Using Vectors to Find Cable Tension

Today I want to show an exercise demonstrating the use of vectors that is so complicated that a graphing utility is required to solve it, and yet, the underlying principle is simple once the basic nature of the problem is understood.

Two cranes are lifting an object that weighs 20,240 pounds. The cable from the crane on the left side makes an angle of 24.3° with the horizontal; the cable from the crane operating on the right side makes an angle of 44.5° with the horizontal. Find the tension, in pounds, exerted on each cable.

The first step is to draw a diagram of the situation:

The magnitude of vector \(\|\overrightarrow{\mathbf{T_{L}}}\|\) represents the tension in the left cable while the magnitude of vector \(\|\overrightarrow{\mathbf{T_{R}}}\|\) represents the tension in the right cable. The vector pointing straight down represents the object being lifted. The weight of the object is 20,240 pounds.

I think it may make things a little clearer if we write down the vectors:

\begin{align*}
\text{Left crane: } \|\overrightarrow{\mathbf{T_{L}}}\| &= \|\mathbf{T_{L}}\| \left(\cos 24.3° \mathbf{i} + \sin 24.3° \mathbf{j}\right) \\
\text{Right crane: } \|\overrightarrow{\mathbf{T_{R}}}\| &= \|\mathbf{T_{R}}\| \left(\cos 44.5° \mathbf{i} + \sin 44.5° \mathbf{j}\right) \\
\text{Resultant: } \|\overrightarrow{\mathbf{T_{L}}}\| + \|\overrightarrow{\mathbf{T_{R}}}\| &= 20240\mathbf{j}
\end{align*}

Here we have what I consider the major insight into this problem: a vector making an angle with the x-axis can be resolved into two components, one along the y-axis and one along the x-axis.

\begin{align*}
\text{Along the y-axis: } \|\mathbf{T_{L}}\| \sin 24.3° + \|\mathbf{T_{R}}\| \sin 44.5° &= 20240 \\
\text{Along the x-axis: } \qquad \qquad \qquad \, \|\mathbf{T_{L}}\| \cos 24.3° &= \|\mathbf{T_{R}}\| \cos 44.5°
\end{align*}

\begin{align*}
\text{Now, a substitution in the y-axis equation: } \left(\frac{\|\mathbf{T_{R}}\| \cos 44.5°}{\cos 24.3°}\right) \sin 24.3° + \|\mathbf{T_{R}}\| \sin 44.5° &= 20240 \\
\text{With a calculator, solve for the} \cos \text{and} \sin \text{functions: } \qquad \left(\frac{\|\mathbf{T_{R}}\| \left(0.713\right)}{0.911}\right) \left(0.411\right) + \|\mathbf{T_{R}}\| \left(0.701\right) &= 20240 \\
\text{And now by hand or again with a calculator, solve for } \, \|\mathbf{T_{R}}\|\text{: } \qquad \qquad \qquad \qquad \|\mathbf{T_{R}}\| &\approx 19786
\end{align*}

We now substitute \(\|\mathbf{T_{R}}\| = 19786 \text{ in: } \, \|\mathbf{T_{L}}\| \cos 24.3° = \|\mathbf{T_{R}}\| \cos 44.5°\) in order to find \(\|\mathbf{T_{L}}\|\).

\begin{align*}
\|\mathbf{T_{L}}\| \cos 24.3° &= \|\mathbf{T_{R}}\| \cos 44.5° \\
\|\mathbf{T_{L}}\| \cos 24.3° &= \left(19786\right) \cos 44.5° \\
\|\mathbf{T_{L}}\| &\approx 15484
\end{align*}

\begin{align*}
\text{The final answer is: tension in the left cable: } &\approx \text{15,484 pounds} \\
\text{right cable: } &\approx \text{19,786 pounds}
\end{align*}

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Quiz #3

In this week’s quiz we continue with vectors:
A tetherball weighing 1 pound is pulled outward from the pole by a horizontal force (vector \(\overrightarrow{\mathbf{u}}\)) until the rope makes a 45 degree angle with the pole. Determine the resulting tension (in pounds) in the rope and also the magnitude of \(\overrightarrow{\mathbf{u}}\).

Here’s a quick diagram of the situation.

Answer next week.

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Answer to Quiz #2

The problem was:
Neglecting air resistance, with what initial velocity must an object be launched upward from a height of \(\mathbf{2\, meters}\) to reach a maximum height of \(\mathbf{200\, meters}\)? The problem is asked in meters, so the acceleration due to gravity is \(\mathbf{-9.8}\) meters per second square.

\begin{align*}
\text{Acceleration function: } a\left(t\right) &= -9.8 \\
\text{Velocity function: } v\left(t\right) &= -9.8\,t + v_{0} \\
\text{Position function: } s\left(t\right) &= -4.9\,t^{2} + v_{0}\,t + 2
\end{align*}

This is my solution to this problem. When the object reaches 200 meters, its velocity is zero:

\begin{align*}
\text{Set the velocity function to zero: } v\left(t\right) = -9.8\,t + v_{0} &= 0 \\
\text{Solving for } t \text{: } \qquad \qquad t &= \frac{v_{0}}{9.8} \\
\text{Set the position function to 200 using the value of } t \text{: }\; -4.9\left(\frac{v_{0}}{9.8}\right)^{2} + v_{0}\left(\frac{v_{0}}{9.8}\right) + 2 &= 200 \\
\text{And solving for } v_{0} \text{: } \qquad \qquad v_{0} &\approx 62.3 \text{ meters per second}
\end{align*}

The following graph shows the \(\boldsymbol{\color{blue}{\text{position function}}}\) and the \(\boldsymbol{\color{red}{\text{velocity function}}}\). An object launched with an initial velocity of \(\approx 62.3\) meters per second from a height of 2 meters reaches a maximum altitude of 200 meters when its velocity is zero.

Once the position and velocity functions are set up, the rest is just algebra.

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