Question

Two cables are tied together at C and loaded as shown in Fig. Knowing that P = 360 N, determine the tension (a) in
cable AC, (b) in cable BC.​

Answer 1

Given: P = 360 N

$\text { Resolve the forces along } x \text {-direction. }$

$\begin{array}{l}T_{A C} \cos (22.62)=360 \cos (36.87) \\T_{A C}=312 \mathrm{~N}\end{array}$

$\text { Resolve the forces along } y \text {-direction. }$

$\begin{array}{l}T_{B C}+T_{A C} \sin (22.62)+360 \sin (36.87)=480 \\T_{B C}+312 \sin (22.62)+360 \sin (36.87)=480 \\T_{B C}=144 \mathrm{~N}\end{array}$

$\text { Thus, the tension in the cables: }$

$\begin{array}{l}\mathbf{T}_{\mathbf{A C}}=\mathbf{3 1 2} \mathbf{N} \\\mathbf{T}_{\mathbf{B C}}=\mathbf{1 4 4} \mathbf{N}\end{array}$