APPLICATION
OF NEWTON’S SECOND LAW
TABLE
5-2
VARIATIONS
OF g WITH LATITUDE AND ELEVATION
Station
|
North
latitude
|
Elevation,
m
|
g,
cm / sec2
|
g,
ft / sec2
|
Canal Zone
|
9o
|
0
|
978.243
|
32.0944
|
Jamaica
|
18o
|
0
|
978.591
|
32.1059
|
Bermuda
|
32o
|
0
|
979.806
|
32.1548
|
Denver
|
40o
|
1638
|
979.609
|
32.1393
|
Cambridge
|
42o
|
0
|
980.398
|
32.1652
|
Standard station
|
980.665
|
32.1740
|
||
Greenland
|
70o
|
0
|
982.534
|
32.2353
|
or
less than the average
density of the earth.
The greatest variation,
however, arises from the rotation
of the earth, and will be discussed in
Chapter 6.
From
a survey of variations in the value
of g, conclusions can be drawn as to
the presence of deposits of ore or oil beneath the earth’s
surface. Hence the precise measurement
of g is one of the methods of geophysical prospecting. Some representative values of g are given in
Table 5-2.
5-8 Applications of
Newton’s second law. We now give a number of examples of the application of Newton’s
second law to specific problems. In all these examples, and in the problems at
the end of the of the chapter, it wil be assumed that the acceleration of gravity is 32.0 ft/sec2 or
9.8m/sec2, unless otherwise specified.
EXAMPLE
1. What is the resultant force
on a body weighing 48 lb when its acceleration is 6 ft/sec2 ?
We must first find the mass of the body. From the relation m =w/g,
.
Hence
EXAMPLE
2. What is the resultant force on a body of mass 48 kgm when its acceleration
is 6 m/sec2 ?
The
mass of the body is given, and hence
EXAMJPLE
3. A block whose mass is 10 Kgm rests on
a horizontal surface. What constant
horizontal force T is required to give it a velocity of 4 m / sec in 2 sec ,starting from rest, if
the friction force between the block and the surface is constant and is equal
to 5 Newton’s ? Assume that all forces act at the center of the block. (
See Fig. 5-4.)
The mass
of the block is given. Its y- acceleration is zero. Its x- acceleration can be found from the data on the velocity acquired
in a given time. Since the forces are constant,
the x- acceleration is
constant and from the equations of motion with constant acceleration,
.
Hence
from Newton’s second law in component
form,
EXAMPLE 4. An elevator and its load weigh a total of
1600 lb. Find the tension T in the supporting cable when the elevator, originally moving downward
at 20 ft/sec, is brought to
rest with constant acceleration in a
distance of 50 ft. ( see Fig. 5-5.)
The
mass of the elevator is
From
the equations of motion with constant acceleration,
The
initial velocity v0 is – 20 ft/sec ;
the velocity v is zero. If we take the origin at the point
where the deceleration begins, then y =
Hence
.
Fig.
5-5. The resultant force has Fig.5-6.
The resultant force has the magnitude T – w . the magnitude P
–w .
The acceleration is therefore positive ( upward ).
From
the free-body diagram ( Fig. 5-5 ) the resultant force is
Hence
T = 1800 lb.
EXAMPLE
5. With what force will the feet of a passenger press downward on the elevator floor when the
elevator has the acceleration above, if
the passenger weighs 160 lb ?
This
example illustrates a problem that is frequently encountered, in which it is
necessary to find a desired force by first computing the force that is the reaction
to the one desired, and then using Newton’s third law. That is, we first
calculate the force with which the elevator floor pushes upward on the
passenger ; the force desired is the reaction to this.
Figure
5-6 shows the force acting on the passenger. The resultant force is P—𝛚, and since the mass of the
passenger is 5 slugs and his acceleration
is the same as that of the elevator,
∑F = ma
P – 160 lb = s slugs x 4
= 20 lb
P= 180 lb.
The passenger exerts an equal and opposite
force downward on the elevator floor.
Example 6
What is the acceleration of a block on a
frictionless plane inclined at an angle ɵ with the horizontal ?
The only forces acting on the block are its weight w and the normal force N exerted by the plane
(fig. 5-7). Since neither the weight nor the mass of the block is given, a
letter must be used to represent one or the other. Let us call the mass m;
the weight is then mg. take exes
parallel and perpendicular to the surface of
the plane and resolve the weight into x- and y- components.
