Question

AA=7,BB=1

1) If the system below is under equilibrium, find the forces passing from cable CA and CB 30 350 AA+BB (kg)

Answer 1

Solution:- the values and figure(diagram) given in the question are as follows:

AA=7 and BB=1

В. 4 o 30° с 35 w W=AA-BB kg

let tension in cable CA and CB is T1 and T2 respectively

W=AA-BB=7-1

W=6 kg

Free Body Diagram-

T1 T2 o 30° с 35 W=6 kg

system is in equilibrium so apply equilibrium conditions-

$\Sigma Fx=0$

-T1*cos30^o+T2*cos35^o=0

T1=T2*(cos35^o/cos30^o)

T1=0.945875*T2 , [Eq-1]

$\Sigma Fy=0$

T1*sin30^o+T2sin35^o=6 kg , [Eq-2]

value of T1 put in equation(2) from equation-(1)

0.94587*T2*sin30^o+T2*sin35^o=6

T2*1.04651=6

T2=5.7333 kgf

[we know that, 1kgf=9.81 N]

tension in cable CB(T2)=5.7333 kgf, 56.2436 N

value of T2 put in equation-(1) and calculate the value of T1

T1=5.7333*0.945875

T1=5.423 kgf, or 53.1996 N

tension in cable CA(T1)=5.423 kgf, or 53.1996 N

[Ans]