Question

2. Three cables are joined at the junction ring C. Determine the tensions caused by the weight of a 30-kg cylinder in the following cables: (Assurne g = 9.81 m/s2.) a) AC b) BC (10 points) (10 points) 45° 15° 30 kg

Answer 1

Tension in cable CD is T1 = 30*9.8 = 294 N

now tension in cable AC is T2 and in cable BC is T3

now Horizontal balance

T2*cos(45 degrees) + T3*cos(60 degrees) = T1*cos(15 dgerees)

=> 0.707*T2 + 0.5*T3 = 294*0.965 = 283.71 ........eq1

and vertical balance

T2sin(45 degrees)+ T1sin(15 degrees) = T3sin(60degrees)

=> 0.707*T2 + 0.258*T1 = 0.866*T3

=> 0.707*T2 + 75.852 = 0.866*T3......eq2

from both equations

283.71 - 0.5*T3 + 75.852 = 0.866*T3

=> T3 = 263.22 N Tension in BC

=> T2 = 283.71 - 0.5*T3 = 152.1 N Tension in AC