College Physics 3.24 – Resultant displacement of a pilot


Suppose a pilot flies 40.0 km in a direction 60º north of east and then flies 30.0 km in a direction 15º north of east as shown in Figure 3.61. Find her total distance R from the starting point and the direction θ of the straight-line path to the final position. Discuss qualitatively how this flight would be altered by a wind from the north and how the effect of the wind would depend on both wind speed and the speed of the plane relative to the air mass.

The vector A has a magnitude of 40 km in a direction 60º  north of east and the vector B has a magnitude of 30.0 km in a direction 15º north of east. The resultant vector is unknown.
Figure 3.61: The given vectors A and B, with the unknown vector R.

Solution:

The pilot’s displacement is characterized by 2 vectors, A and B, as depicted in Figure 3.61. To determine her total displacement R from the starting point, we need to add the two given vectors. To do this, we individually get the x and y components of each vector. This is presented in the table that follows:

Vectorx-componenty-component
A\displaystyle 40\:\cos 60^{\circ} =20\:\text{km}\displaystyle 40\:\sin 60^{\circ} =34.6410\:\text{km}
B\displaystyle 30\:\cos 15^{\circ} =28.9778\:\text{km}\displaystyle 30\:\sin 15^{\circ} =7.7646\:\text{km}
Sum\displaystyle 48.9778\: \text{km}\displaystyle 42.4056 \:\text{km}

The table above indicates east and north as positive components, while west and south indicates negative components. The last row is the sum of the components. This is also the x and y components of the resultant vector.

To calculate the magnitude of the resultant, we simply use the Pythagorean Theorem as follows:

\displaystyle \text{R}=\sqrt{\left(48.9778\:\text{km}\right)^2+\left(42.4056\:\text{km}\right)^2}

\displaystyle \text{R}=64.7847\:\text{km}     ◀

The direction of the resultant is calculated as follows:

\displaystyle \theta =\tan ^{-1}\left(\frac{42.4056}{48.9778}\right)

\displaystyle \theta =40.9^{\circ}     ◀

Therefore, the pilot’s resultant displacement is about 64.7847 km directed 40.9° North of East from the starting island.

Discussion:

If the wind speed is less than the speed of the plane, it is possible to travel to the northeast, but she will travel more to the east than without the wind. If the wind speed is greater than the speed of the plane, then it is no longer possible for the plane to travel to the northeast, it will end up travelling southeast.


College Physics 3.23 – Addition of multiple vectors


In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he is blown along the following straight lines: 2.50 km 45.0º north of west; then 4.70 km 60.0º south of east; then 1.30 km 25.0º south of west; then 5.10 km straight east; then 1.70 km 5.00º east of north; then 7.20 km 55.0º south of west; and finally 2.80 km 10.0º north of east. What is his final position relative to the island?


Solution:

Gilligan’s displacement is characterized by 7 vectors. To determine his final position relative to the starting point, we simply need to add the vectors. To do this, we individually get the x and y components of each vector. This is presented in the table that follows:

Vectorx-componenty-component
\displaystyle -2.5\:\cos 45^{\circ} =-1.7678\:\text{km}\displaystyle +2.5\:\sin 45^{\circ} =+1.7678\:\text{km}
\displaystyle +4.70\:\cos 60^{\circ} =+2.3500\:\text{km}\displaystyle -4.70\:\sin 60^{\circ} =-4.0703\:\text{km}
\displaystyle -1.30\:\cos 25^{\circ} =-1.1782\:\text{km}\displaystyle -1.30\:\sin 25^{\circ} =-0.5494\:\text{km}
\displaystyle +5.1000\:\text{km}\displaystyle 0
\displaystyle +1.70\:\sin 5^{\circ} =+0.1482\:\text{km}\displaystyle +1.70\:\cos 5^{\circ} =+1.6935\:\text{km}
\displaystyle -7.20\:\cos 55^{\circ} =-4.1298\:\text{km}\displaystyle -7.20\:\sin 55^{\circ} =-5.8979\:\text{km}
\displaystyle +2.80\:\cos 10^{\circ} =+2.7575\:\text{km}\displaystyle +2.80\:\sin 10^{\circ} =+0.4862\:\text{km}
Sum\displaystyle 3.2799\:\text{km}\displaystyle -6.5701\:\text{km}

The table above indicates east and north as positive components, while west and south indicates negative components. The last row is the sum of the components. This is also the x and y components of the resultant vector.

