Cable BAC passes through a frictionless ring A and is attached to fixed supports at B and C, while cables AD and AE are both tied to the ring and are attached, respectively, to supports at D and E. Knowing that a 200-lb vertical load P is applied to ring A, determine the tension in each of the

Question

Cable BAC passes through a frictionless ring A and is attached to fixed supports at B and C, while cables AD and AE are both tied to the ring and are attached, respectively, to supports at D and E. Knowing that a 200-lb vertical load P is applied to ring A, determine the tension in each of the three cables.

Accepted Answer

Since [latex]T_{B A C}=[/latex] tension in cable BAC, it follows that

[latex]\begin{gathered}T_{A B}=T_{A C}=T_{B A C} \\mathbf{T}_{A B}=T_{B A C} \boldsymbol{\lambda}_{A B}=T_{B A C} \frac{(-17.5 \,\mathrm{in.}) \mathbf{i}+(60 \,\mathrm{in} .) \mathbf{j}}{62.5 \,\mathrm{in} .}=T_{B A C}\left\lgroup\frac{-17.5}{62.5} \mathbf{i}+\frac{60}{62.5} \mathbf{j} ight group \\mathbf{T}_{A C}=T_{B A C} \boldsymbol{\lambda}_{A C}=T_{B A C} \frac{(60 \,\mathrm{in.}) \mathbf{i}+(25 \,\mathrm{in} .) \mathbf{k}}{65 \,\mathrm{in} .}=T_{B A C}\left\lgroup\frac{60}{65} \mathbf{j}+\frac{25}{65} \mathbf{k} ight group \\mathbf{T}_{A D}=T_{A D} \boldsymbol{\lambda}_{A D}=T_{A D} \frac{(80 \,\mathrm{in.}) \mathbf{i}+(60 \,\mathrm{in.}) \mathbf{j}}{100 \,\mathrm{in.}}=T_{A D}\left\lgroup\frac{4}{5} \mathbf{i}+\frac{3}{5} \mathbf{j} ight group \\mathbf{T}_{A E}=T_{A E} \boldsymbol{\lambda}_{A E}=T_{A E} \frac{(60 \,\mathrm{in} .) \mathbf{j}-(45 \,\mathrm{in} .) \mathbf{k}}{75 \,\mathrm{in.}}=T_{A E}\left\lgroup\frac{4}{5} \mathbf{j}-\frac{3}{5} \mathbf{k} ight group\end{gathered}[/latex]

Substituting into [latex]\Sigma \mathbf{F}_{A}=0[/latex], setting [latex]\mathbf{P}=(-200 \,\mathrm{lb}) \mathbf{j}[/latex], and setting the coefficients of [latex]\mathbf{i}, \mathbf{j}, \mathbf{k}[/latex] equal to [latex]\phi[/latex], we obtain the following three equilibrium equations:

From [latex]\qquad\mathbf{i}:\quad-\frac{17.5}{62.5} T_{B A C}+\frac{4}{5} T_{A D}=0\qquad (1)[/latex]

From [latex]\qquad\mathbf{j}:\left\lgroup\frac{60}{62.5}+\frac{60}{65} ight group T_{B A C}+\frac{3}{5} T_{A D}+\frac{4}{5} T_{A E}-200 \,\mathrm{lb}=0\qquad(2)[/latex]

From [latex]\qquad\mathbf{k}: \quad \frac{25}{65} T_{B A C}-\frac{3}{5} T_{A E}=0\qquad (3)[/latex]

Solving the system of linear equations using convential acgorithms gives:

[latex]T_{B A C}=76.7 \,\mathrm{lb} ; T_{A D}=26.9 \,\mathrm{lb} ; T_{A E}=49.2 \,\mathrm{lb}\blacktriangleleft[/latex]

Question 2.121: Cable BAC passes through a frictionless ring A and is attach...

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