
 Newton's Laws
 First Law. An object at rest remains at rest and an
object moving with velocity v remains moving
with velocity v if and only if no net external
force acts on the object.
 Second Law. The net force acting on an object is
directly proportional to and in the same direction as
the acceleration. The proportionality constant is the
mass of the object.
F_{net}
= ma
 Third Law. When two objects interact in whatever
manner, the force
F_{12}
exerted by the first object on the second is equal
in magnitude and opposite in direction to the force
F_{21}
of the second object on the first.
F_{12}
= F_{21}
 Application of Newton's Laws
 An object hanging motionlessly in Fig. 1 above has
a weight of 800 N. Find the tensions T_{1} and
T_{2} in the massless ropes. For the object
to be at rest, the sum of the forces acting on the object
must be zero. Since the tensions are neither in the
X nor Ydirection, take components:
_{
}
(F_{net})_{x}
= ma_{x} = 0

(F_{net})_{y}
= ma_{y} = 0

T_{2 }cos 20^{o}
– T_{1 }cos 20^{o} = 0

T_{2 }sin20^{o
}+ T_{1 }sin20^{o} –
800N = 0

T_{2 }cos
20^{o} = T_{1 }cos 20^{o}^{}

T_{2 }sin20^{o
}+ T_{1 }sin20^{o} = 800N

T_{2 } =
T_{1 }= T
2T sin20^{o} = 800 N
T = 1.17 x 10^{3}
N = T_{2 } = T_{1}_{}

_{
}
 When you apply Newton's second law, you must be very
careful about your choice of system.
 Imagine two masses m_{1} = 1.0 kg and
m_{2} = 2.0 kg are connected by a massless
string and are pulled along a frictionless surface
by a force F = 9.0 N as shown in Fig. 2a above.
Find (a) the acceleration of the blocks and (b)
the tension in the string.
 In Fig. 2a, I have isolated the entire system
of the two blocks. There are vertical forces
acting on both blocks: the normal force of the
surface and the weight of each block. Since
there is no acceleration in the vertical direction
and there is no friction (we would need to know
the normal force in order to find the frictional
force on a block), we shall not consider forces
in the vertical direction. The only external
horizontal force acting on the system is F =
9.0 N. There are internal horizontal forces
acting on the blocks and the string, but we
are not interested in internal forces because
F_{net external }= m_{system}
a
9.0 N = [(1.0 + 2.0)kg] a
and
a = 3.0 m/s^{2
}
 Now we choose our system as shown in Fig.
2b and 2c above. Again I only look at the horizontal
forces. I have isolated:
 the string experiencing F_{1}_{}_{s},
the force of block 1 on the string and F_{2}_{}_{s},
the force of block 2 on the string
 m_{1} experiencing F_{s}_{}_{1}
= the force of the string on
block 1
 m_{2} experiencing F_{s}_{2},
the force of the string on block 2 and F
= 9.0 N.
 For the string, F_{net external }=
m_{string}a
F_{2}_{}_{s}
 F_{1}_{}_{s}
= (0)3.0 m/s^{2} = 0
So F_{2}_{}_{s}
= F_{1}_{}_{s}
By Newton's Third Law, F_{s}_{2}
= F_{2}_{}_{s}
and F_{s}_{}_{1}
= F_{1}_{}_{s}
Thus, F_{s}_{}_{2}
= F_{2}_{s}
= F_{1}_{}_{s}
= F_{s}_{}_{1}.
We drop all this fancy notation and write
F_{s}_{2}
= F_{2}_{s}
= F_{1}_{}_{s}=
F_{s}_{}_{1}
= T, as shown in Fig. 2c above.
 For m_{1}, F_{net external }=
m_{1}a
T = 1.0 kg(3.0 m/s^{2}) = 3.0 N
For m_{2}, F_{net external }=
m_{2}a
9.0 N – T = 2.0 kg(3.0 m/s^{2})
= 6.0 N
T = 3.0 N
 As you can see from the above example, Newton Third
Law of Motion forces always act on different objects.
In other words, they do NOT act on the same object.
 Sample problems in 107
Problem Set for Newton's Laws: 16.
 Frictional Forces
 The frictional force f = µN, where the coefficient
of friction µ is labeled with a subscript s for
static situations and k for objects in motion. The coefficient
of friction for static situations is always greater
than that for kinetic. The static frictional force can
go from 0 to a value necessary to just about move the
object.
 Finding the frictional force (F_{net})_{y}
= ma_{y} = m(0) = 0
 For Fig. 3a above, N  mg = 0.
N = mg and f = µN = µmg
 For Fig. 3b, N + F sin Θ
 mg = 0.
N = mg  F sin Θ
and f = µN = µmg  F sin Θ)
 For Fig. 3c, N  mg cos Θ=
0.
N = mg cos Θ;
f = µN = µmg cos Θ
 Sample problems in 107
Problem Set for Newton's Laws: 711.
 Gravitational force F_{g} = Gm_{1}m_{2}/r^{2},
where m_{1} and m_{2} are the masses of
two point particles separated by a distance r.
 Near the earth's surface at height h very small compared
to the radius of the earth R_{E}. You can treat
a spherical mass M_{E} as a point mass located
at the center of the earth. The force on an object of
mass m at height h above the earth by the earth is F_{g}
= GmM_{E}/(R_{E} + h)^{2}. Since
R_{E} is much greater than h, drop the h and
F_{g} = GmM_{E}/R_{E}^{2}.
In
general, F_{net}
= ma
 For a freely falling object, GmM_{E}/R_{E}^{2
}= ma
For free fall a = g
GmM_{E}/R_{E}^{2
}= mg = Weight of object
 (mg) is not a mass times an acceleration, it
is a force. To determine the direction of the force
in a problem, decide on a direction to call positive
and then see if mg is in that direction. After
doing this, never substitute
9.8 m/s^{2 } for g in (mg).
Example:
An astronaut puts a bowling ball into a circular orbit
about the Earth at an altitude h of 350 km. Find the
ball's period of motion. The radius and mass of the
earth are M = 5.98 x 10^{24} kg and
R = 6.37 x 10^{6} m, respectively.
Answer
to Example: The radius of the
orbit of the bowling ball r = the radius R of the earth
+ the height h above the earth's surface =
(6.37 + 0.35)10^{6} m, and the constant in Newton's
law of gravitation G = 6.67 x 10^{11} Nm^{2}/kg^{2}.
The gravitational force produces the centripetal acceleration:


