2
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Kinetic Friction
Occurs when surfaces slide. Friction force opposes the relative
motion of the surfaces.
The magnitude of the kinetic friction force is:
fk = µk FN
µk = coefficient of kinetic friction
The kinetic friction force is:
• independent of area of contact between the surfaces
• independent of relative speed of the surfaces for small speeds –
high speed of sliding may cause heating and a change in the
properties of the surfaces.
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Serway, Physics for Scientists and Engineers
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Yahoo search –
The end of two bones which
meet to form the joint are
covered with articular
cartilage, a surface material
much like a tread of a tire. Its
strength comes from tough
fibers called collagen. The
joint surface cartilage is well
lubricated - more slippery
than well-manufactured
ball bearings... Its living cells
are nourished by joint fluid,
called synovial fluid which is
also extremely good
lubrication.
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A block of weight 45 N rests on a horizontal table. Will the
block move and, if so, what is its acceleration?
FN
µs = 0.650, µk = 0.420
As there are only two forces
with vertical components:
P = 36 N
FN = w = 45 N
The maximum static friction force is:
m
Ff
Fs = µsFN = 0.65 × 45 = 29.25 N
w = 45 N
w = mg
This is less than the force applied, so the block starts sliding.
The kinetic friction force is: Fk = µk FN = 0.42 × 45 = 18.9 N
The acceleration is:
a = Fnet /m = (36 − 18.9 N)/(45/g kg) = 3.72 m/s2
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A 20 kg sled is pulled across a horizontal surface at
constant speed.
FN
Fk
F = 80 N
y
30o
x
mg
What is the coefficient of kinetic friction?
Speed is constant, so net force on sled = 0.
x direction: F cos 30◦ = Fk = µk FN
y direction: F sin 30◦ + FN = mg
so, FN = F cos 30◦/µk
and, FN = mg − F sin 30◦
Therefore, FN = F cos 30◦/µk = mg − F sin 30◦
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Therefore, FN = F cos 30◦/µk = mg − F sin 30◦
F cos 30◦
µk =
mg − F sin 30◦
Substitute F = 80 N, m = 20 kg
80 cos 30◦
µk =
= 0.444
◦
20 × 9.8 − 80 sin 30
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A crate is being pushed across the floor at constant
velocity by applying a downward force at an angle θ to
the horizontal.
Show that θ if is greater than a certain value, then it is
impossible to push the crate, no matter how large the
force.
F
Friction force: Fk = µk FN
FN
θ
How large is FN?
µk = 0.41
Forces in vertical direction:
F
k
mg
FN − F sin q − mg = 0
So FN = F sin q + mg
The kinetic friction force is then Fk = µk FN = µk (F sin q + mg)
The force pushing the crate sideways is F cos q
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The kinetic friction force is then Fk = µk FN = µk (F sin q + mg)
The force pushing the crate sideways is F cos q
If the friction force is greater than the sideways pushing force,
then the crate cannot be moved, no matter how great the applied
force.
That is, µk (F sin q + mg) > F cos q
If F is very large, then mg is negligible by comparison, and then:
µk F sin q > F cos q
So, tan q > 1/µk = 1/0.41
Or, q > 68◦
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