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MECHANICS. 3 11
Note.—From these calculations it is seen that it is more
advantageous to apply the force parallel to the incline than
parallel to the base. When force is applied parallel to the in-
cline :
The force required to overcome gravity = 50 pounds.
The force required to overcome friction == 12.99 pounds.
Total force = 62.99 pounds.
When the force is acting parallel to the base :
The force required to overcome gravity = 57.74 pounds.
The force required to overcome friction = 21.62 pounds.
Total force = 79.36 pounds.
II F
R X 2 7T
Weight of the ;load lifted, or force exerted, if the
Screws.
When friction is not considered, the force which may be
exerted by a screw (see Fig. 18) will be
:
F X 7?X2tt r , W X P
P
W
screw acts as a press.
F = Acting force.
R = Radius in inches at which the force acts.
P = Pitch of screw in inches.
FIG. 18
Regarding friction in
screws, the thread of a screw
may be considered as an in-
clined plane, of which the
cos. is the middle circum-
ference of the screw, the
sin. is the pitch, and the
force is acting parallel to
the base. Hence the fol-
lowing formula
:
F= W X P +/ X d - r
dn — fX P X R
F = Force, acting at a radius of R inches.
W= Weight.
P = Pitch of screw in inches.
/ — Coefficient of friction, usually taken as 0.15.
R = Radius in inches at which the force is acting.
r == Middle radius of screw in inches.
d = Middle diameter of screw in inches.
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