Question

For the stepladder shown in the figure, sides AC and CE are each 2.89 m long and hinged at C. Bar BD is a tie-rod 0.555 m long, halfway up. A man weighing 663 N climbs 2.17 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.)

Answer 1

Here we apply concept of static equilibrium and apply condition for static equilibrium.

first we f observe free body divide the ladder into a parte forces on each by drawing diagram o Tension in Ro CD - T= Tension in BD - N R Na are Normal force by floor as no friction floor is smooth). mg = weight of man À 4h are horizontal & vertical component of force by each half on each other. BC = AC-2-897 - 2 2 - CO1.745 m. _ CM wa perpendicular on BD BM = BP = 0.555M = 0.2775m Caso = BM = 0.2775 ВС 1.445 - >10= 78.93 I now for static Equilibrium & Fn =0, Efy = 0 {tro Tatfv = mgo beat-2 No = Fy] - NatNg = ng

Now Torque about to bmg (@-89-2-17) Coes0 +T (BC) sino - M Caccaman counter clock the tve clockwise eve) => 663 (0-72) Cos 73:199+T(1475)Sin19:13.- Na (289) (ca. 18 te?

(1 41897 - 6.555) N = 91.657L Torque about c' T Ć=o P T (CD) Sino - NECE) Coco o =0 T(1143) Sin 78.939 — NECQ.89 ) Cos 7893 =0 ai (1-418) T = (0.55S) Net from © 26 No - Na = 165:15 from L - ® NotNA = 663 NE = 414.1 Newton Ma = 248.93 Newton T = 162 Newton T=162 N NA = 248.93 N (a) *)