To calculate the magnitude of the resultant, we simply use the Pythagorean Theorem as follows:

\displaystyle \text{R}=\sqrt{\left(3.2799\:\text{km}\right)^2+\left(-6.5701\:\text{km}\right)^2}

\displaystyle \text{R}=7.3433\:\text{km}

The direction of the resultant is calculated as follows:

\displaystyle \theta =\tan ^{-1}\left(\frac{6.5701}{3.2799}\right)

\displaystyle \theta =63.47^{\circ}

Therefore, Gilligan is about 7.3433 km directed 63.47° South of East from the starting island.


College Physics 3.19 – Vector Operations


Do Exercise 3.16 again using analytical techniques and change the second leg of the walk to 25.0 m straight south. (This is equivalent to subtracting B from A — that is, finding R’=A – B ) (b) Repeat again, but now you first walk 25.0 m north and then 18.0 m east. (This is equivalent to subtract A from B —that is, to find A=B+C . Is that consistent with your result?)


Solution:

Part A

From the given statement, you first walk 18.0 m straight west and then 25.0 straight south. These vectors are represented by the graph shown below.

The two displacement vectors A and B, and their resultant R directed θ degrees from the x-axis.
The two displacement vectors A and B, and their resultant R directed θ degrees from the x-axis.

To solve for the resultant, we simply need to use the Pythagorean theorem to solve for the hypotenuse of the right triangle formed. That is,

\displaystyle \text{R}=\sqrt{\text{A}^2+\text{B}^2}

\displaystyle \text{R}=\sqrt{\left(18.0\:\text{m}\right)^2+\left(25.0\:\text{m}\right)^2}

\displaystyle \text{R}=30.8\:\text{m}

To solve for the angle, θ, we shall use the tangent function.

\displaystyle \theta =\tan ^{-1}\left(\frac{\text{B}}{\text{A}}\right)

\displaystyle \theta =\tan ^{-1}\left(\frac{25.0\:\text{m}}{18.0\:\text{m}}\right)

\displaystyle \theta =54.25^{\circ}

Thus, the value of 𝜙 can be calculated as

\displaystyle \phi=90^{\circ} -54.25^{\circ}

\displaystyle \phi=35.75^{\circ}

Therefore, the compass direction of the resultant is 35.75° West of South.

Part B

From the statement, you walk 25.0 m north first and then 18.0 m east. This is represented by the figure shown below.

The displacement vectors A and B, and their resultant R directed 𝜙 degrees from the east axis.

So, we have a right triangle with legs 25.0 m and 18.0 m. We are tasked to solve for the value of R, and the angle 𝜙 for the compass direction. The value of R can be solved using the Pythagorean Theorem as in Part A.

\displaystyle \text{R}=\sqrt{\text{A}^2+\text{B}^2}

\displaystyle \text{R}=\sqrt{\left(18.0\:\text{m}\right)^2+\left(25.0\:\text{m}\right)^2}

\displaystyle \text{R}=30.8\:\text{m}

To solve for the angle, θ, we shall use the tangent function.

\displaystyle \theta =\tan ^{-1}\left(\frac{\text{B}}{\text{A}}\right)

\displaystyle \theta =\tan ^{-1}\left(\frac{18.0\:\text{m}}{25.0\:\text{m}}\right)

\displaystyle \theta =35.75^{\circ}

Therefore, the compass direction of the resultant is 35.75° East of North.


Farmer wants to Fence off his Four-Sided Plot with missing Side| Vector Addition and Subtraction| Analytical Method| College Physics| Problem 3.22


A farmer wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in the figure, and then correctly calculates the length and orientation of the fourth side D. What is his result?

