Environmental Engineering Reference
In-Depth Information
where
v
is the velocity and
m
is the inertial mass of the particle. We postulate that
the equation of motion is
d
p
d
t
.
F
=
(2.14)
For a particle of fixed mass this equation reduces to
d
(m
v
)
d
t
F
=
=
m
a
(2.15)
and this is in accord with Eq. (2.12). Of course Eq. (2.14) looks like a more
general equation for it has the capacity to describe systems of variable mass. That
is indeed the case, as we shall explore in the following example. Eq. (2.14) is a
very important equation: it is Newton's Second Law.
Example 2.2.1
Flour falls from a hopper onto a railroad truck at a rate of 30 kg s
−
1
.
What horizontal force is required to pull the truck at a constant speed of 5ms
−
1
.
Solution 2.2.1
The speed of the truck is constant, but the mass of the truck, and
therefore the momentum, changes continuously. Using Eq. (2.14), we obtain
d
p
d
t
F
=
d
(m
v
)
d
t
=
v
d
m
d
t
=
.
v
d
dt
5kgms
−
2
The force needed has magnitude F
150 N
. Note that in
this problem we have considered the horizontal forces only; the vertical force that
stops the sand is provided by the normal reaction of the truck.
=
=
30
×
=
Momentum lies at the heart of Newton's Second Law. For a particle we have
defined it to be the product of mass and velocity and through Newton's Sec-
ond Law we see that a change in momentum can be induced by applying a
force. The bigger the force, the more one can change the momentum. Thus we
understand that the momentum of something expresses how hard it is to stop or
deflect it.
We can also use Eq. (2.14) to fix the scale of inertial mass. Two bodies, subject
to the same force, will experience accelerations
a
1
and
a
2
in the direction of the
applied force. The ratio of the masses of the two bodies is then
m
1
m
2
=
a
2
a
1
.
(2.16)
Choosing a standard mass (a lump of platinum-iridium alloy in the case of the
S.I. system of units) fixes the mass scale. We can then fix the unit of force, using