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Review 1 - REVIEW 1 I0.132 Determine the magnitude and...

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Unformatted text preview: REVIEW 1 I0.132. Determine the magnitude and coordinate direction angles of F3 so that the resultant of the three forces is zero. Eh = 21;; 0 = —180 + 300 cos30°sin40° + 1*; cosa 2 EU = 214;; 0 = 300 cos30°cos40° + F3 c035 15"“ = 21;; O = —300 sin30° + F3 cosy F =180lb l cosza + coszfi + 00527 = 1 Solving : F; = 250 1b Ans y at = 87.0° Ans X [3 = 143° Ans F2=3001b y = 53.1 Ans *2-136. The leg is held in position by the quadriceps AB, which is attached to the pelvis at A. If-the force exerted on this muscle by the pelvis is 85 N, in the direction shown, determine the stabilizing force component acting along the positive y axis and the supporting force component acting along the negative x axis as N % Ex: 85 cos 10° = 83.7N Ans IV; = 85 sin 10° = 14.8N Ans 2-139. If F1 = F; = 301b, determine the angles 9 and 4: so that the resultant force is directed along the positive x axis and has a'magnitude of FR = 20 lb. (30)Z : (30)2 + (20)2 — 2(30)~(20)cos9 (1} = 6 2 705° Ans 3-66. Ihe pipe is held in place by the vice. If the bolt exerts a force of 50 lb on the pipe in the direction shown, determine the forces F A and F3 that the smooth contacts at A and B exert on the pipe. 4 F,i * acos60° — 50(3) = 0 —1«;sin60° + 50(3) = 0 1'} = 346 lb Ans F, 573 lb Ans *3-72. The "scale" consists of a known weight W which is suspended at A from a cord of total length L. Determine the weight w at B if A is at a distance y for equilibrium. Neglect the sizes and weights of the pulleys. +TEF, =0; 2Wsin6—w=0 w = 2Wsi116 m/t—g—YI — (ET = 2W ;—,/(L — y)2 — d2 4/; L-y 9 w: 2W W = y/(L-y)2 —d2 Ans 4-162. Determine the moment of the force F about point 0. The force has coordinate direction angles of a = 60°, B = 120°, 7 = 45°. Express the result as a Cartesian vector. F=20Ib Puma» Victor And Fore: Valor: : rm = ((6-0)i+(l4—0)j+(l0-0)k) in. a (6i+l4j+ 10k) in. F = 20(eos 60°i+cos 120°j+ cos 45°k) lb 3 (10.0i -l0.0j+ H.142“ lb Moment of Force F About Point 0 : Applying 31.4—7. we have Mo = rm xF i 5 k = 6 14 10 10.0 — 10.0 14.142 ={298i+15.lj-2(I)k) lb-in Ans 4-165. Determine the magnitude of the moment of the force FL. about the hinged axis (40 of the door. Position Vector And Force Wants: m, = [[—0.5 — (—0.5)1i + [o — (—Jm‘ + (o — mm m = {U} m t—us—t—zsm Fc =250 N l—o.s—(—z.sn=+ [04-1wn.5mssn“)lP+tu—I 5mm“)? = [159.33i + 183.l5j — 59.75klN Mam-mt of Farce Fc About a » mm.- The unit vector along the a —— z: axis is i. Applying Eq. 4—] I, we have Ma—a = i ' (1245 X Fc) i 0 0 = 0 E 0 15933 18115 “5975 The negative sign indicate-2; that Maw“ is directed toward negative x axis. = l[1(—59.75) — (183.l5)(0)1— 0 + (l MP,x = 59.7 N - m Ans = ~59.7 N-m *4-168. The horizontal 30-N force acts on the handle of the wrench. What is the magnitude of the moment of this force about the z axis? Position Vector And Forte Vectors: r“ = {—0.01i +0.2j] n1 r04 = [(—0.01 —- (Di + (0.2 — OH + (0.05 —- 0)k} m = [—0.01i + 0.2j + 0.05M in F = 30(sin 45°i — cos 45°j) N = (21.113i— 21.7.13le Moment of Fame F About 1 Axis.- Tlie unit Vector along the z axis is k. Applying Bq. 4—11. we have M: = k‘G‘M KF) 0 o 1 = 4101 0.2 0 21.213 —21.213 (1 = o — o + lE(—0.0l )(—21.2i3) — 2| .213(0.2)] = —4.03 N . m Ans 4-170. If the resultant couple moment of the three couples acting on the triangular block is to be zero, determine the magnitudes of forces F and P. Couple moment of the [01b force : The sense of direction of this couple moment is outward perpendicular to the inclined plane. 3 4 =-'+-k “ 5J 5 M =Mu=10(3)@j+§k)={183+24k} Ib- in. Couple moment of P and F : MF = -6F j Resultant moment : MR=0=EM; 0=M+Mp+Mp 0=(18j+24 k)+(-6Fj)+(—6P k) 0=(18—6F)j+(24-6P)k Equaling j, 1: components : (IS-GE')=0 (24—6P)=0 Mp=—6Pk F=31b Ans P=4lb Ans 0r Mz=k'(l'wt XF) O 0 l = —~0.0l 0.2 0.05 21.213 -—2L2l3 0 = o - o + l[(—0.0l)(!21.213) — 21.213(0.2)] = —-4.03 N v m ADS T he negative sign indicates that MI is directed along the negative 2 axis. 2 ...
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