Full resolution (JPEG) - On this page / på denna sida - Mechanics - Inclined plane
<< prev. page << föreg. sida << >> nästa sida >> next page >>
Below is the raw OCR text
from the above scanned image.
Do you see an error? Proofread the page now!
Här nedan syns maskintolkade texten från faksimilbilden ovan.
Ser du något fel? Korrekturläs sidan nu!
This page has never been proofread. / Denna sida har aldrig korrekturlästs.
MECHANICS. 309
Example 1.
The weight, W, (Fig. 16) is 100 pounds; the angle a is 30°.
What force, F, is required to sustain this weight, friction not
considered ?
Solution :
Sin. 30° = 0.5
Thus :
F = W X sin. 30° = 100 X 0.5 = 50 pounds.
Example 2.
What is the perpendicular pressure under conditions stated
in Example 1 ?
Solution
:
P = W X cos. a = 100 X 0.86603 = 86.6 pounds.
Therefore, the frictional resistance between the sliding
body and the inclined plane will be only what is due to 86.6
pounds pressure ; in other words, the force required to over-
come friction will be IV X /* X cos. a.
Example 3.
What force is required to move the body mentioned in
Example 1 when friction is also considered, taking coefficient
of friction, F, as 0.15?
Solution
:
F’
= W (sin. a + cos. a X /)
^ = 100X (0.5 + 0.86603X0.15)= 100 X 0.6290 =62.99 pounds.
Note.—This is the force required for moving the load.
In order to put it in motion more force must be applied, varying
according to velocity, but after motion is commenced the speed
would be, under these conditions, maintained forever by this
force of 62.99 pounds.
When a load is moving down an inclined plane the force
due to W X sin. a will assist in moving the body, and if the
product W X sin. a exceeds the product W X cos. a X f the
body will slide by itself. For instance, in the body mentioned
in the previous example, the force required to overcome gravity,
regardless of friction, is 50 pounds, and the force required to
overcome friction is 12.99 pounds; thus, if the body should be
let down the plane instead of pulled up, it would have to be
held back with a force of 50 — 12.99 = 37.01 pounds.
Note.—When the incline is less than 1 in 35, cosine is so
nearly equal to 1 that it may be neglected, and the force required
to overcome friction may be considered to be the same as on a
level plane. For instance, a horse is pulling a load and ascend-
ing a gradient of 1 in 35 ; if the tractive force required to pull
the load on a level road was 30 pounds and the weight of the
load was 1400 pounds, when ascending the hill, the horse will first
<< prev. page << föreg. sida << >> nästa sida >> next page >>
