Machinery's Handbook, 31st Edition
Force Systems 167 Example: Find the resultant of three coplanar nonconcurrent forces for which the following data are given. ; .; .; ; .; . .; ; .; .; F x y F x y F x y 10 5 1 270 20 4 15 50 30 2 2 60 lbs in in lbs in in lbs in in 1 1 1 1 2 2 2 2 3 3 3 3 ° ° ° i i i = = =− = = = = = = = = =
. . 10 270 10 0 0 20 50 20 0 64279 12 86 30 60 30 0 5000 15 00 lbs. . . cos cos cos # # # ° ° ° = = = = = =
F F F F F F
= = = = = =
x x x
1 2 3
lbs.
lbs.
. 10 270 10 1 10 00 20 50 20 0 76604 15 32 30 60 30 0 86603 25 98 . . . . sin sin sin # # # # # # ° ° ° = − = − = = = = ^ h
lbs. lbs
y
1
.
y y
2 3
lbs.
5 10 # = − − −
. . 1 0 50 4 1532 15 1286 4199 2 2598 2 15 2196 . . . . # # # # # = − − − = = ^ h ^ h
in. lbs.
M M M
O O O
1 2 3
in. lbs.
= =
in. lbs.
Note: When working in metric SI units, pounds are replaced by newtons (N); inches by meters or millimeters, and inch-pounds by newton-meters (N · m) or newton-millimeters (N · mm). Force Coordinates of F Components of F Moment of F about O F x y θ F x F y F 1 = 10 5 − 1 270 ° 0 − 10.00 − 50.00 F 2 = 20 4 1.5 50 ° 12.86 15.32 41.99 F 3 = 30 2 2 60 ° 15.00 25.98 21.96 ∑ F x = 27.86 ∑ F y = 31.30 ∑ M O = 13.95
y
27.86 ( ) 2 = = 41.90lbs
31.30 ( ) 2 +
R
41.90 LBS.
48° 20 ’
31.30 27.86 = ------- = 1.1235
θ tan R
– x
x
48 ° 20 ′ =
θ R
0.33 ”
d 13.95 41.90 = ------- = 0.33 inch measured as shown on the diagram.
– y
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