Question

use k 500N/M and 250mm instead of 340mm

A continuous cable of total length 4 m is wrapped around the small pulleys at A, B, C, and D. If each spring is stretched 340 mm, determine the mass m of each block. Neglect the weight of the pulleys and cords. The springs are unstretched when d = 2 m and k = 632 N/m. M

Answer 1

FBD FBD os pulley A de Fspring=F= F = kxsl ol=250 mm= 0.25 m K = 500nm F=F<=125N (500x0.250) = 125N A E=125n FBD of pulley C FBD of pulley B. Erresive arcoso-mg-o mg = Ro >OTcoO УТ (3) Considero FBD of ci) g (2) Espoing = F = 125 N Solve equations in y direction ZFy - OT tve 125 - Tsino-Tan 0 =0 125- QTaino =0

150- 25 ain o = 0 75 T = - Sino Stoeched-FBD 0 Oiasm CC,=> deflection of Spring a A A, =) deflection of spoing A CC=AA,=0-25m total cable length=4m AB=BC = CO-AD-Im 0.15m tomo-am 0:75m, A --- 7- 0.25 m. Unstrecbed condition Sino = + =) O=sin() 0-90° In unstreched condition 0 = 90° Consider FBD in Streched conclition d=0-asmom then & AC=1.5m. sino = 0.75 5) 0 = sin" (0:79) = 48.595

Substitute, 0= 48.59 in eqD 75 TOON sin (48.59) Consider FBD (3) ► +ve -mg+ at coso = 0 21 COSO 2 100 X COS (48-59) 9.81 3 in = 13-48 kg (Ans)