## 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?

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 $\theta =tan^{-1}\left|\frac{D_x}{D_y}\right|$ $\theta =tan^{-1}\left|\frac{1.12\:km}{2.75\:km}\right|$ $\theta =22.2^{\circ} \:W\:of\:S$

## 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’.

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

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

## Solution:

### Part A

Consider the following figure:

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

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

## 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 $35.75^{\circ} \:West\:of\:North$

## 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. $\phi =90^{\circ} -54.25^{\circ}$ $\phi =35.75^{\circ}$

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

## Solution:

### North Component

The North component is given by $S_N=123\:km\:\cdot sin\left(45^{\circ} \right)=86.97\:km$

### East Component

The east component is $S_E=123\:km\:\cdot cos\:\left(45^{\circ} \right)=86.97\:km$

## College Physics 3.14 – Distance and displacement of a path

#### In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition. Figure 3.58 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.

## Solution:

### Part A

The total distance traveled following Path D is $d=\left(2\times 120\:m\right)+\left(6\times 120\:m\right)+\left(4\times 120\:m\right)+\left(1\times 120\:m\right)$ $d=1560\:m$

### Part B

Treat all motions upward and to the right positive, while downward and to the left negative.

The displacement in the x-direction is $s_x=0+720\:m+0-120\:m=600\:m$

The displacement in the y-direction is $s_y=-240\:m+0+480\:m+0=240\:m$

The total displacement is $s=\sqrt{\left(s_x\right)^2+\left(s_y\right)^2}=\sqrt{\left(600\:m\right)^2+\left(240\:m\right)^2}=646.22\:m$

The direction angle from th x-axis is given by $\theta _x=tan^{-1}\left|\frac{s_y}{s_x}\right|=tan^{-1}\left|\frac{240}{600}\right|=21.8^{\circ} \:North\:of\:East$

Therefore, the displacement is 646.22 m, directed at 21.8 degrees north of east.

## College Physics 3.13 – Distance and Displacement

#### (b) the magnitude and direction of the displacement from start to finish.In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition. Figure 3.58 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.

## Solution:

### Part A

The total distance traveled following Path C is $d=\left(1\times 120\:m\right)+\left(5\times 120\:m\right)+\left(2\times 120\:m\right)+\left(1\times 120\:m\right)+\left(1\times 120\:m\right)+\left(3\times 120\:m\right)$ $d=1560\:m$

### Part B

Treat all motions upward and to the right positive, while downward and to the left negative.

The displacement in the x-direction is $s_x=0+600+0-120+0-360=120\:m$

The displacement in the y-direction is $s_y=120+0-240+0+120+0=0\:m$

The total displacement is $s=\sqrt{\left(s_x\right)^2+\left(s_y\right)^2}$ $s=\sqrt{\left(120\:m\right)^2+\left(0\:m\right)^2}$ $s=120\:m$

The direction angle from th x-axis is given by $\theta _x=tan^{-1}\left|\frac{s_y}{s_x}\right|$ $\theta _x=tan^{-1}\left|\frac{0\:m}{120\:m}\right|$ $\theta _x=0^{\circ}$

Therefore, the displacement is 120 m, east.