I have been concentrating on vectors lately because I need to obtain a certain level of proficiency with them in order to study college-level physics. The physics textbook I am using gets heavy on vectors right on chapter one.
So, here’s another problem using vectors: a 5,000-pound load is lifted by 2 cables. What is the tension in each cable supporting the load? Refer to the following drawing.
It is quite obvious that to solve this problem we should use the same method as in the last problem involving cable tension. However, in that problem we were given the angles the cables made with the horizontal. Luckily, getting that information is almost trivial using a little bit of Geometry and Trigonometry.
Let’s add some markings in \(\mathbf{\color{red}{ red }}\) to the drawing:
Our first task is to find the angles the cables make with the horizontal. The dashed horizontal line where the two cables meet just above the 5,000 weight and the solid line near the top where the cables hang from are parallel to each other; the cables are the ‘transversals’. Using these transversals, we can say that \(\mathbf{\theta_{R}}\) are ‘alternate interior angles’, by definition they have the same value. Likewise for the angles \(\mathbf{\theta_{L}}\). The way to proceed now is to find their values.
\begin{align*}
\text{Tangent is define as } \frac{\text{opposite side}}{\text{adjacent side}}. \text{Therefore, } \tan \theta_{R} &= \frac{24}{20} = \frac{6}{5} \\
\text{What is the angle whose tangent is….}\, \theta_{R} &= \arctan \left(\frac{6}{5}\right) \text{ ?} \\
\text{Using a calculator: } \boldsymbol{\theta_{R}} &= \mathbf{50.2^o} \\
\text{Similarly for the left cable: } \tan \theta_{L} &= \frac{24}{10} = \frac{12}{5} \\
\text{And…. }\, \theta_{L} &= \arctan\left(\frac{12}{5}\right) \\
\text{Again using a calculator: } \boldsymbol{\theta_{L}} &= \mathbf{67.4^o}
\end{align*}
This next drawing shows the angles we just found. Also, for clarity, it has some of the additional \(\mathbf{\color{red}{ red }}\) markings removed.
Now we proceed as in the previous problem dealing with cable tension:
\begin{align*}
\text{Along the horizontal (x-axis): } \qquad \qquad \qquad \|\mathbf{C_{L}\|} \cos 67.4^{o} &= \|\mathbf{C_{R}\|} \cos 50.2^{o} \\
\text{Along the vertical (y-axis): } \|\mathbf{C_{L}\|} \sin 67.4^{o} + \|\mathbf{C_{R}\|} \sin 50.2^{o} &= 5000 \\
\text{We have a system of two equations with two unknowns.} \quad \\
\text{We solve for } \|\mathbf{C_{L}\|} \text{ in terms of } \|\mathbf{C_{R}\|} \text{: } \quad \|\mathbf{C_{L}\|} &= \frac{\|\mathbf{C_{R}\|} \cos 50.2^{o}}{\cos 67.4^{o}} \\
\text{Replace } \|\mathbf{C_{L}}\| \text{ in the y-axis equation: } \left(\frac{\|\mathbf{C_{R}}\| \cos 50.2^{o}}{\cos 67.4^{o}}\right) \sin 67.4^{o} + \|\mathbf{C_{R}}\| \sin 50.2^{o} &= 5000 \\
\text{Solve for } \|\mathbf{C_{R}}\| \text{ using a calculator: } \: \qquad \qquad \|\mathbf{C_{R}}\| &\approx 2168 \text{ pounds} \\
\text{Use the value for } \|\mathbf{C_{R}}\| \text{ in the x-axis equation: } \quad \|\mathbf{C_{L}}\| \cos 67.4^{o} &= \left(2168\right) \cos 50.2^{o} \\
\text{Solve for } \|\mathbf{C_{L}}\| \text{ using a calculator: } \: \qquad \qquad \|\mathbf{C_{L}}\| &\approx 3611 \text{ pounds}
\end{align*}
The tension in the left cable is 3,611 pounds; tension in the right cable is 2,168 pounds. A little bit of work but, once the technique is mastered, similar problems shouldn’t appear difficult.
Quiz #4
I think everyone has had enough of vectors, including myself! For this week’s quiz let’s try something different.
A particle is moving along the x-axis. The position of the particle at time \(\boldsymbol{t}\) is given by:
\[\mathbf{x\left(t\right) = t^{3} -\, 6t^{2} + 9t\, -\, 2, \quad 0 \le t \le 5}\]
Find the total distance the particle travels in 5 units of time.
Notice the problem asks for ‘total distance’ and not ‘displacement’.
Answer to Quiz #3
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 (vector \(\overrightarrow{\mathbf{t}}\)) and also the magnitude of vector \(\overrightarrow{\mathbf{u}}\).
Here’s a quick diagram of the situation.
To solve this problem, I admit that I had to get help to get me started in the right direction.
There are three vectors: \(\overrightarrow{\mathbf{u}}\), \(\overrightarrow{\mathbf{w}}\), and \(\overrightarrow{\mathbf{t}}\). Let’s breakdown each one in turn:
\begin{align*}
\text{Horizontal force: } \quad \overrightarrow{\mathbf{u}} &= \|\mathbf{u}\| \cdot \mathbf{i} \\
\text{Vertical force (weight):} \quad \overrightarrow{\mathbf{w}} &= -\mathbf{j} \\
\text{Rope: } \quad \overrightarrow{\mathbf{t}} &= \|\mathbf{t}\| \cdot \left(\cos \left(135^{o}\right) \cdot \mathbf{i} + \sin \left(135^{o}\right) \cdot \mathbf{j}\right) \\
\text{The three forces are balanced, therefore } \qquad \qquad \overrightarrow{\mathbf{u}} + \overrightarrow{\mathbf{w}} + \overrightarrow{\mathbf{t}} &= 0 \\
\text{which implies that } \quad \|\mathbf{u}\| + \|\mathbf{t}\| \cdot \cos 135^{o} &= 0 \\
\text{and } \quad -\, 1 + \|\mathbf{t}\| \cdot \sin 135^{o} &= 0 \\
\text{Solve for the magnitude of vector } \overrightarrow{\mathbf{t}}\text{: } \quad \qquad \qquad \qquad \|\mathbf{t}\| &= \frac{1}{\sin 135^{o}} = \sqrt{2} \approx 1.4142 \\
\text{Use this value and solve for the magnitude of vector } \overrightarrow{\mathbf{u}}\text{: } \quad \|\mathbf{u}\| + \sqrt{2} \cdot \cos 135^{o} &= 0 \\
\|\mathbf{u}\| &= -\, \sqrt{2} \cdot \cos 135^{o} = 1
\end{align*}
The final answer is: tension on the rope \(\approx\) 1.4142 pounds. Horizontal pull, that is, magnitude of vector \(\overrightarrow{\mathbf{u}} = \) 1 pound, same as the weight of the ball.
If you find any errors in this post, or have some thoughts about it, please, email me at [email protected] .