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Physics LibreTexts

3.7: Algebra of Vectors Examples

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Example 3.7.1: Analytical Computation of a Resultant

Three displacement vectors A, B, and C in a plane (Figure 2.3.6) are specified by their magnitudes A = 10.0, B = 7.0, and C = 8.0, respectively, and by their respective direction angles with the horizontal direction α = 35°, β = −110°, and γ = 30°. The physical units of the magnitudes are centimeters. Resolve the vectors to their scalar components and find the following vector sums:

  1. R = A + B + C,
  2. D = AB, and
  3. S = A − 3 B + C.
Strategy

First, we use Equation 2.4.13 to find the scalar components of each vector and then we express each vector in its vector component form given by A=Axˆi+Ayˆj. Then, we use analytical methods of vector algebra to find the resultants.

Solution

We resolve the given vectors to their scalar components:

{Ax=Acosα=(10.0cm)cos35o=8.19cmAy=Asinα=(10.0cm)sin35o=5.73cm

{Bx=Bcosβ=(7.0cm)cos(110o)=2.39cmBy=Bsinβ=(7.0cm)sin(110o)=6.58cm

{Cx=Ccosγ=(8.0cm)cos(30o)=6.93cmCy=Csinγ=(8.0cm)sin(30o)=4.00cm

For (a) we may substitute directly into Equation 2.6.7 to find the scalar components of the resultant:

{Rx=Ax+Bx+Cx=8.19cm2.39cm+6.93cm=12.73cmRy=Ay+By+Cy=5.73cm6.58cm+4.00cm=3.15cm

Therefore, the resultant vector is R=Rxˆi+Ryˆj=(12.7ˆi+3.1ˆj)cm. For (b), we may want to write the vector difference as

D=AB=(Axˆi+Ayˆj)(Bxˆi+Byˆj)=(AxBx)ˆi+(AyBy)ˆj.

Hence the difference vector is D=Dxˆi+Dyˆj=(10.6ˆi+12.3ˆj)cm.

For (c), we can write vector S in the following explicit form:

S=A3B+C=(Axˆi+Ayˆj)3(Bxˆi+Byˆj)+(Cxˆi+Cyˆj)=(Ax3Bx+Cx)ˆi+(Ay3By+Cy)ˆj.

Then, the scalar components of S are

{Sx=Ax3Bx+Cx=8.19cm3(2.39cm)+6.93cm=22.29cmSy=Ay3By+Cy=5.73cm3(6.58cm)+4.00cm=29.47cm

The vector is S=Sxˆi+Syˆj=(22.3ˆi+29.5ˆj)cm.

Significance

Having found the vector components, we can illustrate the vectors by graphing or we can compute magnitudes and direction angles, as shown in Figure 3.7.1. Results for the magnitudes in (b) and (c) can be compared with results for the same problems obtained with the graphical method, shown in Figure 2.3.7 and Figure 2.3.8. Notice that the analytical method produces exact results and its accuracy is not limited by the resolution of a ruler or a protractor, as it was with the graphical method used in Example 2.3.2 for finding this same resultant.

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.
Figure 3.7.1: Graphical illustration of the solutions obtained analytically.
Exercise 2.8

Three displacement vectors A, B, and F (Figure 2.3.6) 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 F = A + 2 BF. Verify that G = 28.15 cm and that θG = −68.65°.

Example 3.7.2: The Tug-of-War Game

Four dogs named Astro, Balto, Clifford, and Dug play a tug-of-war game with a toy (Figure 3.7.2). Astro pulls on the toy in direction α = 55° south of east, Balto pulls in direction β = 60° east of north, and Clifford pulls in direction γ = 55° west of north. Astro pulls strongly with 160.0 units of force (N), which we abbreviate as A = 160.0 N. Balto pulls even stronger than Astro with a force of magnitude B = 200.0 N, and Clifford pulls with a force of magnitude C = 140.0 N. When Dug 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 Dug 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. Astro is pulling at an angle alpha which is 55 degrees clockwise from the plus x (east) direction. Balto is pulling at an angle beta which is 60 degrees clockwise from the plus y (north) direction. Clifford is pulling at an angle gamma which is 55 degrees counterclockwise from the plus y (north) direction. Dug is pulling in an unspecified direction in the third quadrant.
Figure 3.7.2: 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 Example 3.7.1, we have to resolve the three given forces — A (the pull from Astro), B (the pull from Balto), and C (the pull from Clifford)—into their scalar components and then find the scalar components of the resultant vector R = A + B + C. When the pulling force D from Dug balances out this resultant, the sum of D and R must give the null vector D + R = 0. This means that D = R so the pull from Dug must be antiparallel to R.

