VM - Vector Applications Lesson

Vector Applications

Now that you know about vectors, how can they be used in real applications?

Source: FuseSchool - Global Education. (2020, February 16). Resultant Forces | Force & Motion | Physics | FuseSchool. YouTube.

Running Example:

²A person runs 9 blocks east and 5 blocks north. Draw and picture and then find the final displacement. What does this represent?

Answer:

Answer to the example problem

Component form of the resultant vector: <9, 5>

To get the magnitude of the resultant, we will use the Pythagorean theorem.

$LaTeX: 9^2+5^2=c^2\\c\approx10.3^\circ$ $9^2+5^2=c^2\\c\approx10.3^\circ$

To get the direction of the resultant we will use the tangent ratio.

$LaTeX: \tan\;\theta=\frac59\\\tan^{-1}(\frac59)\;=\;29.1^\circ$ $\tan\;\theta=\frac59\\\tan^{-1}(\frac59)\;=\;29.1^\circ$

The displacement is 10.3 blocks at an angle 29.1^o north of east. This means that if the person took a direct path from the initial point to the final destination, that would be the path.

Camping Example:

Two campers in a national park hike from their cabin to the same spot on a lake, each taking a different path, as illustrated below. The total distance traveled along Path 1 is 7.5 km, and that along Path 2 is 8.2 km. What are the rectangular components of force (horizontal and vertical displacement) of each camper?

This problem is actually going to be worked out in the reverse order as our last one. We are given the angle measure and the magnitude of the resultant vector.

Let's find the horizontal displacement:

$LaTeX: \tan\;40=\frac5x\\x=5/\tan\;40\\x\approx5.96\\$ $\tan\;40=\frac5x\\x=5/\tan\;40\\x\approx5.96\\$

Then we can use the pythagorean theorem to find the vertical displacement:

$LaTeX: {5.96}^2+\;b^2=5^2\\Solve\;for\;b.\\b\approx3.24\\\\\\$ ${5.96}^2+\;b^2=5^2\\Solve\;for\;b.\\b\approx3.24\\\\\\$

If the displacement is 5 km at 40 degrees North of East, the component form of the vector is <5.96, 3.24>, which means the horizontal displacement is 5.96 and the vertical displacement is 3.24.

Wagon Example:

⁴Jake is towing his friend on a wagon, using a rope which makes an angle of 25 degrees with the ground. If Jake is pulling with a force of 70 Newtons, what horizontal force is he exerting on the wagon?

Forces Example:

The magnitude and direction of two forces acting on an object are 110 pounds, N15°E and 180 pounds, N70°E. Find the magnitude (to the nearest pound) and the direction angle (to the nearest tenth of a degree) of the resultant force F.

Force is represented by the magnitude of the vector.

F1 represents the force on the object at 110 pounds, N15°E (vector magnitude is 110)

F2 represents the force on the object at 180 pounds, N70°E (vector magnitude is 180)

We want F1+F2=F (resultant vector)

On the image above, F is drawn in using the Parallelogram Method. (See previous lesson)

First, find the direction angles (from the positive x-axis) for the forces F1 and F2. To find the direction angle for the forces we would subtract their bearing from 90°

F1: 90° - 15° = 75°

F2: 90° - 70° = 20°

In order to add F1 and F2, we need the component form. Remember when using the component form <x, y>, the x is the horizontal displacement and y is the vertical displacement.

The formula for the component form of a vector is <m*cos θ, m*sin θ> with m being the magnitude of the vector. In the image below we can see that vector m is component form is <m*cos θ, m*sin θ>.

components with trig

F1 = <|| F1 || cos θ, || F1 || sin θ>

F1 = <110 cos 75°, 110 sin 75° >

F1 ≈ <28.47,106.25>

F2 = <|| F2 || cos θ, || F2 || sin θ>

F2 = <180 cos 20°, 180 sin 20°>

F2 ≈ <169.14, 61.56>

Resultant Force:

F = F1 + F2

F ≈ <28.47,106.25>+<169.14, 61.56> ≈ <197.61,167.81

Magnitude of resultant force:

|| F1 ||= $LaTeX: \sqrt{{197.61}^2+167.81^2}\approx259\\\\\\$ $\sqrt{{197.61}^2+167.81^2}\approx259\\\\\\$

Find the direction angle of the resultant force:

$LaTeX: \tan\;\theta=\;(\frac{167.81}{197.61})\\\theta\approx40.3^\circ\\\\\\$ $\tan\;\theta=\;(\frac{167.81}{197.61})\\\theta\approx40.3^\circ\\\\\\$

Boat Example:

A boat travels 30 mph due west if there is a 7 mph current at N30 degrees W. Find the actual speed and direction of the boat.

