# The Nature of Friction by nikeborome

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```									  The Nature of Friction

Friction (Ff) is a force that works
against the motion of an object!
Friction involves objects that are in
contact with one another!
Different Types of Friction
• Static friction- the force you must overcome
to get an object sliding.
• Sliding friction- the force of friction that
exists as one surface slides past another.
Frictionstatic > Frictionsliding
because of inertia
• Rolling friction- wheels on the ground
• Fluid friction- air resistance, water drag
Characteristics of Sliding
Friction
1) Friction acts parallel to the surface and
opposite of the motion!
2) Friction depends on the type of materials in
contact!
3) Sliding friction is always less than
starting(static) friction!
4) Friction is independent of the surface area
in contact!
5) Friction depends upon the Normal Force!
Measuring Sliding Friction
The amount of friction between two surfaces is
represented by the Coefficient of Sliding Friction.
Coefficient of Sliding Friction (µ) depends
directly upon the amount of friction force (Ff) and
inversely upon the force pressing the surfaces
together (N).

Ff
µ=
N
Examples of Friction force diagrams:
A box pulled horizontally across a level
floor at constant speed:
Balanced Forces?
N
Yes- constant speed!
Fa
Ff                         Fa = Ff

Fw = N
Fw
A crate sliding down an incline at constant
speed:
N             Balanced Forces?
Ff       Yes- constant speed!

Fp                     Ff = Fp
ø
N = Fn
ø   Fn
Fw
A box weighing 450 N is pulled along a level floor
at constant speed by a rope that makes an angle of
30.0˚ with the floor. If the force in the rope is 260
N, what is the coefficient of sliding friction?
Fw= 450 N
Fa = 260 N
ø = 30.0˚        N
Fa
F               ø    Fv
f

Fh
Fw
Balanced Forces? Yes- constant speed!
x direction:          Fh = Ff

y direction:        Fw = N + Fv

Fh = Ff
µ = Ff
cosø = Fh/Fa
N
Fh= cosø(Fa)
= cos(30.0˚)(260 N) = 225N
N = Fw - Fv     sinø = Fv/Fa
Fv = sinø(Fa)
= sin(30.0˚)(260 N) = 130 N

N = 450 N - 130 N = 320 N

µ = Ff   = 225 N       = .703
N      320 N
A force of 225 N is applied horizontally to a box
of mass 40.0 kg and it produces an acceleration of
2.50 m/s2. What must be the coefficient of
friction?
Fa = 225 N
N
m = 40.0 kg
Fw = mg = 392 N
Ff             Fa
a = 2.50 m/s2
µ=?
Fw
Balanced Forces?        No- Acceleration!

x direction:      Fa > Ff
Fa - Ff = F

y direction: Fw = N              F = ma
Ff = Fa - F   = (40.0 kg)(2.50m/s2)
µ = Ff           = 225 N - 100 N        = 100 N
= 125 N
N
= 125 N
= .319
392 N
A rope is used to pull a 75.0 kg crate up a 35.0˚
incline at constant speed. If the coefficient of
friction between the crate and the plane is .480,
what must the tension in the rope be?
N                           m = 75.0 kg
Fa
Fw = mg = 735 N
Fp
Ff                              ø = 35.0˚
ø                 µ = .480
ø              Fa = ?
Fn
Fw
Balanced Forces?     Yes- constant speed!

Fa = Ff + Fp      N = Fn

Fp = sinø(Fw)= sin(35.0)(735N)
= 422N
Fn = cosø(Fw)= cos(35.0)(735N)
= 602 N
Ff = µN = (.480)(602 N) = 289N

Fa = 422 N + 289 N = 711 N
If the rope breaks while the crate is at rest on the
ramp, how far down the ramp will the crate be
when it speed hits 5.25 m/s?

N              vi = 0
Ff    vf = 5.25 m/s
Fp                      ∆d = ?

ø                  Balanced?
No!- Acceleration!
ø    Fn
Fw
Fp > Ff             N = Fn
Fp - Ff = F

F   = 133 N
∆d = vf2 - vi2       a=
m     75.0 kg
2a
= 1.77 m/s2
∆d = (5.25 m/s)2
2(1.77 m/s2)     F = Fp - Ff
= 422 N - 289 N = 133 N
= 7.79 m
A box of mass 125 kg is accelerated at 1.50 m/s2
across a level floor by a rope tied at a 28.5˚ angle
to the horizontal. If the tension in the rope is
found to be 555 N, what is µ?
m = 125 kg
N                      Fw = mg = 1230 N
Fa            a = 1.50 m/s2
Ff                       Fv ø = 28.5˚
ø
Fa = 555 N
Fh
µ=?
Fw
Balanced? No- Acceleration!
Fh > Ff             Fw = N + Fv
Fh - Ff = F
Ff = Fh - F
µ = Ff
Ff = 488 N - 188 N = 300 N
N

F = ma
Fh = cosø(Fa)
= cos(28.5˚)(555 N)     =(125 kg)(1.50 m/s2)
= 188 N
= 488 N
N = Fw - Fv          Fv = sinø(Fa)
= sin(28.5˚)(555 N) = 265N

N = 1230 N - 265 N = 965 N

µ = Ff        = 300 N
= .311
N          965 N
A 56.0 kg box is accelerated across a level floor
by a force applied horizontally to the box. If the
coefficient of friction is .340 and the box is
accelerated at 3.00 m/s2, what force is being used
to move the box?
A 565 N crate is pulled up a 25.0˚ incline at
constant speed. If the coefficient of sliding
friction is measured as .250, what must be the
tension in the rope used to pull the crate up the
incline?
A 3.00 kg wood box slides from rest down a 35.0˚
inclined plane. How long does it take the box to
reach the bottom of the 4.75 m incline if the
coefficient of friction is .350?

A 60.0 kg crate is attached to a weight by a cord
that passes over a frictionless pulley. A) If the
coefficient of friction is .500, what weight will
keep the crate moving up a 20.0˚ incline at
constant speed? B) If the cord is cut when the
crate is at rest at the top, how far will the crate
have slid down the incline by the time its speed
reaches 7.50 m/s?

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