Solutions to Engineering Mechanics "RESULTANT OF ANY FORCE SYSTEM" 3rd Edition by Ferdinand Singer

Problem 260 The effect of a certain non-concurrent force system is defined by the following data: ΣFx = +90 kN, 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.

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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

The y-intercept of the resultant is 6 ft above the origin.

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Problem 262 Determine completely the resultant of the forces acting on the step pulley shown in Fig. P-262.

Solution 262

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.

Solution 263

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

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

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

Righting moment

Overturning moment

Moment at the toe (downstream side - point B)

Location of Ry 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 inclination with the horizontal, and where it intersects AB.

Solution 267

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 inFig. 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.

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.

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Problem 269 Repeat Prob. 268 is 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

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 yintercept 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 Ry = ΣY to compute T.

Solution 270 For horizontal resultant, Ry = 0 and Rx = R

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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, Rx = 0 and Ry = R

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