Find the velocity vector of the boat.
Find the velocity vector of the current.
Find the actual velocity vector.
Find the actual speed of the boat.
Find the actual direction of the boat.

Answer:

Graph

$LaTeX: b=<30\;\cos\;180^\circ,30\;\sin\;180^\circ>=<-30,0>\\$ $b=<30\;\cos\;180^\circ,30\;\sin\;180^\circ>=<-30,0>\\$
$LaTeX: c=<7\;\cos\;120^\circ,\;7\;\sin\;120^\circ>=<-3.5,\;6.06>\\$ $c=<7\;\cos\;120^\circ,\;7\;\sin\;120^\circ>=<-3.5,\;6.06>\\$
$LaTeX: \overset\rightharpoonup{b+c}=<-33.5,\;6.06>\\$ $\overset\rightharpoonup{b+c}=<-33.5,\;6.06>\\$
$LaTeX: \left\|\overset\rightharpoonup{b+c}\right\|=\sqrt{33.5^2+6.06^2}=34.04\;mph\\$ $\left\|\overset\rightharpoonup{b+c}\right\|=\sqrt{33.5^2+6.06^2}=34.04\;mph\\$

graph

$LaTeX: -33.5=34.04\;\cos\;\theta\\\frac{-33.5}{34.04}=\cos\;\theta\\\cos^{-1}\;(\frac{33.5}{34.04})=\;\theta\\\\\;\theta=10.21^\circ\\180-10.21^\circ=169.7^\circ\\\\\\$ $-33.5=34.04\;\cos\;\theta\\\frac{-33.5}{34.04}=\cos\;\theta\\\cos^{-1}\;(\frac{33.5}{34.04})=\;\theta\\\\\;\theta=10.21^\circ\\180-10.21^\circ=169.7^\circ\\\\\\$

Above I took out the negative, keeping in mind we are working in the 2nd quadrant. Then once I found the inverse cosine, I subtracted it from 180.

$LaTeX: N79.7^\circ W\;or\;bearing\;280.3^\circ\\\\\\$ $N79.7^\circ W\;or\;bearing\;280.3^\circ\\\\\\$

Plane Example:

³A plane flies through wind that blows with a speed of 35 miles per hour in the direction N 43° W.
In still air: the plane has a speed of 550 miles per hour; the plane is headed in the direction N 60° E.

Find the true velocity of the plane as a vector.
Find the true speed and direction of the plane.

Answer:

Let's use W for Wind and P for Plane in still air.

airplane speed direction table

Sketch a picture for both so we can find the standard direction.

airplane convert standard

Then use our previous knowledge that

airplane problem workout

Let's see what this looks like:

Airplane sketch

True Velocity as a vector:

airplace true velocity

True Speed:

airplane sketch Truespeed

This is because the wind slowed the plane down.

True Direction:

The true direction of the plane is the direction of the true velocity vector.

True direction

Or if we are asked to find the true direction as bearing we can also do that:

$LaTeX: 90^\circ-33.6^\circ=56.4^\circ\\\\N\;56.4^\circ\\$ $90^\circ-33.6^\circ=56.4^\circ\\\\N\;56.4^\circ\\$

You try:

If a small boat crossing a river has a velocity of 5 km/h due north in still water and the water has a current of 2 km/h due west (see the following figure), what is the velocity of the boat relative to shore? What is the angle $LaTeX: \Theta$ $\Theta$ that the boat is actually traveling?

Figure of small boat crossing the river

²Hamm, K. (2020, August 1). 4.2 Vector Addition and Subtraction – Biomechanics of Human Movement Links to an external site.. Pressbooks.

³Vector Application: Finding True Speed and Direction Links to an external site.. (n.d.).

⁴PrepAnywhere. (2015, December 16). 3. Force as a Vector, pulling a Toboggan ex2. YouTube. https://www.youtube.com/watch?v=nDDyixApcLw

^{5OpenStax University Physics.
Links to an external site.}Authored by: OpenStax CNX. Located at: Links to an external site. License: CC BY: Attribution.

⁶Libretexts Links to an external site.. (2022, September 7). 12.8: Chapter 12 Review Exercises. Mathematics LibreTexts.