Vectors are used all the time in Physics and I am still only learning how to use them. I will try to provide a good example of their usefulness.
Suppose you spent a nice afternoon fishing and now you’re getting ready to go back home. A force of 600 pounds is required to pull your boat and trailer up the ramp. This ramp is inclined at 15° from the horizontal. What is the combined weight of the boat and trailer?
After reading the problem carefully, all textbooks always recommend to draw a diagram of the situation:
Yes, this is a very amateurish diagram, but it should be good enough. It shows the boat and trailer on a ramp. This ramp has a 15° inclination. I have also drawn three vectors:
- Vector \(\overrightarrow{\mathbf{BA}}\) represents the combined weight of the boat and trailer; what we are asked to find.
- Vector \(\overrightarrow{\mathbf{AC}}\) is the force required to pull the boat and trailer up the ramp (600 lbs).
- Vector \(\overrightarrow{\mathbf{BC}}\) is the force against the ramp
Notice that vector \(\overrightarrow{\mathbf{AC}}\) and the ramp are parallel to each other and perpendicular to vector \(\overrightarrow{\mathbf{BC}}\). Also, angles WDB and C are right angles; therefore, BWD and ABC are similar triangles. This makes ABC a 15° angle, like the ramp.
How do we use this configuration to find the combined weight of the boat and trailer? We have a right triangle (ABC) and, using Trigonometry, we write an expression relating the 15° angle and the magnitude of vector \(\overrightarrow{\mathbf{AC}}\), (opposite side) to find the magnitude of vector \(\overrightarrow{\mathbf{BA}}\) (hypotenuse).
\begin{align*}
\sin 15° = \frac{opp.}{hyp.} = \frac{\|\overrightarrow{\mathbf{AC}}\|}{\|\overrightarrow{\mathbf{BA}}\|} &= \frac{600}{\|\overrightarrow{\mathbf{BA}}\|} \\
\text{Using algebra: } {\|\overrightarrow{\mathbf{BA}}\|} &= \frac{600}{\sin 15°} \approx 2318\,lbs
\end{align*}
Answer: the combined weight of the boat and trailer is about \(\mathbf{2318\,lbs}\).
I think it is obvious that learning vectors and basic Trig makes this kind of problem almost easy!
I am adding a new feature to these posts – a weekly quiz! So, here it is:
Quiz #1
Getting the answer to the problem given above required getting the value of the quotient \(\boldsymbol{\frac{600}{\sin 15°}}\). A calculator is normally used for this type of calculation, giving us an approximation. However, it is not absolutely necessary. How would you make this computation with pencil and paper, without a calculator and obtain an EXACT value?
Answer next week.
If you find any errors in this post, or have some thoughts about it, please, email me at [email protected] .