resultant of non concurrent force system
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RESULTANT OF NON CONCURRENT FORCE SYSTEM
Problem 260
The effect of a certain non-concurrent force system is defined by the following data: ΣFx = +90kN, ΣFy = -60 kN, and ΣMO = 360 kN·m counterclockwise. Determine the point at which the
resultant intersects the x-axis.
Solution 260
The x-intercept is at 6 m to the left of the origin. answer
Problem 261 In a certain non-concurrent force system it is found that ΣFx = -80 lb, ΣFy = +160 lb, and ΣMO =
480 lb·ft in a counterclockwise sense. Determine the point at which the resultant intersects the y-
axis.
Solution 261
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The y-intercept of the resultant is 6 ft above the origin. answer
Problem 262 Determine completely the resultant of the forces acting on the step pulley shown in Fig. P-262.
Solution 262
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Thus, R = 1254.89 lb downward to the right at θx = 44.21° and passes through the axle.
Problem 263 Determine the resultant of the force system shown in Fig. P-263 and its x and y intercepts.
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Solution 263
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Thus, R = 161.314 lb upward to the right at θx = 21.69° and intercepts at (1.668, 0) and (0, -
0.671).
Problem 264 Completely determine the resultant with respect to point O of the force system shown in Fig. P-
264.
Solution 264
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Thus, R = 544.68 N upward to the right at θx = 28.25°. The intercepts of R are (-4.57, 0) and (0,2.46).
Problem 265 Compute the resultant of the three forces shown in Fig. P-265. Locate its intersection with X and
Y axes.
Solution 265
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Thus, R = 957.97 lb downward to the right at θx = 32.19°. The x-intercept is at 2.90 ft to the right
of O and the y-intercept is 1.83 ft above point O.
Problem 266 Determine the resultant of the three forces acting on the dam shown in Fig. P-266 and locate its
intersection with the base AB. For good design, this intersection should occur within the middle
third of the base. Does it?
Solution 266
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Righting moment
Overturning moment
Moment at the toe (downstream side - point B)
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Location of R y as measured from the toe
(within the middle third)
Thus, R = 27 424.02 lb downward to the right at θx = 79.91° and passes through the base at 8.44
ft to the left of B which is within the middle third.
Problem 267 The Howe roof truss shown in Fig. P-267 carries the given loads. The wind loads are
perpendicular to the inclined members. Determine the magnitude of the resultant, its inclinationwith the horizontal, and where it intersects AB.
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Solution 267
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Thus, R = 10 778.16 N downward to the right at θx = 68.2° passing 4.46 m to the right of A.
Problem 268 The resultant of four forces, of which three are shown in Fig.P-268, is a couple of 480 lb·ft clockwise in sense. If each
square is 1 ft on a side, determine the fourth force completely.
Solution 268
Let F4 = the fourth force and for couple resultant, R is zero.
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Thus,
Assuming F4 is above point O
d is positive, thus, the assumption is correct that F4 is above point O.
Therefore, the fourth force is 200 lb acting horizontally to the left at 5.8 ft above point O.
answer
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Problem 269 Repeat Prob. 268 if the resultant is 390 lb directed down to the right at a slope of 5 to 12 passing
through point A. Also determine the x and y intercepts of the missing force F.
Solution 269 Let F4 = the fourth force
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Resolve F4 into components at the x-axis
Resolve F4 into components at the y-axis
Thus, F4 = 219.32 lb downward to the right at θx = 43.15° with x-intercept ix = 3.27 to the right
of O, and y-intercept iy = 3.06 ft above point O.
Problem 270 The three forces shown in Fig. P-270 are required to cause a horizontal resultant acting through
point A. If F = 316 lb, determine the values of P and T. Hint: Apply MR = ΣMB to determine R,then MR = ΣMC to find P, and finally MR = ΣMD or R y = ΣY to compute T.
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Solution 270 For horizontal resultant, R y = 0 and R x = R
answer
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answer
Problem 271 The three forces in Fig. P-270 create a vertical resultant acting through point A. If T is known to
be 361 lb, compute the values of F and P.
Solution 271 For vertical resultant, R x = 0 and R y = R