Question

Required Information The box shown has cardboard sides and wood strips along the edges and from corner to corner. The strength of the box is provided primarily by the wood strips, and a truss model with 18 members for the wood strips is also shown. Let the truss be supported by rollers at points A and B. a socket at point C, and a short link at Ethat is parallel to the y direction 252 mm 240 mm 275 mm For the FG-100 N and FE-150 N forces, determine the force supported by the six members that emanate from joint A The force supported by the six members that emanate from joint A is as follows: TAB-CON TAD-DIN TAE- TAC-O TAF-LO TAHLN

Answer 1

GOLULTon Given that FG = loon FE = TSON from the given Figure draw the free body diagram. Ey 1150N 252mond FM 100N se uomo yo ziston py moment abou6.20-axis E Marco -Ey (0.252) + By Co_240) + Cz C0.240) : - 40000.252) = 0 - Ey 20.252) + By (0.240). +C2C0.240). = 25.2 moments about y carcis Az C0.275) – B70.275) + ISO C0.275)50

AqtB z = 150 moment about Z-001's E Mizo Ey (0.275) + Cac (0.240) = 0 forces along 5C racing Efs=0 (x=0 Eu (0.275) + Cbc (0.240) 20 Ey (0-275) -0 Ey = 0 - Eu Cơ, G ) + 3 Co., 2 H0) + c Co, 240) = 2s 2 Lo + By C0.240) + C7.CO.240) – 25.2 By (9.240) +cą CO.240) = 25.2 forces along y -axis Ey-cy + 150 = 0 0-cy + 150 =0 C4 -150N

forces along Z-AXIS Efzco Az.tB z tlz = 150 Azt Bz = 150 Cza 0 ByC0.240) + Cyc0.240) = 25.2 Bz z 25.2 0.240 BZ = 105N Az + B z. = 150 Az +105 = 150 Az = H5N joint B on Resolve forces. along 4-axis I Fy zo along x-axis 2 Ex=0 [Toc = o)

on Joint o Forceg along 3-asis { Foc=0 TAD =0 on jointe forces along Z-axis I Fz=6 252mm TTAF TAH -The -150=0 SCELTAF ** TAD $0 TAE E-150 N ITACTO 240mm TAB consider joint TAB, TAC, TAD, TAE, TAF, TAH Sise forces that emanate from soin A 0.240 kane = 0.275 0 = tans' Comune 02419 a. 252 tan. pl. 2.0240 x = 46.40

tan p = 0.252 0.275 B = 42.50 on 0-100is EF g = 0 :- TAD - SAH COS 42.5 - TAC COG 41° = 0 OSTAM COS 42.5 - TAC COS H19 = 0 TAM (0.737) & TAC (0.755) = 0 TAH =-1.024 TAC on y - axis Στη Tas it Tac sin (Hi) + TAF COS (46.4) 20. o + Tac sin (410) + TAF.COS (46.4) - 0.656 TAC + 0.689 TAF = 0 TAF -0,951 TAC

on Z-axis Az + Tae + Tan sin (42.5.) + Taf sin (46.4) =0 45-150 +TAH Sin 42.5 t Taf Sin (464) =0 ~ 105 +0.676 TAH +0.74 TAF=0 0.676 PAH +0.724 TAF = 105 0.676 (-1.024 TAC) 70.724 -0.951 TAC) 210S -0.692 TAC -0.689PACE',105 TAC = -log 1-381 FAC --76.03N) TAH = -1.024 676.03) TAH : +77.85 NI TAH = 77.85N. TAF = -0.51 €76.03) TE+72.30 N TAP = 72.30 N