For the system shown calculate the force in the cables AC an

For the system shown, calculate the force in the cables AC and BC due to the 200 lb weight shown. AC = 448.01b BC = 292.4 lb AC = 428.0 lb BC = 92.4 lb AC = 292.4 lb BC= 448.0 lb AC = 342.5 lb BC= 142.5 lb

Solution

suppose tension in AC is T1 and tension in BC is T2 .

system is equilibrium hence Fnet at C will be zero.

in horizontal,

T2 cos40 - T1sin30 = 0

T1 = 1.532 T2 ........(i)

In vertical,

T1cos30 - T2sin40 - 200 = 0

putting T1 from (i)

1.532T2 cos30 - T3 sin40 - 200 = 0

0.684 T2 = 200

T2 = 292.4 lb .

T1 = 1.531 x 292.4 = 448 lb

AC = T1 = 448 lb

Bc = T2 = 292.4 lb

Ans(A)