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Chapter 4
Dynamics: Newton’s
Laws of Motion
The Tension Force
Example: A window washer standing in a bucket pulls herself up with
a rope over a pulley as shown. The bucket+window washer have a
combined mass of 65 kg. Starting from rest at the bottom, how much
constant force must she exert on the rope to reach a window 160 m
above her in one-half of a minute?
+y
T
T
!
ay
m= 65 kg
160 m
W
∑F
y
= 2T − W = may
The Tension Force
∑F
y
mg
= 2T − W = may
⇒T =
m ( ay + g)
2
Find ay using constant acceleration equation:
y = v0 y t + ay t
1
2
2
y = 160 m, v0 y = 0, t = 30 s
2y 2(160)
2
y = 0 + ay t ⇒ ay = 2 =
=
0.36
m/s
2
t
30
m(ay + g) 65(0.36 + 9.80)
∴T =
=
= 330 N
2
2
≈ 74 lbs
1
2
2
Types of Forces: An Overview
Examples of Nonfundamental Forces -All of these are derived from the electroweak force:
normal or support forces
friction
tension in a rope
Static and Kinetic Frictional Forces
When an object is in contact with a surface there is a force
acting on that object. The component of this force that is
parallel to the surface is called the
frictional force.
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“cold welds”
Static and Kinetic Frictional Forces
When the two surfaces are
not sliding across one another
the friction is called
static friction.
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Static and Kinetic Frictional Forces
The magnitude of the static frictional force can have any value
from zero up to a maximum value.
fs ≤ f
f
MAX
s
0 < µs < 1
MAX
s
= µ s FN
Not a vector equation!
fS is parallel to the surface,
FN is perpendicular to
the surface.
is called the coefficient of static friction.
Static and Kinetic Frictional Forces
Note that the magnitude of the frictional force does
not depend on the contact area of the surfaces.
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Static and Kinetic Frictional Forces
Static friction opposes the impending relative motion between
two objects.
Kinetic friction opposes the relative sliding motion that actually
does occur.
f k = µ k FN
0 < µ ks < 1
is called the coefficient of kinetic friction.
Static and Kinetic Frictional Forces
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Usually, µs > µk
Static and Kinetic Frictional Forces
Example. A sled and a rider are moving at a speed of 4.0 m/s along a
horizontal stretch of snow. The snow exerts a kinetic frictional force on the
runners of the sled, so the sled slows down and eventually comes to a stop.
The coefficient of kinetic friction is 0.050 and the mass of the sled and rider
is 40 kg. Find the kinetic frictional force and the displacement, x, of the sled.
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Static and Kinetic Frictional Forces
1.  Use Newton’s 2nd law in x and y directions.
ΣFx = -fk = max --> ax = -fk/m = -µkFN/m
(since fk = µkFN)
ΣFy = FN - W = FN - mg = may = 0 --> FN = mg
fk = µkFN = µkmg = (0.050)(40)(9.8) = 20 N
Static and Kinetic Frictional Forces
ax = -µkFN/m = -µkmg/m = -µkg = -(0.050)(9.8) = -0.49 m/s2
2. Solve for x using ax and kinematic equations.
x
?
v0x
4.0 m/s
vx
0 m/s
ax
-0.49 m/s2
t
independent
vx2 = v0x2 + 2axx --> x = (vx2 - v0x2)/(2ax)
of mass of
2
2
= (0 - 4.0 )/(2(-0.49)) = 16 m sled+rider
Example. Block 1 (mass m1 = 8.00 kg) is moving on a 30.0o incline with
a coefficient of kinetic friction between the block and incline of 0.300.
This block is connected to block 2 (mass m2 = 22.0 kg) by a massless
cord that passes over a massless and frictionless pulley. Find the
acceleration of each block and the tension in the cord.
fk
fk
fk
m1 = 8.00 kg
m2 = 22.0 kg
Find a, T and T’
µk = 0.300
Block 1:
Σ  Fx = -fk - W1 sin 30.0o + T = m1a
fk
Σ  Fy = FN - W1 cos 30.0o = 0 è FN = W1 cos 30.0o
Block 2:
Σ  Fy = T’ - W2 = m2(-a)
We also know:
T = T’ since the pulley and cord are massless
fk = µkFN = µkm1g cos 30.0o = (0.300)(8.00)(9.80)(0.866) = 20.4 N
Equations we’re left with to solve for a and T:
-fk - W1 sin 30.0o + T = m1a
T - W2 = -m2a è
T = W2 - m2a
Substituting for T in the first equation:
-fk - W1 sin 30.0o + W2 - m2a = m1a
a = (-fk - W1 sin 30.0o + W2)/(m1 + m2)
= (-20.4 - (8.00)(9.80)(0.500) + (22.0)(9.80))/(8.00 + 22.0)
= 5.20 m/s2
T = W2 - m2a = (22.0)(9.80) - (22.0)(5.20) = 101 N