A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AC is 60 N, determine the weight of the plate.

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Accepted Answer

We note that the weight of the plate is equal in magnitude to the force P exerted by the support on Point A.

[latex]\Sigma F=0: \quad \mathbf{T}_{A B}+\mathbf{T}_{A C}+\mathbf{T}_{A D}+P \mathbf{j}=0[/latex]

We have:

[latex]\begin{array}{ll}\overrightarrow{A B}=-(320 \mathrm{~mm}) \mathbf{i}-(480 \mathrm{~mm}) \mathbf{j}+(360 \mathrm{~mm}) \mathbf{k} & A B=680 \mathrm{~mm} \\overrightarrow{A C}=(450 \mathrm{~mm}) \mathbf{i}-(480 \mathrm{~mm}) \mathbf{j}+(360 \mathrm{~mm}) \mathbf{k} & A C=750 \mathrm{~mm} \\overrightarrow{A D}=(250 \mathrm{~mm}) \mathbf{i}-(480 \mathrm{~mm}) \mathbf{j}-(360 \mathrm{~mm}) \mathbf{k} & A D=650 \mathrm{~mm}\end{array}[/latex]

Thus:

[latex]\begin{aligned}& \mathbf{T}_{A B}=T_{A B} \lambda_{A B}=T_{A B} \frac{\overrightarrow{A B}}{A B}=\left\lgroup-\frac{8}{17} \mathbf{i}-\frac{12}{17} \mathbf{j}+\frac{9}{17} \mathbf{k} ight group T_{A B} \& \mathbf{T}_{A C}=T_{A C} \lambda_{A C}=T_{A C} \frac{\overrightarrow{A C}}{A C}=(0.6 \mathbf{i}-0.64 \mathbf{j}+0.48 \mathbf{k}) T_{A C} \& \mathbf{T}_{A D}=T_{A D} \lambda_{A D}=T_{A D} \frac{\overrightarrow{A D}}{A D}=\left\lgroup\frac{5}{13} \mathbf{i}-\frac{9.6}{13} \mathbf{j}-\frac{7.2}{13} \mathbf{k} ight group T_{A D}\end{aligned}[/latex]

Substituting into the Eq. [latex]\Sigma F=0[/latex] and factoring [latex]\mathbf{i}, \mathbf{j}, \mathbf{k}[/latex] :

[latex]\begin{aligned}& \left\lgroup-\frac{8}{17} T_{A B}+0.6 T_{A C}+\frac{5}{13} T_{A D} ight group \mathbf{i} \& +\left\lgroup-\frac{12}{17} T_{A B}-0.64 T_{A C}-\frac{9.6}{13} T_{A D}+P ight group \mathbf{j} \& +\left\lgroup\frac{9}{17} T_{A B}+0.48 T_{A C}-\frac{7.2}{13} T_{A D} ight group \mathbf{k}=0\end{aligned}[/latex]

Setting the coefficient of [latex]\mathbf{i}, \mathbf{j}, \mathbf{k}[/latex] equal to zero:

[latex]\begin{array}{ll}\mathbf{i}: & -\frac{8}{17} T_{A B}+0.6 T_{A C}+\frac{5}{13} T_{A D}=0 & (1)\\\mathbf{j}: & -\frac{12}{7} T_{A B}-0.64 T_{A C}-\frac{9.6}{13} T_{A D}+P=0 & (2) \\\mathbf{k}: \quad & \frac{9}{17} T_{A B}+0.48 T_{A C}-\frac{7.2}{13} T_{A D}=0 & (3)\end{array}[/latex]

Making [latex]T_{A C}=60 \mathrm{~N}[/latex] in (1) and (3):

[latex]\begin{gathered}-\frac{8}{17} T_{A B}+36 \mathrm{~N}+\frac{5}{13} T_{A D}=0 & (1′)\\frac{9}{17} T_{A B}+28.8 \mathrm{~N}-\frac{7.2}{13} T_{A D}=0 & (3′)\end{gathered}[/latex]

Multiply (1′) by 9, (3′) by 8, and add:

[latex]554.4 \mathrm{~N}-\frac{12.6}{13} T_{A D}=0 \quad T_{A D}=572.0 \mathrm{~N}[/latex]

Substitute into (1') and solve for [latex]T_{A B}[/latex] :

[latex]T_{A B}=\frac{17}{8}\left\lgroup 36+\frac{5}{13} imes 572 ight group \quad T_{A B}=544.0 \mathrm{~N}[/latex]

Substitute for the tensions in Eq. (2) and solve for P:

[latex]\begin{aligned}P & =\frac{12}{17}(544 \mathrm{~N})+0.64(60 \mathrm{~N})+\frac{9.6}{13}(572 \mathrm{~N}) \& =844.8 \mathrm{~N} & ext{Weight of plate}=P=845 \mathrm{~N}\blacktriangleleft\end{aligned}[/latex]

Question 2.109: A rectangular plate is supported by three cables as shown. K...

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