F_{net} = ma



GMm/r^{2} = mv^{2}/r
(Equation
1)


Since v = 2πr/T,

GMm/r^{2} = m(2πr/T)^{2}/r


or 
T^{2} = 4π^{2}
r^{3}/GM and


T = 2 π
(r^{3}/GM)^{1/2}^{}



= 2{(6.72
x 10^{6} m)^{3}/6.67 x 10^{11}
Nm^{2}/kg^{2} x 5.98 x 10^{24}
kg}^{1/2}^{}


= 55
x 10^{2} s


 Free Body Diagrams
 Force Diagrams such as those shown in Fig. 4 below
are called Freebody diagrams. They are extremely important
in solving Newton Second Law of Motion problems.
 In each situation shown in Fig. 4 above, one or more
forces act upon an object. All drawings are in a vertical
plane and friction is negligible except in (b) and (d).
Draw free body diagrams for the figures, scale the forces
as close as possible. Label all the forces acting on
the objects. If the object has an acceleration, show
its direction. If there is no acceleration, indicate
that it is zero.
 In Fig. 4' above, (a) the only force that acts on
the object is its weight mg. Its acceleration
a is down. (b) For a constant velocity, the net
force acting on the object must be zero. The upward
frictional force equals the weight of the object. (c)
There is no net force perpendicular to the plane. The
normal force is equal to the component of the weight
perpendicular to the plane. The component of the weight
parallel to the plane gives the acceleration down the
incline. (d) Now in addition to the forces talked about
in (c) there is a frictional force up the plane that
is equal in magnitude to the component of the weight
parallel and down the plane. There is no acceleration.
(e) and (f) The only force acting on an object in projectile
motion (neglecting a frictional force) is the weight
of the object down. The acceleration is down.
 Separating Objects in Free Body Problems
Example:
A rectangular block of mass m_{1} rests on an
inclined plane of mass m_{2}, as illustrated
in Fig. 5a above. There is neither friction between
the block and the incline nor between the incline and
the surface on which it moves. Find an expression for
F in terms of m_{1}, m_{2}, and angle
of the incline Θ
so that the block remains at rest with respect to the
incline.
Solution:
In Fig. 5b above the forces acting on the incline of
mass m_{2} are the normal force of the block
on the incline N_{1 on 2}, the normal
force of the floor on the wedge N_{floor on wedge},
the attraction of the earth for the incline, its
weight, m_{2}g, and the applied force F.
In Fig. 5c above, the forces acting on the block of
mass m_{1} are the normal force of the incline
on the block N_{2 on 1} and the attraction of
the earth for the block, its weight, m_{1}g.
Using the principle of superposition we divide the problem
into the forces in the X and Y direction.
For m_{1},
X 
Y 
(F_{net})_{x}
= m_{1}a
N_{2 on 1 }sin Θ_{1}
= m_{1}a
(Equation
1) 
(F_{net})_{y}
= m_{1}(0) = 0
N_{2 on1} cos Θ
m_{1}g = 0
N _{2 on1} cos Θ=
m_{1}g
(Equation
3) 
For m_{2},
X 
Y 
(F_{net})_{x}
= m_{2}a
F  N_{1 on 2} sin Θ_{1}
= m_{2}a
(Equation
2) 
(F_{net})_{y}
= m_{1}(0) = 0
N_{floor on 2}  N_{1 on 2} cos
Θ m_{2}g
= 0 
By Newton's third law of motion,
N_{2 on 1 }= N_{1 on 2}
We may rewrite Eq. (2) as:
F  N_{2 on 1} sin Θ_{1}
= m_{2}a (Equation
2')
Substitute Eq. (1) into Eq. (2'):
F  m_{1}a = m_{2}a or F
= (m_{1} + m_{2})a (Equation
4)
Divide Eq. (1) by Eq. (3):
or
tan Θ= a/g
and a = g tan Θ (Equation
5)
Substitute Eq (5) into Eq. (4):
F = (m_{1} + m_{2}) g tan
Θ
 Sample problems in 107
Problem Set for Newton's Laws: 1224.


