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3M MECHANICS.
The table on page 313 was calculated by the following
formulas
:
When friction is not considered
:
n, . , , , , Pitch in inches Pitch in inches
Force to balance load = = —
—
12 X 2 7T 75.4
When both one pound of load and friction are considered,
Force — ( P*tch *n i"ches + middle circum. X/ \ s/ / middle radius \
^middle r.irr.nm. — nitch in inches X f ^
middle circum. — pitch in inches X f 12
calculations by table on preceding page.
Example 1.
A j ack screw, as shown in
Fig. 19, is 1)4" diameter, three
threads per inch. What tan-
gential force is required to act
with a leverage of 18 inches in
order to lift 5000 pounds ? Co-
efficient of friction in the thread
is assumed as 0.16. Tangen-
tial force absorbed in friction
by the collar at a is assumed
to be equal to force absorbed
by friction in the thread of
the screw, and may, therefore,
be taken from the thirteenth
column in the table.
Solution
:
Tangential force per pound at 1 foot radius = 0.0133
Tangential force absorbed by friction in collar = 0.0089
Total force per pound of load at 1 foot radius = 0.0222
The tangential force is acting with 18 inches leverage =
\y
2 feet, and the load is 5000 pounds; therefore, the required
force will be,
F = 0.0222 X 5000
= 74 pounds.
Example 2.
A load of 16,000 pounds rests on a slide and is moved back
and forth on a horizontal plane by a screw. The coefficient of
friction between slide and plane is 0.1, and the screw should
not be loaded with more than 400 pounds per square inch of
projected area or thread. Find the suitable diameter of screw.
If a pulley of 20-inch diameter is attached to the end of the
screw, also find the tangential force required to act at the rim
of the pulley in order to turn the screw.
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