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29O MECHANICS.
The force required to obtain a given velocity in a given
time, when both resistance due to inertia and resistance due to
friction is considered, is calculated by the formula
:
Force —
^
Clty
X Mass ) + (weight X coefficient of friction).
V Time /
which may be written :
Force = / Velocity
x Mass \ _j_ (resistance due to friction).
V, Time /
Important.—Always calculate the force required to over-
come the resistance due to inertia and the force required to
overcome the resistance due to friction separately, and add the
two forces in order to obtain the total force required.
It is sometimes assumed that adding so much to the mass,
as 3
:
2
- of the product of weight and coefficient of friction, should
give the result in one operation; but such an assumption is
erroneous, because the correct value for the required force is
:
F= v X W 4- WXf
which cannot be transposed to
F _ v X W+ WXf
TXg
_F= Required force ; v = velocity ; T= time ; IV= weight
of moving body in pounds ;
g-= acceleration due to gravity, or
32.2; f= coefficient of friction.
Example. 1.
A railroad train weighing 225,400 pounds is started from
rest to a velocity of 50 feet per second ; the road is straight and
level ; the resistance due to friction is assumed to remain con-
stant and to be 1000 pounds. What average constant pull in
pounds must be exerted by the locomotive at the draw-bar in
order to bring the train up to this speed in 40 seconds ?
Solution
:
For the inertia,
velocity X mass _ 50 x
225400
Force
time
4()
-
= 8750 lbs.
For friction the force — 1000
Total force, 9750 lbs.
Note.—This constant force of 9750 pounds has been acting
under a uniformly increasing velocity from rest or nothing at
the start, to 50 feet per second at the end of 40 seconds ; therefore,
the average velocity has been half of the final velocity, or 25 feet
a second. The average work of the locomotive in starting the
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