Question

Consider the diagram below: Note: On your FBD draw all internal forces in the positive direction (tension). RN QN E SN PN TNB а т т а т т а т т т а т COLLAPSE IMAGES P = 113N Q = 312 N R = 129 N S = 82 N T = 81 N = 45 deg a = 2 m Determine the force in member EB N ENTER 3 tries remaining. 1 point(s) possible

Answer 1

129N 312N 182 N 11BN * X 480 SiN Cyr km km to am lant Daking moment about e, 2M = 0, Ayx 8 = 113 x 8 +3 12 x 6 + 129 x 4 + 8184 at 8 2 2 2 Ay = 472.5 N (1) 2 ty=0, Ant cu = 113 + 312+129 +82+84 | 5 - 7 - 172 = 244.5 N(9)

consider joint A Fassino C = compression T = Tension . 713N V I AGCOSO 472.5N. { Fy = 0 F Sin 450+472.5 = 113 - Free - 553.55 N. = 553.55 N (c). I FR0, 7 to coso ao AB AG - B - (-553.55) cos 45° = 420.928 (T).

consider Joint G. Felino FGE factus Fucsok .CO "GB" f sino TT 5 t = 10 2 y- sino = 312 4 to Sing - 553.55 sino GE Frefo -553.55 + for foB = -73.14 - O I for o caso + f caso = -553.55 coso Fore+faB -- 553.55 - 0 GB

0 +0 FGEF/= - 73.14 F GE & For = - 558.55 GE= - 313.35 N FoR = -553-55 +813-35 - 240.2 N. consider joint E 129 N. - 313.35cos s otto faccoso -313.95 14-313-3 sind F ES I ty = 0 , & #313.35 sin 45+ 129=0 & FER = 313.35 880450 - 129 EB = 74.5 N I FEB = 74.50.1 (Tensile).