18 RESULTANTS OF FORCE SYSTEMS (Chap. whence EXERCIO'S #1 F. *-*-*300* 3.61 = 249 lb and F, = -166 lb 203. The components of a certain force are defined by F, = 300 lb and F, -200 lb. Determine the magnitude, inclination with the X axis, and pointing ( the force. Solution: The magnitude of the force is found by applying the first of Eq. (2-2) F=361 lb Am [P = (F) + (P,})}] F=(300)+(200) The inclination with the X axis is determined by the second part of Eq. (2-2) tan 0 = [tan.-] 200 tan 0, = = 0.667 300 033.7 An Note particularly that by neglecting the given signs of the components the angle found is the acute angle between the force and the X axis. The direction of the force is found by sketching a tip-to-tail summation of the components as shown in Fig. 2-5, or by visualizing it mentally Note that the minus sign of F, indicates it to be directed downward. Hence the force F points down to the right. FR 0=33.7 FIG. 2-5. X This technique of determining a force eliminates the necessity of remembering certain arbitrary conventions. For ex- ample, a mathematical convention de fines an angle as always measured in a counterclockwise sense from the X axis. Accordingly in the given example, 0, might be defined as -33.7 or as +326.3. PROBLEMS 204. Determine the X and Y components of each of the forces shown in Fig. P-204. F=390 lb Y 12 Ans. T ==== -307 lb; Ty = -257 lb Y T=722 lb P=200lb 2 60 X $30 40 2 P=300lb T= 400 lb Fia. P-204. F 448 lb Fia. P-205. 2 Art. 2-3] Resultant of Three or More Concurrent Forces 19 205. Compute the X and Y components of each of the forces shown in Fig. P-205. Ans. T-400 lb; T, = +600 lb L 13 1600 6 in., resolve each load into com- 206. The triangular block shown in Fig. P-206 is subjected to the loads P lb and F = 600 lb. If AB = 8 in. and BC ponents normal and tangential to AC. 10% Ans. P1280 lb \; P 960 lb/; F360 lb ; F = 480 lb/ 207. Rework Prob. 206 if 0 = 60. 208. The horizontal and vertical components = -200 lb and A of several forces are: (a) Ph: An Lng P, 100 lb; (b) F = 300 lb and F, = -200 lb; (c) Th =-50 lb and T. mine each force. = -90 lb. Deter- = 26.6 150 lb and Ans. (a) P =224 lb up to the left at 0 209. Repeat Prob. 208 if (a) Ph == fth P. = -200 lb; (b) Fr = -240 lb and F. = 360 tlb; (c) T-500 lb and T, LOW == -300 lb. Ans. (c) T ==== A F=600 lb P=1600 lb FIG. P-206 and P-207. 583 lb down to the left at On == 31 all 210. In Fig. P-210, the X component of the force P is 140 lb to the left. De- cate termine P and its Y component. th P P ford ring ex de in 20 4 X [30 xis .. FIG. P-210. FIG. P-211. 211. The body on the 30 incline in Fig. P-211 is acted upon by a force P in- clined at 20 with the horizontal. If P is resolved into components parallel and perpendicular to the incline and the value of the parallel component is 400 lb, com- 7pute the value of the perpendicular component and that of P. Fig 2-3. Resultant of Three or More Concurrent Forces The determination of the resultant of three or more concurrent forces that are not collinear requires determining the sum of three or more vectors. There are two ways of accomplishing the addition of three or more vectors: graphically and analytically. Graphically. Two vectors can be added to give a resultant; this resultant in turn can be added to a third vector, etc., until all the vectors have been added together to give an overall resultant. These vectors can be added in any order.