Resultants and Equilibrium of Force Systems: Determining Angles and Components, Essays (high school) of Computer science

Download Resultants and Equilibrium of Force Systems: Determining Angles and Components and more Essays (high school) Computer science in PDF only on Docsity! Chapter 1 - Resultants of Force System 1.)Completely determine the resultant with respect to point O of the force system shown in figure 1. ¥ Fy = M1 A(1/ v2) + 260(5/13) — 2400830 — 300cos60 = 257.831b. ¥ 1414 Ib Ep. 300 Ib R= fS Fre +h) Fy? 1200 Ib R= y((AT9.79)? + (257.83)? = 544.68lb 300 Ib 5 7 13 30 r tan ox = 30 Fa! 3 Fy x ex = arctan(257.83/479.79) = 28.25 Figure 1. SD Mo = 141.4(1/V2)(3) + 300cos60(4) + 300 sinG0(4) + 260(12/13)(1)— 260(5 //T3)(4) = 1779.19 — tb 2.) Replace the system of forces shown in figure 2 by an equivalent force through 0 and a couple acting through A and B. Solve if the forces of the couple are (a.) horizontal and (b.) vertical. So Fe = 141 A(i/ vB) + 224(2/ V5) — 361(2/ vT3) = 100.0828( Right) 3 Fy = ML A(1/y2)—224(1/ V5) —361(8/ vT3) = —300.561b( Downward) 141.4 Ib A T 4 R= \/(100.08)? + (—300.56)? = 316.79/b(downward, right) oe = aretan(300.56/100.08) = 71.58 2 e B 361 Ibi v | Aa 224 ib o (a.) 8 Ma a 1414 Ib 3F = 361(2/ 13) (1) +361(3¥19)(1) +141. 4(1/ y2)(2)-141.4(1/ v2)(3)— he 224(2/V5)(1) — 224(1/¥5)(1) 3F = 100.1 F = 33.37lb 2 t B ol Y 2 a8 Ib (b.)4P = LOOLL BN esuitant of ad-eertain systenp efzfomgas has the X and Y components shown in figure 3. Determine the components of this resultant with 1 Chapter 1 - Resultants of Force System respect to the H and Y axes rotated 30 degrees counterclockwise relative to the X and Y axes. R = ,/(300)2 + (48072 = 566.041 ox = arctan(480/300) ox = arctan(480/300) = 57.99 @ = 57.99 — 30 = 27.99 Re = 566.04sin27.99 = 265.6515 Fleure 3. Rh = 566.04cos27.99 = 499.816 4.) Compute the value of the resultant of the resultant of the concurrent forces shown in figure 4. YO Fx = 150(2/1? + 22) + 470cas30 — 340cos60 — 230cos45 = 104 AOIbs. Yo Fy = 150(1/ V5) + 340sin60 — 470sin30 — 230sind5 = —36.10bs R= (COLAO + (36.10 = 110.46lbs ox = arctan(36.10/104.40) = 19.07 340 Ib |¥ 230 Ib 470 Ib Figure 4, 5.)A concurrent force system is shown in figure 5. Determine the resultant. Y Fr = 410cos30 + 240(3/5) — 300cos60 = 349.07Ibs. Ti Fy = 410sin30 + 240(4/5) — 300sin60 = 137.19tbs. R= (4007) + (137.19) = 375.061bs ox = arctan(137.19/349.07) = 21.46