Then
∑Fy =N- mg
cos q
∑Fz =N- mg sin
q
But we know that ay = 0, so from the equation ∑Fy = may we find that N= mg cos q
. From the equation ∑Fx = max , we
have
mg
sin q
=
max,
ax = g sin
q
The mass does not appear in the final
result, which mean that any block, regardless of its mass, will slide on a
frictionless inclined plane with an acceleration down the plane of g sin q . (note that the velocity is not
necessarily down the plane).
Example 7
Refer to fig 2-3 (chapter 2). Let the
mass of the block be 4 kgm and that of the rope be 0.5 kgm. If the force F1
is 9 Newton’s what are the forces F1´
, F2 dan F2´ ? the surface on which the block
moves is level and frictionless.
We know from Newton’s third law that F1=
F1´ and that F2= F2´.
Hence F2´=
9 Newton’s. The force F2
could be computed by applying Newton’s second law to the block, if its
acceleration were known, or the force F2´could be
computed. By applying this law to the rope if its acceleration were known. The acceleration in not given, but it can be
found by considering the block and rope together as a single system. The vertical forces on this system need not
be considered. Since there is not friction, the resultant external force acting on the system is the force F1. (The
forces F2 and F2´.are internal forces when
we consider block and rope as a single
system, and the force F1´ does not act on the system, but on the man) then, from Newton’s second
law,
∑F =ma
9 n=
( 4 kgm + 0.5 kgm ) x a
a = 2
.
We can
now apply Newton’s second law to the block.
∑F =ma
F2 = 4 kgm x 2
.= 8 newtons.
Considering the rope alone, the resultant
force on it is..
∑F = F1- F2´
= 9 n - F2´.
And from the second law
9 n - F2´ = 0.5 kgm
x 2
= 1 n
F2´ = 8 Newton
In agreement with Newton’s Third law,
which was tacitly used when the forces F2 and F2´
were omitted in considering the system as a whole, we find that F2 and F2´ are equal in
magnitude. Notice, however, that the forces F1 and F2 are
not equal and opposite (the rope is not in equilibrium) and that these forces
are not an action- reaction pair.
Example 8
In fig.5-8, a block of weight w1
= 16 lb moves on a level
frictionless surface, connected by a light flexible cord passing over a
small frictionless pulley to a second hanging block of weight w2
= 8 lb. what is the
acceleration of the system, and what is
the tension in the cord connecting the two blocks ?
The diagram shows the forces acting on
each block. The forces exerted on the blocks by the cord can be considered an
action-reaction pair, so we have used the block on the surface,
∑Fx =T =
0.5 slugs x a
∑Fy =N-w1
= N – 16 lb =0
For the hanging block ,
∑Fy = w1-
T = 8 lb - T
=0.25 slugs X a
Since a is the same for both blocks, on solving the first and third equations simultaneously, we obtain
a = 10.7
,
T = 5.3 lb
Note carefully that although the earth
pulls on the hanging block with a force
of 8 lb, the force exerted on the 16 lb block is only 5.3 lb. it is the connecting cord which pulls on the 16 lb
block, and the tension in this cord must
be less than 8 lb or the hanging block would not accelerate downward.
Example 9
Up to this point, applications of Newton’s second law
have been limited to cases where
the resultant force acting on a
body was constant, thereby imparting to the body a constant acceleration. Such
cases are very important, but make only
light demands on one’s
mathematical knowledge. When the resultant force is variable, however, the acceleration
in not apply. We conclude this section
with three examples of motion under the action of a variable force.
An automobile of mass 50 slugs is traveling at 30 ft/sec.
the driver applies the brakes in such
a way that velocity decreases
to zero according to the
relation.
v
= v0 –kt2
where v0 = 30 ft/ sec ,
k = 0.30 ft/sec3, and t is the time in seconds after the brakes are applied. Find the resultant force
decelerating the automobile, 5 sec after the brakes are applied
the mass of automobile is given. To find
the resultant force on it from Newton’s second law we must first compute its
acceleration. This is given by
a
=
=
(v0 – kt2) =
-2 kt
hence when t = 5
sec,
a
= -2 X 0.30
X 5 sec
= -3
therefore at this instant
∑F
= ma =
50 slugs X (-3
) = -150 lb.
(what is the significance of the negative sign)
Example 10
An automobile of mass 50 slugs
accelerates from rest. During the first 10 sec, the resultant force acting on
it is given by ∑F = F0 – kt, where F0 =200 lb, k = 10 lb/sec, and t is the time in seconds after the start. Find
the velocity at the end of 10 sec, and the distance covered in this time.
