Question

For the stepladder shown in the figure, sides AC and CE are each 2.60 m long and hinged at C. Bar BD is a tie-rod 0.778 m long, halfway up. A man weighing 823 N climbs 1.95 m along the ladder. Assuming that the floor is frictionless and neglecting the mass of the ladder, find (a) the tension in the tie-rod and the magnitudes of the forces on the ladder from the floor at (b) A and (c) E. (Hint: Isolate parts of the ladder in applying the equilibrium conditions.)

В. А Е

Answer 1

26mlong. en 0-778m bng, B22N, 1.95m Snce the ladder is in equilibrions, the Vatical Comportants of the forces must hum to zero FutFA - W=0 Horizontal Grepenent Sin to za T-Fh=0 ogresor Then FALGs O-W(44) Cose - T(42) 100 Man is with the length & ladda for top so FE-FV=0, F-T-O and FELCOS - T(4) Sin 8 = So equations formed are (en A r t Euth-wool T-F2-0 - FALC88-w(44) Gs8 - T(42) 568-0 B It-tvo - FEL COSE-T(1/2) Sine = 0 - FA = A-FU Fv=FE - 9th epatia 7-F1,0 NLGESE- FEL GSE - W (4/) 458-T (4.)u6=0 FELCOSE-T(42) Sino=o.

Fe = T Sn E = (1/2 tane 1 (3w/4 tan o) 2- 26/2= 13 hypotenure and 1-0H8/2 - 0.389m Vatical Site has a longth of (138-(0-389)?- noum ton = (1-2470.389) - 31846 +- 3 (823N) _ 3(823) 4 tane 4(3-1876) (193.64N Ex= 36/8 (3) Since & FE, FE Ertin- WO 2650 Sine & 4. W. - W 3W/8 = 50 = 5 394-375 N - 5(82) 69060345 (9) Bs3up> FE= 3x93) = 18 8,625 N] v