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

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

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.

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.

Vector Addition and Subtraction|College Physics| Problem 3.5

Suppose you first walk 12.0 m in a direction 20º west of north and then 20.0 m in a direction 40.0º south of west. 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 3.56, then this problem finds their sum R = A + B .)

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