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example
F-F12 = m1 a
56 - m1 a = F12
56 -12 x1.75 =35 N
22 May 2017
Norah Ali Al-moneef
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If objects in motion tend to stay in motion, why don’t
moving objects keep moving forever?
Things don’t keep moving forever because there’s almost always
an unbalanced force acting upon it.
A book sliding across a table slows down and stops because of
the force of friction.
Why Doesn’t Gravity Make the Box Fall?
Force of Floor acting on Box
Force from floor on box cancels
gravity.
If the floor vanished, the box would
begin to fall.
Force of Earth acting on Box (weight)
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Force on box
by person
Force on floor by box
Force on person
by box
Force on box by floor
It’s the sum of all the forces that determines the acceleration.
Every force has an equal & opposite partner.
What’s missing in this picture?
A pair of forces acting between person and floor.
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Friction Mechanism
Corrugations in the surfaces grind when things
slide.
Lubricants fill in the gaps and let things slide
more easily.
Don’t all forces then cancel?
• How does anything ever move
(accelerate) if every force has an
opposing pair?
• The important thing is the net force
on the object of interest
Net Force
on box
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Friction Forces
When two surfaces are in contact, friction forces oppose
relative motion or impending motion.
F
Friction forces are parallel to the
surfaces in contact and oppose
motion or impending motion.
Static Friction: No relative
motion.
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Kinetic Friction: Relative
motion.
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Example
When you push a book against a wall, the static friction between
the wall and the book can prevent it from falling.
If you press harder, the friction force will be:
A. Larger than before
B. The same
C. Smaller than before.
For the book not to fall
down, fS = W
Nbook,hand
Pushing harder (increasing Nbook,hand)
increases Nbook,wall and therefore fS MAX
increases, but not the actual value of fS
that we had, which needs to continue to
be exactly W !
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fS,book,wall
Nbook,wal
l
Wbook,Earth
7
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Friction and the Normal Force
8N
4N
12 N
N
N2 N
N
4N
6N
The force required to overcome static or kinetic friction is
proportional to the normal force,N
fs = msN
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fk = mkN
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Friction forces are independent of area.
4N
4N
If the total mass pulled is constant, the same force (4 N) is
required to overcome friction even with twice the area of
contact.
For this to be true, it is essential that ALL other variables
be rigidly controlled.
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The Static Friction Force
When an attempt is made to move an object on a
surface, static friction slowly increases to a MAXIMUM
value.
N
fs
n
F
W
In this module, when we use the following equation,
we refer only to the maximum value of static friction
and simply write:
fs = msN
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Constant or Impending Motion
For motion that is impending and for motion at
constant speed, the resultant force is zero and SF = 0.
(Equilibrium)
F
fs
Rest
F– fs = 0
fk
F
Constant Speed
F – fk = 0
Here the weight and normal forces are balanced
and do not affect motion.
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Friction and Acceleration
When F is greater than the maximum fs the resultant
force produces acceleration.
a
fk
F
Constant Speed
fk = mkN
Note that the kinetic friction force remains constant
even as the velocity increases.
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Friction force opposes motion by acting opposite the object’s velocity and
direction of travel.
“When 2 things are in contact with each other, there will be
friction acting between them”
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Coefficient of Friction
• µ is determined from experiment.
• Determine how much force is required, f, to just get an object moving on a level
surface.
• Weigh the object to determine the contact force, N. (Level surface N = weight = mg)
• Divide f/N. This is the coefficient of friction and characterizes the slickness of the
surface.
Two Kind of Coefficients
• Static Coefficient of Friction, µs-for objects at rest.
• Kinetic Coefficient of Friction, µk-for objects in motion.
• µk < µs ( Wood on wood, µs = 0.06, µk = 0.04 Rubber of dry concrete, µs = 1.2, µk = 0.9)
• Two factors govern the magnitude of the force or maximum static friction or kinetic
friction in any situation: the coefficient of friction, represented by the small Greek letter
mu (m), and the normal (perpendicular) reaction force (N).
F = mN
•
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Static Friction...
• Opposes motion except here a = 0 is the constant
x:
Fnet fS = 0
y:
N = mg


While the block is static: fS Fnet (unlike kinetic friction)
fs is NOT fixed in magnitude
N
y
Fnet
x
fS
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mg
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Static Friction...
• The maximum possible force that the friction between two
objects can provide is fMAX = mSN, where ms is the “coefficient
of static friction”.
– So fS  mS N.
– As one increases F, fS gets bigger until fS = mSN and the
object “breaks loose” and starts to move.
N
y
F
x
fS
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mg
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Static Friction...
