Problem

In trying to move across a slippery icy surface, a 175-lb man uses two ropes AB and AC. Kn...

In trying to move across a slippery icy surface, a 175-lb man uses two ropes AB and AC. Knowing that the force exerted on the man by the icy surface is perpendicular to that surface, determine the tension in each rope.

Fig. P2.90

Step-by-Step Solution

Solution 1

Given that, weight of the \(\operatorname{man}, W=175 \mathrm{lb}\)

\(\therefore\) From the figure we have

$$ \begin{aligned} &O=(0,0,0) \\ &A=(30,-16,12) \\ &B=(0,8,44) \\ &C=(0,4,0) \\ &\overrightarrow{A B}=\overline{O B}-\overline{O A}=-30 \bar{i}+24 \bar{j}+32 \bar{k} \\ &\overline{A C}=\overline{O C}-\overline{O A}=-30 \bar{i}+20 \bar{j}-12 \bar{k} \end{aligned} $$

\(\therefore\) Let \(T_{A B}\) and \(T_{A C}\) are the tensions in the ropes. Let \(F\), be the force exerted by the icy surface on the man.

\(\therefore\) From the equilibrium condition at 'A'

$$ \Rightarrow \overline{F_{A B}}+\overline{F_{A C}}+\bar{W}+\overline{F_{s}}=0 \quad \rightarrow(1) $$

\(\therefore \overline{F_{A n}}=T_{A B} \lambda_{A B}\)

\(\Rightarrow \overline{F_{A B}}=T_{A B} \frac{\overline{A B}}{|\overrightarrow{A B}|}\)

\(\Rightarrow \overline{F_{A B}}=T_{A B} \cdot \frac{(-30 \bar{i}+24 \bar{j}+32 \bar{k})}{\sqrt{(-30)^{2}+(24)^{2}+(32)^{2}}}\)

\(\therefore \overline{F_{A B}}=T_{A B} \cdot(-0.6 \bar{i}-0.48 \bar{j}+0.64 \bar{k})\)

\(\therefore \overline{F_{A C}}=T_{A C}, \lambda_{A C}\)

\(\Rightarrow \overline{F_{A C}}=T_{A C} \cdot \frac{\overline{A C}}{|\overrightarrow{A C}|}\)

\(\Rightarrow \overline{F_{A C}}=T_{A C} \cdot \frac{(-30 \bar{i}+20 \bar{j}-12 \bar{k})}{\sqrt{(-30)^{2}+(20)^{2}+(-12)^{2}}}\)

\(\therefore \overline{F_{A C}}=T_{A C} \cdot(-0.789 \bar{i}+0.526 \bar{j}-0.316 \bar{k})\)

\(\therefore \bar{W}=-175 \bar{j}\)

\(\therefore\) Let \(D=(30,-16,0)\) be the point on the icy surface

$$ \therefore \overline{O D}=30 \bar{i}-16 \bar{j} $$

\(\therefore\) Vector normal to the icy surface, \(\bar{N}=\overline{O A} \times \overline{O D}\)

$$ \begin{aligned} &\Rightarrow \bar{N}=\left|\begin{array}{ccc} \bar{i} & \bar{j} & \bar{k} \\ 30 & -16 & 12 \\ 30 & -16 & 0 \end{array}\right| \\ &\therefore \bar{N}=192 \bar{i}+360 \bar{j} \end{aligned} $$

\(\therefore\) Unit vector normal to the icy surface, \(\lambda_{N}=\frac{\bar{N}}{|\bar{N}|}\)

$$ \begin{aligned} &\Rightarrow \lambda_{N}=\frac{(192 \bar{i}+360 \bar{j})}{\sqrt{(192)^{2}+(360)^{2}}} \\ &\therefore \lambda_{\mathrm{N}}=(0.470 \bar{i}+0.882 \bar{j}) \\ &\therefore \bar{F}_{s}=F_{s} \cdot \lambda_{N} \\ &\therefore \bar{F}_{s}=F_{s}(0.47 \bar{i}+0.882 \bar{j}) \end{aligned} $$

\(\therefore\) From equation (1) we get

$$ \begin{aligned} \Rightarrow &\left(-0.6 T_{A B}-0.789 T_{A C}+0.470 F_{s}\right) \bar{i} \\ &+\left(0.48 T_{A B}+0.526 T_{A C}+0.882 F_{s}-175\right) \bar{j} \\ &+\left(0.64 T_{A B}-0.316 T_{A C}\right) \bar{k}=0 \end{aligned} $$

\(\therefore\) From the above equation we get

$$ \begin{aligned} &0.6 T_{A B}+0.789 T_{A C}-0.470 F_{x}=0 \\ &0.48 T_{A B}+0.526 T_{A C}+0.882 F_{s}=175 \\ &0.64 T_{A B}-0.316 T_{A C}=0 \end{aligned} $$

\(\therefore\) By solving the above 3 equations we get

$$ T_{A B}=30.90 \mathrm{lb} $$

\(T_{A C}=62.51 \mathrm{lb}\)

\(F_{s}=144.34 \mathrm{lb}\)

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