Solution

The direction angles are θA = α = −55°, θB = 90° − β = 30°, and θC = 90° + γ = 145°, and substituting them into Equation 2.4.13 gives the scalar components of the three given forces:

{Ax=AcosθA=(160.0N)cos(55o)=+91.8NAy=AsinθA=(160.0N)sin(55o)=131.1N

{Bx=BcosθB=(200.0N)cos30o=+173.2NBy=BsinθB=(200.0N)sin30o=+100.0N

{Cx=CcosθC=(140.0N)cos145o=114.7NCy=CsinθC=(140.0N)sin145o=+80.3N

Now we compute scalar components of the resultant vector R=A+B+C:

{Rx=Ax+Bx+Cx=+91.8N+173.2N114.7N=+150.3NRy=Ay+By+Cy=131.1N+100.0N+80.3N=+49.2N

The antiparallel vector to the resultant R is

D=R=RxˆiRyˆj=(150.3ˆi49.2ˆj)N.

The magnitude of Dug's pulling force is

D=D2x+D2y=(150.3)2+(49.2)2N=158.1N.

The direction of Dug's pulling force is

θ=tan1(DyDx)=tan1(49.2N150.3N)=tan1(49.2150.3)=18.1o.

Dug pulls in the direction 18.1° south of west because both components are negative, which means the pull vector lies in the third quadrant (Figure 2.4.4).

Exercise 2.9

Suppose that Balto in Example 3.7.2 leaves the game to attend to more important matters, but Astro, Clifford, and Dug continue playing. Astro and Clifford’s pull on the toy does not change, but Dug runs around and bites on the toy in a different place. With how big a force and in what direction must Dug pull on the toy now to balance out the combined pulls from Clifford and Astro? Illustrate this situation by drawing a vector diagram indicating all forces involved.

Example 3.7.3: Vector Algebra

Find the magnitude of the vector C that satisfies the equation 2 A − 6 B + 3 C = 2 ˆj, A = ˆi − 2ˆk and B = − ˆj + ˆk2 .

Strategy

We first solve the given equation for the unknown vector C. Then we substitute A and B; group the terms along each of the three directions ˆi, ˆj, and ˆk; and identify the scalar components Cx, Cy, and Cz. Finally, we substitute into Equation 2.5.6 to find magnitude C.

Solution

2A6B+3C=2ˆj3C=2ˆj2A+6BC=23ˆj23A+2B=23ˆj23(ˆi2ˆk)+2(ˆj+ˆk2)=23ˆj23ˆi+43ˆk2ˆj+ˆk=23ˆi+(232)ˆj+(43 +1)ˆk=23ˆi43ˆj+73ˆk

The components are Cx = 23, Cy = 43, and Cz = 73, and substituting into Equation 2.5.6 gives

C=C2x+C2y+C2z=(23)2+(43)2+(73)2=233.

Example 3.7.4: Displacement of a Skier

Starting at a ski lodge, a cross-country skier goes 5.0 km north, then 3.0 km west, and finally 4.0 km southwest before taking a rest. Find his total displacement vector relative to the lodge when he is at the rest point. How far and in what direction must he ski from the rest point to return directly to the lodge?

Strategy

We assume a rectangular coordinate system with the origin at the ski lodge and with the unit vector ˆi pointing east and the unit vector ˆj pointing north. There are three displacements: D1, D2, and D3. We identify their magnitudes as D1 = 5.0 km , D2 = 3.0 km , and D3 = 4.0 km . We identify their directions are the angles θ1 = 90°, θ2 = 180°, and θ3 = 180° + 45° = 225°. We resolve each displacement vector to its scalar components and substitute the components into Equation 2.6.5 to obtain the scalar components of the resultant displacement D from the lodge to the rest point. On the way back from the rest point to the lodge, the displacement is B = − D. Finally, we find the magnitude and direction of B.

Solution

Scalar components of the displacement vectors are

{D1x=D1cosθ1=(5.0km)cos90o=0D1y=D1sinθ1=(5.0km)sin90o=5.0km

{D2x=D2cosθ2=(3.0km)cos180o=3.0kmD2y=D2sinθ2=(3.0km)sin180o=0

{D3x=D3cosθ3=(4.0km)cos225o=2.8kmD3y=D3sinθ3=(4.0km)sin225o=2.8km

Scalar components of the net displacement vector are

{Dx=D1x+D2x+D3x=(03.02.8)km=5.8kmDy=D1y+D2y+D3y=(5.0+02.8)km=+2.2km

Hence, the skier’s net displacement vector is D = Dx ˆi + Dy ˆj = (−5.8 ˆi + 2.2 ˆj)km . On the way back to the lodge, his displacement is B = − D = −(−5.8 ˆi+ 2.2 ˆj)km = (5.8 ˆi − 2.2 ˆj)km. Its magnitude is B = B2x+B2y = (5.8)2+(2.2)2 km = 6.2 km and its direction angle is θ= tan−1(2.25.8) = −20.8°. Therefore, to return to the lodge, he must go 6.2 km in a direction about 21° south of east.

