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310 MECHANICS.
have to exert a force of 30 pounds, which is all due to friction,
but beside that he must also exert a force of ^ times 1400 = 40
pounds ; thus the total pull exerted by the horse will be 70
pounds.
Inclined Plane With the Force Acting Parallel to the Base.
When the pressure is continually
acting in a line parallel to the base of ^/\ T
the incline, as F, (see Fig. 17) which
is frequently the case in mechanical
movements, as for instance, in screws,
some kinds of cam motions, etc., it ^^ ? ^_
will require more force to move the
body than it would if the force was
acting parallel to the incline. When
force acts parallel to the base, as in Fig. 17, the force required
to move the body, if friction is not considered, will be
:
F = ^Xsin.* = WXtzng.a
cos. a
Example 1.
What force is required to move 100 pounds upward an
incline of 30°, as in Example 1, excepting that the force is acting
parallel to the base instead of parallel to the incline ?
Solution :
F— W X tang. 30°
F=^’X100X 0.57735 — 57.74 pounds.
When both the friction and the weight of the body are con-
sidered, the force required to move the body will be :
F = W X sm- a + (/ x cos- a)
cos.a — (f X sin.rt)
Example 2.
What force is required to move 100 pounds upward an in-
cline of 30° (as in Example 1) if the force is acting parallel to
the base line instead of parallel to the incline ; coefficient of
friction is supposed to be 0.15 ?
Solution :
F = 100 X
sin- 30° + (0 - 15 X cos- 3Q°)
cos. 30° — (0.15 X sin. 30°)
F _ 100 x 0.5 + (0.15 X 0.86603)
0.86603 — (0.15 X 0.5)
F= 10Q
0.5 + 0.1277045
0.86603 — 0.075
F= 100 X 0.7936 = 79.36 pounds.
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