Tag Archives: Components of a Displacement

College Physics by Openstax Chapter 3 Problem 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 Figure 3.60, and then correctly calculates the length and orientation of the fourth side D. What is his result?

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 Figure 3.60, and then correctly calculates the length and orientation of the fourth side D.
Figure 3.60

Solution:

For the four-sided plot to be closed, the resultant displacement of the four sides should be zero. The sum of the horizontal components should be zero, and the sum of the vertical components should also be equal to zero.

We need to solve for the components of each vector. Take into consideration that rightward and upward components are positive, while the reverse is negative.

For vector A, the components are

Ax=(4.70 km)cos⁡7.5∘Ax=4.6598 km\begin{align*} A_x & = \left( 4.70 \ \text{km} \right) \cos 7.5^\circ \\ A_x & = 4.6598 \ \text{km} \end{align*}
Ay=−(4.70 km)sin⁡7.5∘Ay=−0.6135 km\begin{align*} A_y & = -\left( 4.70 \ \text{km} \right) \sin 7.5^\circ \\ A_y & = -0.6135 \ \text{km} \end{align*}

The components of vector B are

Bx=−(2.48 km)sin⁡16∘Bx=−0.6836 km\begin{align*} B_x & =-\left( 2.48 \ \text{km} \right) \sin 16^\circ \\ B_x & = -0.6836 \ \text{km} \end{align*}
By=(2.48 km)cos⁡16∘By=2.3839 km\begin{align*} B_y & =\left( 2.48 \ \text{km} \right) \cos 16^\circ \\ B_y & =2.3839 \ \text{km} \end{align*}

For vector C, the components are

Cx=−(3.02 km)cos⁡19∘Cx=−2.8555 km\begin{align*} C_x & = -\left( 3.02 \ \text{km} \right) \cos 19^\circ \\ C_x & = -2.8555 \ \text{km} \end{align*}
Cy=(3.02 km)sin⁡19∘Cy=0.9832 km\begin{align*} C_y & = \left( 3.02 \ \text{km} \right) \sin 19^\circ \\ C_y & = 0.9832 \ \text{km} \end{align*}

Now, we need to take the sum of the x-components and equate it to zero. The x-component of D is unknown.

Ax+Bx+Cx+Dx=04.6598 km−0.6836 km−2.8555 km+Dx=01.1207 km+Dx=0Dx=−1.1207 km\begin{align*} A_x+B_x+C_x+D_x & =0 \\ 4.6598 \ \text{km}-0.6836 \ \text{km}-2.8555 \ \text{km}+ D_x & =0 \\ 1.1207 \ \text{km} +D_x & =0 \\ D_x & = -1.1207 \ \text{km} \end{align*}

We also need to take the sum of the y-component and equate it to zero to solve for the y-component of D.

Ay+By+Cy+Dy=0−0.6135 km+2.3839 km+0.9832 km+Dy=02.7536 km+Dy=0Dy=−2.7536 km\begin{align*} A_y +B_y+C_y+D_y & =0 \\ -0.6135 \ \text{km}+2.3839 \ \text{km}+0.9832 \ \text{km}+ D_y & =0 \\ 2.7536 \ \text{km} +D_y & =0 \\ D_y & = -2.7536 \ \text{km} \end{align*}

To solve for the distance of D, we shall use the Pythagorean Theorem.

D=(Dx)2+(Dy)2D=(−1.1207 km)2+(−2.7536 km)2D=2.97 km  (Answer)\begin{align*} D & = \sqrt{\left( D_x \right)^2+\left( D_y \right)^2} \\ D & = \sqrt{\left( -1.1207 \ \text{km} \right)^2+\left( -2.7536 \ \text{km} \right)^2} \\ D & = 2.97 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

Then we can solve for θ using the tangent function. Since it is taken from the vertical axis, it can be solved by:

θ=tan⁡−1∣DxDy∣θ=tan⁡−1∣−1.1207 km−2.7536 km∣θ=22.1∘  (Answer)\begin{align*} \theta & = \tan^{-1} \left| \frac{D_x}{D_y} \right| \\ \theta & = \tan^{-1} \left| \frac{-1.1207 \ \text{km}}{-2.7536 \ \text{km}} \right| \\ \theta & = 22.1 ^ \circ \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}
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College Physics by Openstax Chapter 3 Problem 21

You fly 32.0 km in a straight line in still air in the direction 35.0º 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 45.0º west of north. These are the components of the displacement along a different set of axes—one rotated 45º.

Solution:

Part A

Consider the illustration shown.

The south and west components of the 32.0 km distance are denoted by DS and DW, respectively. The values of these components are solved below:

DS=(32.0 km)sin⁡35.0∘DS=18.4∘  (Answer)\begin{align*} D_S & = \left( 32.0\ \text{km} \right) \sin 35.0 ^\circ \\ D_S & = 18.4^\circ \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}
DW=(32.0 km)cos⁡35.0∘DW=26.2∘  (Answer)\begin{align*} D_W & = \left( 32.0\ \text{km} \right) \cos 35.0 ^\circ \\ D_W & = 26.2^\circ \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

Part B

Consider the new set of axes (X-Y) as shown below. This new set of axes is rotated 45° from the original axes. Thus, axis X is 45° south of west, and axis Y is 45° west of north. First, we can obviously see that θ has a value of 10°.

Therefore, the components of the 32.0 km distance along X and Y axes are:

DX=(32.0 km)cos⁡10∘DX=31.5∘  (Answer)\begin{align*} D_X & = \left( 32.0 \ \text{km} \right) \cos 10^\circ \\ D_X & = 31.5^\circ \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}
DY=(32.0 km)sin⁡10∘DY=5.56∘  (Answer)\begin{align*} D_Y & = \left( 32.0 \ \text{km} \right) \sin 10^\circ \\ D_Y & = 5.56^\circ \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}
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College Physics by Openstax Chapter 3 Problem 18

You drive 7.50 km in a straight line in a direction 15º 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 illustration shown.

Let DE be the east component of the distance, and DN be the north component of the distance.

DE=7.50 sin⁡15∘DE=1.9411 kmDE=1.94 km  (Answer)\begin{align*} D_E & = 7.50 \ \sin 15^\circ \\ D_E & = 1.9411 \ \text{km} \\ D_E & = 1.94 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}
DN=7.50 cos⁡15∘DN=7.2444 kmDN=7.24 km  (Answer)\begin{align*} D_N & = 7.50 \ \cos 15^\circ \\ D_N & = 7.2444\ \text{km} \\ D_N & = 7.24 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

Part B

It can be obviously seen from the figure below that you still arrive at the same point if the east and north legs are reversed in order.

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College Physics by Openstax Chapter 3 Problem 3

Find the north and east components of the displacement for the hikers shown in Figure 3.50.

Figure 3.50

Solution:

Refer to Figure 3-3-A for the north and east components of the displacement s of the hikers.

Figure 3-3-A

Considering the right triangle formed. The north component is computed as

snorth=(5.00 km)sin⁡40∘snorth=3.21 km  (Answer)\begin{align*} \text{s}_{\text{north}} & = \left( 5.00 \ \text{km} \right)\sin 40^\circ \\ \text{s}_{\text{north}} & = 3.21 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\ \end{align*}

Using the same right triangle, the east component is computed as follows.

seast=(5.00 km)cos⁡40∘seast=3.83 km  (Answer)\begin{align*} \text{s}_{\text{east}} & = \left( 5.00 \ \text{km} \right)\cos 40^\circ \\ \text{s}_{\text{east}} & = 3.83 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\ \end{align*}
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