• mS is discovered by increasing F until the block starts to slide:
x:
FMAX mSN = 0
y:
N = mg
mS  FMAX / mg
N
FMAX
mSmg
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Active Figure
y
x
mg
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 There is some maximum static frictional force, fsmax.
Once the applied force exceeds it, the book moves
f
max
S
 mS N
Magnitudes not
vectors
• ms is the coefficient of static friction, it is a
dimensionless number, different for each
surface-object pair (wood-wood, wood-metal);
also depends on surface preparation
• ms does not depend on the mass or surface
area of the object
•If no applied vertical force
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f Smax  m S mg
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Forces of Friction
• Static friction, fs
• Kinetic friction, fk
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The following empirical laws hold true about friction:
- Friction force, f, is proportional to normal force, n.
f s  ms N
f k  mk N
- ms and mk: coefficients of static and kinetic friction, respectively
- Direction of frictional force is opposite to direction of relative
motion
- Values of ms and mk depend on nature of surface.
- ms and mk don’t depend on the area of contact.
- ms and mk don’t depend on speed.
- ms, max is usually a bit larger than mk.
- Range from about 0.003 (mk for synovial joints in humans) to 1 (ms
for rubber on concrete). See table 5.2 in book.
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Kinetic friction
When the relative motion between the two objects is not zero. It “slows
down” the sliding motion.
Experimentally, it is observed that the kinetic friction force between two
surfaces:
• Is parallel to the surface (and thus perpendicular to the normal)
• Its magnitude does not depend on the speed (except when v = 0) or
on the area of the contact surfaces.
• Its magnitude is proportional to the magnitude of the normal force
between the two surfaces:
normal force between the two surfaces:
Use the correct one!!! (N is NOT always mg)
fk = μ k N
μk = coefficient of kinetic friction
(depends on the materials)
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EXAMPLE : If mk = 0.3 and ms = 0.5, what horizontal
pull F is required to just start a 250-N block
moving?
1. Draw sketch and free-body diagram as
shown.
2. List givens and label what is to be found
N
mk = 0.3; ms = 0.5; W = 250 N
Find: F = ? to just start
F
fs
F– fs = 0
+
W
3. Next we find fs from:
fs = msN = 0.5 (250 N)
4.For this case:
F– fs = 0
F = fs = 0.5 (250 N)
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EXAMPLE ms = 0.5, W = 250 N. Find F to overcome fs (max).
Now we know N = 250 N.
6. Next we find fs from:
fs = msN = 0.5 (250 N)
7. For this case:
fs
N
F
+
F– fs = 0
F = fs = 0.5 (250 N)
250 N
ms = 0.5
F= 125 N
This force (125 N) is needed to just start motion. Next we
consider F needed for constant speed.
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If mk = 0.3 and ms = 0.5, what horizontal pull F is required
to move with constant speed? (Overcoming kinetic friction)
SFy = m ay = 0
mk = 0.3
N
F
fk
N= W
fk = mkN = mk W
N -W=0
+
mg
Now:
SFx = 0;
F - fk = 0
F = fk = mkW
F = 75.0 N
F = ( 0.3)(250 N)
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The direction of the friction force, F, is always opposite of
the direction of possible motion
Possible motion
W
y
F
x
F
y
F
 0; N  W
No motion:
Impending motion:
Motion:
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N
F
F
F
x
 0; P  F  Fmax  ms N
x
 0; P  F  Fmax  ms N
x
 0; P  Fmax  ms N ; F  Fk  mk N
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The direction of the friction force, F, is always opposite of
the direction of possible motion
Possible motion
F
y
Fy
W
Fx
x
F
y
F
 0; N  W  Py
No motion:
Impending motion:
Motion:
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N
F
F
F
x
 0; Px  F  Fmax  ms N
x
 0; Px  F  Fmax  ms N
 0; P  Fmax  ms N ; F  Fk  mk N
x Norah Ali Al-moneef
x
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y
x
W
Wy
Wx
F
F
y
 0; N  Wy
N
Small P
 Fx  0; Wx  P  F  Fmax  ms N
No motion:
Impending motion:  Fx  0; Wx  P  F  Fmax  ms N
Motion:
 Fx  0; Wx  P  Fmax  ms N ; F  Fk  mk N
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y
x
W
Wy
Wx
F
F
y
 0; N  Wy
N
No motion:
Impending motion:
Motion:
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F
F
F
x
 0; Wx  F  Fmax  ms N
x
 0; Wx  F  Fmax  ms N
x
 0; Wx  Fmax  ms N ; F  Fk  mk N
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y
x
W
Wy
Wx
F
F
y
 0; N  Wy
N
Large P
 Fx  0; P  Wx  F  Fmax  ms N
No motion:
Impending motion:  Fx  0; P  Wx  F  Fmax  ms N
Motion:
 Fx  0; P  Wx  Fmax  ms N ; F  Fk  mk N
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A 10-kg box is being pulled across the table to the
right at a constant speed with a force of 50N.