8

SOLUTION:

We know that the sum of the 4 vectors is zero. So, Vector D is calculated as

D=-A-B-C

We solve for the x-component first. The x-component of vector D is

D_x=-A_x-B_x-C_x

D_x=-\left(4.70\:km\right)cos\left(-7.50^{\circ} \right)-\left(2.48\:km\right)cos106^{\circ} -\left(3.02\:km\right)cos161^{\circ}

D_x=-1.12\:km

We solve for the y-component.

D_y=-A_y-B_y-C_y

D_y=-\left(4.70\:km\right)sin\left(-7.50^{\circ} \right)-\left(2.48\:km\right)sin106^{\circ} -\left(3.02\:km\right)sin161^{\circ}

D_y=-2.75\:km

Since we already know the x and y component of vector D, we can finally solve for the distance of vector D.

D=\sqrt{\left(D_x\right)^2+\left(D_y\right)^2}

D=\sqrt{\left(-1.12\:km\right)^2+\left(-2.75\:km\right)^2}

D=2.97\:km

The corresponding direction of vector D is

\theta =tan^{-1}\left|\frac{D_x}{D_y}\right|

\theta =tan^{-1}\left|\frac{1.12\:km}{2.75\:km}\right|

\theta =22.2^{\circ} \:W\:of\:S


College Physics 3.21 – Vector Addition and Subtraction


You fly 32.0 km in a straight line in still air in the direction 35.0° south of west.

(a) Find the distances you would have to fly straight south and then straight west to arrive at the same point. (This determination is equivalent to finding the components of the displacement along the south and west directions.)

(b) Find the distances you would have to fly first in a direction 45.0º south of west and then in a direction 45.0° west of north. These are the components of the displacement along a different set of axes—one rotated 45.0°. 


Solution:

Part A

The component along the south direction is

D_s=Rsin\theta =\left(32.0\:km\right)sin\:35^{\circ} =18.4\:km

The component along the west direction is

D_w=Rcos\theta =\left(32.0\:km\right)cos35^{\circ} =26.2\:km

Part B

Consider the following figure with the rotated axes x’-y’.

7

The component along the southwest direction is

D_{SW}=Rcos\theta '=\left(32.0\:km\right)cos10^{\circ} =31.5\:km

The component along the northwest direction is

D_{NW}=Rsin\theta '=\left(32.0\:km\right)cos10^{\circ} =5.56\:km


College Physics 3.20 – Triangular Piece of Flat Land


A new landowner has a triangular piece of flat land she wishes to fence. Starting at the west corner, she measures the first side to be 80.0 m long and the next to be 105 m. These sides are represented as displacement vectors A from B in Figure 3.61. She then correctly calculates the length and orientation of the third side C. What is her result?

The displacement vectors A, B, and C formed a closed triangle. A is 80 m directed 21° clockwise from the positive x-axis, vector B has a magnitude of 105 m and is directed 11° from the positive y-axis. Find for the magnitude and direction of vector C.
Figure 3.61: The displacement vectors A, B, and C.

Solution:

Since the figure is closed, we know that the sum of the three vectors is equal to 0. So, we have 

\vec{A}+\vec{B}+\vec{C}=0

Solving for C, we have

\vec{C}=-\vec{A}-\vec{B}

Next, we solve for the x and y components of Vectors A and B. 

For Vector A, the x and y components are

A_x=\left(80\:m\right)cos\left(21^{\circ} \right)=74.6864\:m

A_y=-\left(80\:m\right)sin\left(21^{\circ} \right)=-28.6694\:m

For Vector B, the x and y components are

B_x=-\left(105\:m\right)sin\:11^{\circ} =-20.0349\:m

B_y=\left(105\:m\right)cos\:11^{\circ} =103.0709\:m

Solve for the components of Vector C. 