The mass of the automobile and the
resultant force acting on it are both given. The acceleration can be found from
Newton’s second law and the principles
of kinematics can then be used to find the velocity and the position. From the
second law,
a
=
=
.
therefore
dv =
.
∫ dv =
.
v =
(5-7)
Where C1 is a constant of
integration. Since v = 0 when t =
0, C1 is zero also.
Hence when t = 10 sec.
v =
=
To find the position. Return to Eq (5-7)
v
=
∫ dx =
x
=
if the origin is at the starting point, x = 0 when t = 0 and
hence C2 =0. When t = 10 sec,
x
=
=
200 ft – 33 ft = 167 ft
Example 11
Discuss the motion of a freely falling
body (or a body projected vertically upward), taking into account the variation
of the gravitational force on the body
with its distance from the earth’s center. Neglect air resistance.
The gravitational force on the body at a
distance r from the earth’s center is GmmE
/ r2 , and from Newton’s second
law its acceleration is
Where the positive direction is upward (or better, radially
outward).
It was shown in Eq (4-8) that the
acceleration can be expressed as
Then
Where v1
and v2 are the velocities at the radial distances r1 and r2. It follows that
v22 - v12 = 2 GmE (1/r2 1/r2) (5-8)
as an illustration, let us find the
initial velocity v1
required to project a body vertically upward so that it rises to a height above the earth’s surface equal to the earth’s radius R .then v2 = 0, r1 = R,
r2= 2 R, and
v12 = GmE/
R (5-9)
let g0
represent the acceleration of gravity at the
earth’s surface, where r =
R. then from Eq. (5-6),
GmE = g0
R2
and Eq. (5-9) can be written
v12 = g0
R. (5-10)
how does this compare with the velocity
that would be required if the acceleration had the constant value g0 ?
5-9 The equal-arm analytical balance. As
with many other physical quantities, the mass of a body can be measured in
several different ways. One is to use the relation by which the quantity is
defined, which in this case is the ratio of the force on the body to its
acceleration. A measured force is applied to the body, its acceleration is
measured, and the unknown mass is obtained by dividing the force by the
acceleration. This method is used exclusively to measure masses of anatomic
particles.
The second method consists of finding by
trial some other body whose mass (a) is equal to that of the given body, and
(b) is already known. Consider first a method of determining when two masses
are equal. It will be recalled that at the same point on the earth’s surface
all bodies fall freely with the same acceleration g. since the weight 𝛚 of a body equals the product of its mass m and the acceleration g,
it follows that if, at the same point, the weights of two bodies are equal,
their masses are equal also. The equal-arm
balance is an instrument by means of which one can determine very precisely
when the weight of two bodies are equal, and hence when their masses are equal.
The essential feature of the balance,
shown schematically in Fig. 5-9, is a light, rigid beam on which are firmly
mounted tree equally spaced agate knife-edges, parallel to one another and
perpendicular to the length of the beam. The central knife-edge at O rests on a polished plane agate plate
supported from the floor of the balance case. The scale pans are hung from two
similar plates at A and B. A pointer fastened to the beam swings
in front of the scale S. the
knife-edges and plates act as practically
Frictionless pivots, since the scale
pans can swing freely about their supporting knife-edges, the center of gravity
of the pans and of any bodies placed on them will always be directly below the
knife-edges. The weight of the beam is w, and its center of gravity is directly
below the central knife-edge when the beam is horizontal. The weights of the
scale pans are equal
In
using the balance, the body whose mass m1
is desired is placed in the left scale pan, and a combination of known masses m2 is placed in the right
pan. If the combined weights of bodies
and scale pans, w1 and w2, are exactly equal, the
beam remains in stable equilibrium in a horizontal position as in Fig 5-8,
under the action of the four parallel forces w1, w2,w, and the upward force P exerted at the central knife-edge.
Then the weight of the unknown body equals that of the bodies in the right scale
pan, and hence their masses m1 and m2
are equal also.
There
remains the question as to how a set of bodies of known mass (usually referred
to as a set “weights”) is obtained in the first place. This is also done with
the equal-arm balance. Let us start with a standard kilogram and make two
bodies (a) whose masses are equal, as
determined by the balance, and (b) which when combined will just balance the
standard kilogram. Each of these then has a mass of one-half kilogram, or 500
grams. Other submultiples, or multiples, of the standard can be prepared by a
similar procedure.