a) Calculate the Force of Friction
b) Calculate the Force Normal
FN
Ff
Fa
Fa  F f  50 N
mg  N  (10)(9.8)  98 N
mg
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Suppose the same box is now pulled at an angle of 30 degrees
above the horizontal.
a) Calculate the Force of Friction
Fax  Fa cos   50 cos 30  43.3N
b) Calculate the Force Normal
F f  Fax  43.3N
FN
Ff
Fay
30
Fax
mg
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Fa
N  mg!
N  Fay  mg
N  mg  Fay  (10)(9.8)  50 sin 30
N  73N
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example
A hockey puck is given an initial speed
of 20.0 m/s. It slides 115 m before
coming to rest.
(a) Determine the coefficient of kinetic
friction between the puck and the ice.
Applying Newton's Second Law, ΣFx = -Ffk = max
ΣFy = N - mg = 0
Since Ffk = μkN
ΣFx = -μkN = -μkmg = max
ax = -μkg
vxf2 = vxi2 + 2ax(xf - xi)
0 = vxi2 - 2μkgxf
μk = 0.177
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Applying Newton's Second Law,
ΣFx = -Ffk = max
ΣFy = N - mg = 0
Since Ffk = μkN
ΣFx = -μkN = -μkmg = max
ax = -μkg
vxf2 = vxi2 + 2ax(xf - xi)
0 = vxi2 - 2μkgxf
Solving for μk ,
μk = 0.177
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example
A 1200 kg car moving at 100 km/h coasts to a stop in 25 seconds.
What is the value of the static coefficient of friction?
m = 1200 kg
vo = 100km/h = 27.8 m/s
t = 25 s
µ=F/N
where F = ma & N=mg
= ma/mg
=(m(|v – vo|)/t)/mg
vo
f
From v = vo + at,
a = (v-vo)/t
=(27.8m/s/25 s )/ 9.8 m/s2)
µ = 0.113
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Example
What is the weight of an object that is being pulled at a
constant velocity by a force of 25 N if the coefficient of sliding
friction between the object and the surface is 0.75?
∑ Fy = 0
N - W =0
∑ Fx = ma
(a =0
∑ Fx = 0
Fk= F = 25 N
Fk= μk N
N = Fk / μk
W = N = 25 / 0.75 = 33.3 N
W = 33.3 N
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N=W
constant velocity )
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example
A 1200 kg car moving at 100 km/h coasts to a stop in 25 seconds.
What is the value of the static coefficient of friction?
m = 1200 kg
vo = 100km/h = 27.8 m/s
t = 25 s
µ=f/N,
where f = ma & N=mg
= ma/mg
=(m(|v – vo|)/t)/mg
vo
f
From v = vo + at,
a = (v-vo)/t
=(27.8m/s/25 s )/ 9.8 m/s2)
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µ = 0.113
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example
Suppose 35 kg crate was not moving at a constant speed,
but rather accelerating at 0.70 m/s2. Calculate the
applied force. The coefficient of kinetic friction is still
0.30.
N
Fa
Ff
mg
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Example
A box slides down an incline with angle θ = 45°. The coefficient of kinetic friction
between the box and the plane is μk = 0.2. Find the acceleration of the box.
∑ Fx = ma
ma
fk –Wx= - ma
∑ Fx =0
N – Wy = 0
mg sinθ – μk mg cosθ = ma
N – mg cosθ = 0
N
fk
mg sinθ – μk N =
a = g (sinθ – μk cosθ ) = 5.5 m/s2
Wx
Wy
θ
W
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θ
43
Example
A skier is pulled up a slope at a constant velocity by a tow bar.
The slope is inclined at 25.0° with respect to the horizontal. The
force applied to the skier by the tow bar is parallel to the slope.
The skier’s mass is 55.0 kg , and the coefficient of kinetic friction
between the skis and the snow is 0.120. Find the magnitude of
the force that the tow bar exerts on the skier.
