A boom AB is supported as shown in the figure by a cable runs from C over a small smooth pulley at D.

Compute the tension T in cable and reaction at A.Neglect the weight of the boom and size of the pulley.

#### Solution

Given : Beam AB is supported by a cable

To find : Tension T in cable

Reaction at A

Solution :

∠GCA = ∠BAF = θ

∠TCG = α

∠TCA = α + θ

= 63.4349o + 53.16°

= 116.5651°

∠TCB = 180o - 116.5651°

= 63.4349°

AC = AE+EC = 0.6 + 0.6 = 1.2

AB = AC + CB = 1.2 + 0.6 = 1.8

AF = ABcos θ = 1.8cos53.13 = 1.08

AH = AEcos θ = 0.6cos53.13 = 0.36

BEAM AB IS INDER EQUILIBRIUM

Applying conditions of equilibrium

Σ M_{A} = 0

-445 X AF – 890 X AH + Tsin63.4349 X AC = 0

T X 0.8944 X 1.2 = 445 X 1.08 + 890 X 0.36

T = 746.2877 N

Σ FX = 0

H_{A} - Tcos63.4349 = 0

H_{A}=333.75 N

Σ F_{Y} = 0

V_{A} + Tsin63.4349 – 890 – 445 = 0

V_{A} = 667.5 N

`R_A=sqrt(H_23^2+V_A^2)`

`R_A=sqrt((333.75)^2+(667.5)^2)`

`R_A=746.2877N`

Ф =`tan^-1((V_A)/H_A)`

Ф= `tan^-1((667.5)/333.75)`

**Ф =63.4395°Tension in cable = 746.2877 N (63.43949° in second quadrant)Reaction at A = 746.2877 N (63.4395° in first quadrant)**