
 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 or 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 below. Find (a) the acceleration of the blocks
and (b) the tension in the string.
 In Fig. 2a above, 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_{1s},
the force of block 1 on the string and F_{2s},
the force of block 2 on the string
 m_{1} experiencing F_{s1}
= the force of the string on block 1
 m_{2} experiencing F_{s2},
the force of the string on block 2 and F = 9.0
N.
 For the string, F_{net external }= m_{string}a
F_{2s}
 F_{1s}
= (0)3.0 m/s^{2} = 0
So F_{2s}
= F_{1s}
By Newton's Third Law, F_{s2}
= F_{2s}
and F_{s1}
= F_{1s}
Thus, F_{s2}
= F_{2s}
= F_{1s} =
F_{s1}.
We drop all this fancy notation and write
F_{s2}
= F_{2s}
= F_{1s}=
F_{s1}
= 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 
9.0
N  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.
 Practice problems in 104
Problem Set for Newton's Laws: 1  8.
 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
= µ(mg  F sin Θ)
 For Fig. 3c, N  mg cos Θ= 0. N = mg cos Θ;
f = µN = µmg cos Θ
 Practice problem in 104
Problem Set for Newton's Laws: 10.
 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. 4a 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. 4b 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.
 Practice problems in 104
Problem Set for Newton's Laws: 10  14, 16,
18, and 21.
 Uniform Circular Motion
 An object has uniform circular motion when it moves in a
circle with constant speed. The speed v is constant, but the
velocity v is not constant because v is always tangent to
the path so it continually changes direction. In Fig. 5 below,
v_{1}= v_{2} = v_{3}
= v_{4} = v, but this is not a case of
constant velocity because the direction of v changes.
 In Fig. 6 below, the two angles labeled Θare equal
because v_{2} an v_{1} are perpendicular to
their respective radii. Since v_{1}  =
v_{2} = v,_{ }the triangle with
the radii and the triangle with the velocities are similar
because they are isosceles triangles and they have angles
that are equal. Thus Δv/v = Δr/r or Δv = Δr(v/r).
Divide both sides by Δt and take limit as Δt approaches
zero
(a) = (v)(v/r) or a = v^{2}/r. Since a = dv/dt,
a has the same direction as Δv or into
the center of the circle.
 The magnitude of the acceleration is constant: a_{1}
=
a_{2} = a_{3}
= a_{4} = a = v^{2}/r, but
this is not a case of constant acceleration because the
direction of a is not constant.
 Since the acceleration is into the center, for a centripetal
acceleration there must be a net force into the center.
This force is perpendicular the to the tangent or the
direction of the velocity. There is no component of the
force in the direction of the velocity and, therefore,
there is no change in the magnitude of the velocity.
 Descriptive Terms
 Period T = time for one complete rotation = 2 πr/v
 Frequency f = reciprocal of period = 1/T = v/2 πr
 Practice problems in 104
Problem Set for Newton's Laws: 15, 17.
 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).
 Practice Problem I. A classics student of mass 70.0
kg is 0.50 m from a physics student of mass 50.0 kg. Calculate
the approximate magnitude of the gravitational force that
each exerts on the other. Explain why your result is only
approximately equal to the magnitude of the actual gravitational
force.
Solution.
The approximate magnitude of the gravitational force that
each exerts on the other is found from the equation for
pointlike masses.
F = Gm_{c}m_{p}/r^{2}
= (6.67 x 10^{11} Nm^{2}/kg^{2})(70.0
kg)(50.0 kg)/(0.50 m)^{2}
= 9.3 x 10^{7} N.
The force is approximate because students are not pointmasses.
 Practice Problem II. 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.
Solution.
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:
GMm/r^{2} = mv^{2}/r(1) 
GMm/r^{2} = m(2πr/T)^{2}/r 
Since v =
2 πr/T, 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 