 Given: m = 55.0 kg, mk = 0.120,  = 25.0°
since velocity is constant, ax=0; also ay=0
since skier remains on slope  equilibrium

 F  0   Fx  0,  Fy  0
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F
y
0
N  mg cos   0
N  mg cos 
F
x
0
f k  F  mg sin   0
F  f k  mg sin 
 m k N  mg sin 
 m k mg cos   mg sin 
 mg ( m k cos   sin  )
 286 N
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Example
A 5-kg block sits on a 30 degree incline. It is attached to string
that is thread over a pulley mounted at the top of the incline.
A 7.5-kg block hangs from the string.
• a) Calculate the tension in the string if the acceleration of the
system is 1.2 m/s/s
• b) Calculate the coefficient of kinetic friction.
T
FN
m2gcos30
30
T
Ff
m2g
m1
30
m2gsin30
m1g
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FNET  ma
m1 g  T  m1a
m1 g  m1a  T
(7.5)(9.8)  (7.5)(1.2)  T
T  64.5 N
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Example: The figure shows a wooden block B at rest on another blocks
are connected by astringe which goes round a fixed smooth pulley .the
kinetic friction between block A and the table is 14 N and that between
blocks B and A is 10 N . What is the horizontal force P which is required to
move block A with uniform velocity?
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Example
A metal block of mass 4Kg is on a plane inclined at an angle of 300 to
the horizontal. The block slides down the plane from rest and travels a
distance of 1.0 m in 2.0 s. (assume g = 9.81ms-2)
a) What is the friction between the block and the inclined plane when
the block is sliding ?
b) What is the coefficient of kinetic friction ?
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50
µ
µ
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If the box moves with constant velocity
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example
A person pushes a 30-kg shopping cart up a 10 degree incline with
a force of 85 N. Calculate the coefficient of friction if the cart
is pushed at a constant speed.
Fa
mg cos 
N

Ff
mg
mg sin 
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
Example
Block A, with mass 4.0 kg, sits on a frictionless table. Block B,
with mass 2.0 kg, hangs from a rope connected through a pulley
to block A. What is the acceleration of block A?
+
NA = 39.2 N
T
aA ≡ a
T
WA = 39.2 N
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aB ≡ a
Norah Ali Al-moneef
WB = 19.6 N
+
59
For mass A (4 kg),
(+T) = (4 kg)(+a)
For mass B (2 kg),
(+T)+(-19.6 N) = (2 kg)(-a)
We now have two equations in two unknowns (a and
T). We must solve these simultaneously. Substitute
the first into the second,
(4 kg)(+a) + (-19.6 N) = (2 kg)(-a)
Solving: a
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= (+19.6 N)/(6 kg) = 3.27 m/s2
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Example :Pulling against friction.
A 10.0-kg box is pulled along a horizontal surface by a force
of 40.0 N applied at a 30.0° angle above horizontal. The
coefficient of kinetic friction is 0.30. Calculate the
acceleration.
: Lack of vertical motion gives us the normal force
(remembering to include the y component of FP), which
isFy= 78 N. The frictional force is= μk N = 23.4 N,
and the horizontal component of FP is Fx=34.6 N,
Fx- Fk =ma
34.6- 23.4 = 10 a
a = 11.2/ 10 = 1.1 m/s2
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Example : To push or to pull a sled?
Your little sister wants a ride on her sled.
If you are on flat ground, will you exert
less force if you push her or pull her?
Assume the same angle θ in each case.
: Pulling decreases the normal
force, while pushing increases
it. Better pull
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Example : Two boxes and a pulley.
Two boxes are connected by a cord running over a
pulley. The coefficient of kinetic friction between box
A and the table is 0.20. We ignore the mass of the
cord and pulley and any friction in the pulley, which
means we can assume that a force applied to one
end of the cord will have the same magnitude at the
other end. We wish to find the acceleration, a, of the
system, which will have the same magnitude for
both boxes assuming the cord doesn’t stretch. As
box B moves down, box A moves to the right.
For box A, the normal force equals the weight; the magnitude of the
frictional force, but not of the tension in the cord.
Fk =μk N = μk mg = 0.20 x 5 x 9.8
Box B has no horizontal forces; the vertical forces on it are its weight and
the tension in the cord .Solving the two free-body equations for a gives 1.4
m/s2 (and the tension as 17 N).
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Example : The skier.
This skier is descending a 30° slope, at constant speed. What
can you say about the coefficient of kinetic friction?
FG = mg is the force of gravity (weight) on the skier.
Since the speed is constant, there is no net force in any direction.
