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Problem 263: Problem 263 - Resultant of Non-Concurrent Force System

The resultant of the non-concurrent force system shown in Fig. P-263 is 161.314 lb upward to the right at an angle of 21.69 degrees from the horizontal. Its x-intercept is 1.688 ft to the right of the origin and its y-intercept is 0.671 ft below the origin.

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Gavin Raymond Briones
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737 views3 pages

Problem 263: Problem 263 - Resultant of Non-Concurrent Force System

The resultant of the non-concurrent force system shown in Fig. P-263 is 161.314 lb upward to the right at an angle of 21.69 degrees from the horizontal. Its x-intercept is 1.688 ft to the right of the origin and its y-intercept is 0.671 ft below the origin.

Uploaded by

Gavin Raymond Briones
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Problem 263 | Resultant of Non-Concurrent 59.613a=100.

598
Force System a=1.688 ft to the right of the origin

Problem 263 Rxb=MO

Determine the resultant of the force system shown in 149.895b=100.598

Fig. P-263 and its x and y intercepts. b=0.671 ft below the origin
Thus, R = 161.314 lb upward to the right at θx = 21.69°
and intercepts at (1.668, 0) and (0, -0.671).
Rx=ΣFx
Rx=
Problem 264 | Resultant of Non-Concurrent
300sin30∘−224(2/√5)+361(2/√13)
Force System
Rx=
149.895 lb to the right Problem 264
Completely determine the resultant with respect to point
O of the force system shown in Fig. P-264.

Ry=ΣFy

Ry=
300cos30∘+224(1/√5)
−361(3/√13)
Ry=
59.613 lb upward
√R=Rx2+Ry2
√R=149.8952+59.6132
R=161.314 lb

tanθx=Ry/Rx
tanθx=59.613149.895
θx=21.69∘

MO=ΣM
MO=−(300sin30∘)(2)+224(15√)(2)+361(213√)(1)
MO=100.598 lb⋅ft counterclockwise

Rya=MO
Problem 265 | Resultant of Non-
Concurrent Force System

Problem 265
Compute the resultant of the three forces
shown in Fig. P-265. Locate its intersection
with X and Y axes.

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