2.3 Algebra of vectors (Page 3/8)

University physics volume 1 Page 3 / 8

Vector R has magnitude 13.11. The angle between R and the positive x direction is theta sub R equals 13.9 degrees. The components of R are R sub x on the x axis and R sub y on the y axis. Vector D has magnitude 16.23. The angle between D and the positive x direction is theta sub D equals 49.3 degrees. The components of D are D sub x on the x axis and D sub y on the y axis. Vector S has magnitude 36.95. The angle between S and the positive x direction is theta sub S equals 52.9 degrees. The components of S are S sub x on the x axis and S sub y on the y axis. — Graphical illustration of the solutions obtained analytically in [link] .

Check Your Understanding Three displacement vectors $\vec{A}$ , $\vec{B}$ , and $\vec{F}$ ( [link] ) are specified by their magnitudes A = 10.00, B = 7.00, and F = 20.00, respectively, and by their respective direction angles with the horizontal direction $α = 35 °$ , $β = −110 °$ , and $φ = 110 °$ . The physical units of the magnitudes are centimeters. Use the analytical method to find vector $\vec{G} = \vec{A} + 2 \vec{B} - \vec{F}$ . Verify that
G = 28.15 cm and that $θ_{G} = −68.65 °$ .

$\vec{G} = (10.25 \hat{i} - 26.22 \hat{j}) cm$

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The tug-of-war game

Four dogs named Ang, Bing, Chang, and Dong play a tug-of-war game with a toy ( [link] ). Ang pulls on the toy in direction

α = 55 °

south of east, Bing pulls in direction

β = 60 °

east of north, and Chang pulls in direction

γ = 55 °

west of north. Ang pulls strongly with 160.0 units of force (N), which we abbreviate as A = 160.0 N. Bing pulls even stronger than Ang with a force of magnitude B = 200.0 N, and Chang pulls with a force of magnitude C = 140.0 N. When Dong pulls on the toy in such a way that his force balances out the resultant of the other three forces, the toy does not move in any direction. With how big a force and in what direction must Dong pull on the toy for this to happen?

Illustration of 4 dogs pulling on a toy. The toy is at the origin of a coordinate system, with plus x aligned with east and plus y with north. Ang is pulling at an angle alpha which is 55 degrees clockwise from the plus x (east) direction. Bing is pulling at an angle beta which is 60 degrees clockwise from the plus y (north) direction. Chang is pulling at an angle gamma which is 55 degrees counterclockwise from the plus y (north) direction. Dong is pulling in an unspecified direction in the third quadrant. — Four dogs play a tug-of-war game with a toy.

Strategy

We assume that east is the direction of the positive x -axis and north is the direction of the positive y -axis. As in [link] , we have to resolve the three given forces—

\vec{A}

(the pull from Ang),

\vec{B}

(the pull from Bing), and

\vec{C}

(the pull from Chang)—into their scalar components and then find the scalar components of the resultant vector

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

. When the pulling force

\vec{D}

from Dong balances out this resultant, the sum of

\vec{D}

and

\vec{R}

must give the null vector

\vec{D} + \vec{R} = \vec{0}

. This means that

\vec{D} = − \vec{R}

, so the pull from Dong must be antiparallel to

\vec{R}

Solution

The direction angles are

θ_{A} = − α = −55 °

θ_{B} = 90 ° - β = 30 °

, and

θ_{C} = 90 ° + γ = 145 °

, and substituting them into [link] gives the scalar components of the three given forces:

\begin{matrix} {\begin{array}{l} A_{x} = A cos θ_{A} = (160.0 N) cos (−55 °) = + 91.8 N \\ A_{y} = A sin θ_{A} = (160.0 N) sin (−55 °) = −131.1 N \end{array} \\ {\begin{array}{l} B_{x} = B cos θ_{B} = (200.0 N) cos 30 ° = + 173.2 N \\ B_{y} = B sin θ_{B} = (200.0 N) sin 30 ° = + 100.0 N \end{array} \\ {\begin{array}{l} C_{x} = C cos θ_{C} = (140.0 N) cos 145 ° = −114.7 N \\ C_{y} = C sin θ_{C} = (140.0 N) sin 145 ° = + 80.3 N \end{array} \end{matrix} .

Now we compute scalar components of the resultant vector $\vec{R} = \vec{A} + \vec{B} + \vec{C}$ :

{\begin{matrix} R_{x} = A_{x} + B_{x} + C_{x} = + 91.8 N + 173.2 N - 114.7 N = + 150.3 N \\ R_{y} = A_{y} + B_{y} + C_{y} = −131.1 N + 100.0 N + 80.3 N = + 49.2 N \end{matrix} .

The antiparallel vector to the resultant $\vec{R}$ is

\vec{D} = − \vec{R} = − R_{x} \hat{i} - R_{y} \hat{j} = (−150.3 \hat{i} - 49.2 \hat{j}) N .

The magnitude of Dong’s pulling force is

D = \sqrt{D_{x}^{2} + D_{y}^{2}} = \sqrt{{(−150.3)}^{2} + {(−49.2)}^{2}} N = 158.1 N .

The direction of Dong’s pulling force is

θ = {tan}^{−1} (\frac{D_{y}}{D_{x}}) = {tan}^{−1} (\frac{−49.2 N}{−150.3 N}) = {tan}^{−1} (\frac{49.2}{150.3}) = 18.1 ° .

Dong pulls in the direction $18.1 °$ south of west because both components are negative, which means the pull vector lies in the third quadrant ( [link] ).

Check Your Understanding Suppose that Bing in [link] leaves the game to attend to more important matters, but Ang, Chang, and Dong continue playing. Ang and Chang’s pull on the toy does not change, but Dong runs around and bites on the toy in a different place. With how big a force and in what direction must Dong pull on the toy now to balance out the combined pulls from Chang and Ang? Illustrate this situation by drawing a vector diagram indicating all forces involved.

D = 55.7 N; direction $65.7 °$ north of east

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Source: OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5

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