Fk = mg sin 300
μkN = mg sin 300
μkmg cos 300 = mg sin 300
μkcos 300 = sin 300 / cos 300
μk = 0.5 / 0.886 = 0.58
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example
In the figure, block B has a mass of 8.0 kg and moves up the incline, where
the coefficient of kinetic friction is 0.4. Mass A is 20. kg. Assume an ideal
pulley with no mass or friction.
a ) Determine the friction force acting on block B.
b ) Determine the acceleration of each block.
c ) Determine the tension in the connecting rope
a
b
c
a) Since B is moving up the incline, the force of
friction points down the incline with magnitude
fk = μkN
Since N = mBg cosθ,
fk = μkmBg cosθ
Substituting,
fk = (0.4)(8.0 kg)(9.8 m/s2)cos 30
fk = 27 N
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The forces acting on block B along incline are
T – mB g sinθ - fk = mB a
Since fk = μk mB g cos θ,
T – mB g sinθ – μkmB g cos θ = mB a (1)
The forces acting on block A are
T - mA g = - mA a
Solving for T,
T = mA g – mA a (2)
Substituting (2) into (1) :
mA g – mA a – mB g sinθ – μkmB g cos θ = mB a
Moving a's to same side,
mAg - mBg sinθ - μkmBg cosθ = mBa + mAa
Factoring out a,
mAg - mBg sinθ - μkmBg cosθ = a(mB + mA)
Solving for a,
a = (mAg - mBg sinθ - μkmBg cosθ)/(mB + mA)
Substituting values :
a = [(20)(9.8) - (8.0)(9.8)(0.500) - (0.4)(8.0)(9.8)(0.866)]/(28)
a = 4.6 m/s2
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C ) Since the forces acting on block A are
T - mAg = - mAa
Solving for T,
T = m Ag - m Aa
Substituting values,
T = (20) (9.8) - (20) (4.6)
T = 104 N
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Example : A ramp, a pulley, and two boxes.
Box A, of mass 10.0 kg, rests on a surface inclined at 37° to the horizontal. It
is connected by a lightweight cord, which passes over a massless and
frictionless pulley, to a second box B, which hangs freely as shown. (a) If the
coefficient of static friction is 0.40, determine what range of values for mass
B will keep the system at rest. (b) If the coefficient of kinetic friction is 0.30,
and mB = 10.0 kg, determine the acceleration of the system.
Pick axes along and perpendicular to the slope. Box A has no movement
perpendicular to the slope; box B has no horizontal movement. The
accelerations of both boxes, and the tension in the cord, are the same. Solve
the resulting two equations for these two unknowns.
(a)The mass of B has to be between 2.8 and 9.2 kg (so A doesn’t slide either up
or down).
(b) 220.78
m/s2
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example
Two blocks of masses m1 and m2, with m1 > m2, are placed
in contact with each other on a frictionless, horizontal
surface, as shown below. A constant horizontal force F is
applied to m1 as shown.
a ) Find the magnitude of the acceleration of the system.
b) Determine the magnitude of the contact force between the two blocks.
a
a ) Fx(net) = (m1 + m2) ax
so,
ax = F/(m1 + m2)
b )The contact force will be
called P
Fx(net) = P12 = m2ax
since (from part a),
ax = F/(m1 + m2)
P12 = [m2/(m1 + m2)] F
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Example 3- 29
A block is at rest on an inclined plane the coefficient of static fraction is µs .what is
s the
the maximum possible angle of inclination θ ( max ) of the surface for which
block will remain at rest
Fs = w sin θ
N = w cos θ
tanθ




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example
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Summary of Chapter 3
• Newton’s
first law: If the net force on an object is zero, it will remain
either at rest or moving in a straight line at constant speed.
• Newton’s second law:
• Newton’s third law:
• Weight is the gravitational force on an object W= m g.
• Free-body diagrams are essential for problem-solving. Do one object
at a time, make sure you have all the forces, pick a coordinate system
and find the force components, and apply Newton’s second law along
each axis.
•Kinetic friction:
Fk = μkN
• Static friction:
Fs ≤ μsN .
• Fk < Fs
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Newton’s First Law is also called the Law of Inertia
Inertia:
the tendency of an object to resist changes in its
state of motion
The First Law states that all objects have inertia. The more
mass an object has, the more inertia it has (and the harder it
is to change its motion).
• Inertia is a term used to measure the ability of an object to resist
a change in its state of motion.
• An object with a lot of inertia takes a lot of force to start or stop;
an object with a small amount of inertia requires a small amount
of force to start or stop.
• The word “inertia” comes from the Latin word inertus, which can
be translated to mean “lazy.”
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• Identifying Newton’s third law pairs
– Each force has the same magnitude
– Each force acts along the same line but in
opposite directions
– Each force acts at the same time
– Each force acts on a different object
– Each force is of the same type
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