C_x=-A_x-B_x=-74.6864\:m\:-\left(-20.0349\:m\right)=-54.6515\:m

C_y=-A_y-B_y=-\left(-28.6694\:m\right)-\left(103.0709\:m\right)=-74.4015\:m

Therefore, the length of Vector C is

C=\sqrt{\left(C_x\right)^2+\left(C_y\right)^2}

C=\sqrt{\left(-54.6515\:m\right)^2+\left(-74.4015\:m\right)^2}

C=92.3167\:m

The direction of Vector C is

\theta =tan^{-1}\left(\frac{C_y}{C_x}\right)=tan^{-1}\left(\frac{74.4015\:m}{54.6515\:m}\right)=53.7^{\circ} \:\:South\:of\:West


College Physics 3.18 – Driving in a straight line vs east and north distances


You drive 7.50 km in a straight line in a direction 15° east of north.

(a) Find the distances you would have to drive straight east and then straight north to arrive at the same point. (This determination is equivalent to find the components of the displacement along the east and north directions.)

(b) Show that you still arrive at the same point if the east and north legs are reversed in order.


Solution:

Part A

Consider the following figure:

North and East Components of the given displacement
North and East Components of the given displacement

The east distance is the component in the horizontal direction.

D_E=7.50\:km\:\cdot sin\:\left(15^{\circ} \right)

D_E=1.94\:km

The north distance is the vertical component

D_E=7.50\:km\cdot cos\left(15^{\circ} \right)

D_E=7.24\:km

Part B

3.19

Based from the figure, we can easily see that the order is reversible in the addition of vectors. We say that D_E+D_N=D_N+D_E


College Physics 3.17 – Reverse addition of vectors


Repeat Problem 3.16 using analytical techniques, but reverse the order of the two legs of the walk and show that you get the same final result. (This problem shows that adding them in reverse order gives the same result—that is, B+A=A+B) Discuss how taking another path to reach the same point might help to overcome an obstacle blocking your other path.

Reverse addition of the two vectors A and B.
Reverse addition of the two vectors A and B.

Solution:

Consider the right triangle formed by the legs A, B, and R. We know that A is 18 m, B is 25 m, and we are solving for the magnitude of R. We can do this by using the Pythagorean Theorem. That is

R=\sqrt{A^2+B^2}

R=\sqrt{\left(18\:m\right)^2+\left(25\:m\right)^2}

R=\sqrt{324+625}

R=\sqrt{949}

R=30.8\:m

So, the distance is about 30.8 meters from the starting point. To solve for the value of the unknown angle, φ, we can use the tangent function. That is 

tan\:\phi =\frac{A}{B}

tan\:\phi =\frac{18\:m}{25\:m}

\phi =tan^{-1}\left(\frac{18}{25}\right)

\phi =35.75^{\circ}

Therefore, the compass angle is 35.75^{\circ} \:West\:of\:North


College Physics 3.16 – Resultant of two vectors


Solve the following problem using analytical techniques: Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B , as in Figure 3.60, then this problem asks you to find their sum R = A + B .)

The resultant R and θ are shown in the figure. We are given the vectors A and B, and the right triangle is formed.
Figure 3.60: The two displacements A and B add to give a total displacement R having magnitude R and direction θ.

Note that you can also solve this graphically. Discuss why the analytical technique for solving this problem is potentially more accurate than the graphical technique.


Solution:

Consider the right triangle formed by the legs A, B, and R. We know that A is 18 m, B is 25 m, and we are solving for the magnitude of R. We can do this by using the Pythagorean Theorem. That is

R=\sqrt{A^2+B^2}

R=\sqrt{\left(18\:m\right)^2+\left(25\:m\right)^2}

R=\sqrt{324+625}

R=\sqrt{949}

R=30.8\:m

So, the distance is about 30.8 meters from the starting point. To solve for the value of the unknown angle, θ, we can use the tangent function. That is 

tan\:\theta =\frac{B}{A}

tan\:\theta =\frac{25\:m}{18\:m}

\theta =tan^{-1}\left(\frac{25}{18}\right)

\theta =54.25^{\circ}

So, the compass reading, can be solved by taking the complimentary angle, \phi as shown in the figure.

The angle 𝜙 is the complimentary angle of 𝜃.
The angle ɸ is the complimentary angle of θ.

\phi =90^{\circ} -54.25^{\circ}

\phi =35.75^{\circ}

Therefore, the compass angle is 35.75^{\circ} \:West\:of\:North