Significance

Notice that no figure is needed to solve this problem by the analytical method. Figures are required when using a graphical method; however, we can check if our solution makes sense by sketching it, which is a useful final step in solving any vector problem.

Example 3.7.5: Displacement of a Jogger

A jogger runs up a flight of 200 identical steps to the top of a hill and then runs along the top of the hill 50.0 m before he stops at a drinking fountain (Figure 3.7.3). His displacement vector from point A at the bottom of the steps to point B at the fountain is DAB = (−90.0 ˆi + 30.0 ˆj)m. What is the height and width of each step in the flight? What is the actual distance the jogger covers? If he makes a loop and returns to point A, what is his net displacement vector?

A coordinate system is shown with positive x to the right and positive y up. A jogger is at point A at the bottom of steps which lead up and to the left. The top of the steps is labeled as point T. At the top of the steps is a flat section extending from point T to the fountain at point B. The distance between T and B is 50 meters.
Figure 3.7.3: A jogger runs up a flight of steps.

Strategy

The displacement vector DAB is the vector sum of the jogger’s displacement vector DAT along the stairs (from point A at the bottom of the stairs to point T at the top of the stairs) and his displacement vector DRB on the top of the hill (from point T at the top of the stairs to the fountain at point B). We must find the horizontal and the vertical components of DTB. If each step has width w and height h, the horizontal component of DTB must have a length of 200w and the vertical component must have a length of 200h. The actual distance the jogger covers is the sum of the distance he runs up the stairs and the distance of 50.0 m that he runs along the top of the hill.

Solution

In the coordinate system indicated in Figure 3.7.3, the jogger’s displacement vector on the top of the hill is DRB = (−50.0 m) ˆi. His net displacement vector is

DAB=DAT+DTB.

Therefore, his displacement vector DTB along the stairs is

DAT=DABDTB=(90.0ˆi+30.0ˆj)m(50.0m)ˆi)=[(90.050.0)hati+30.0ˆj)]m=(40.0ˆi+30.0ˆj)m.

Its scalar components are DATx = −40.0 m and DATy = 30.0 m. Therefore, we must have

200w=|40.0|m and 200h=30.0m.

Hence, the step width is w = 40.0m200 = 0.2 m = 20 cm, and the step height is w = 30.0m200 = 0.15 m = 15 cm. The distance that the jogger covers along the stairs is

DAT=D2ATx+D2ATy=(40.0)2+(30.0)2m=50.0m.

Thus, the actual distance he runs is DAT + DTB = 50.0 m + 50.0 m = 100.0 m. When he makes a loop and comes back from the fountain to his initial position at point A, the total distance he covers is twice this distance,or 200.0 m. However, his net displacement vector is zero, because when his final position is the same as his initial position, the scalar components of his net displacement vector are zero (Equation 2.4.4).

In many physical situations, we often need to know the direction of a vector. For example, we may want to know the direction of a magnetic field vector at some point or the direction of motion of an object. We have already said direction is given by a unit vector, which is a dimensionless entity—that is, it has no physical units associated with it. When the vector in question lies along one of the axes in a Cartesian system of coordinates, the answer is simple, because then its unit vector of direction is either parallel or antiparallel to the direction of the unit vector of an axis. For example, the direction of vector d = -5 m ˆi is unit vector d = -ˆi. The general rule of finding the unit vector V of direction for any vector V is to divide it by its magnitude V:

ˆV=VV

We see from this expression that the unit vector of direction is indeed dimensionless because the numerator and the denominator in Equation ??? have the same physical unit. In this way, Equation ??? allows us to express the unit vector of direction in terms of unit vectors of the axes. The following example illustrates this principle.

Example 3.7.6: The Unit Vector of Direction

If the velocity vector of the military convoy in Example 2.6.1 is v = (4.000 ˆi + 3.000 ˆj + 0.100 ˆk)km/h, what is the unit vector of its direction of motion.

Strategy

The unit vector of the convoy’s direction of motion is the unit vector ˆv that is parallel to the velocity vector. The unit vector is obtained by dividing a vector by its magnitude, in accordance with Equation ???.

Solution

The magnitude of the vector v is

v=v2x+v2y+v2z=4.0002+3.0002+0.1002km/h=5.001km/h.

To obtain the unit vector ˆv, divide v by its magnitude:

ˆv=vv=(4.000ˆi+3.00ˆj+0.100ˆk)km/h5.001km/h=(4.000ˆi+3.000ˆj+0.1100ˆk)5.001=4.0005.001ˆi+3.0005.001ˆj+0.1005.001ˆk=(79.98ˆi+59.99ˆj+2.00ˆk)×102.

Significance

Note that when using the analytical method with a calculator, it is advisable to carry out your calculations to at least three decimal places and then round off the final answer to the required number of significant figures, which is the way we performed calculations in this example. If you round off your partial answer too early, you risk your final answer having a huge numerical error, and it may be far off from the exact answer or from a value measured in an experiment.

Exercise 2.10

Verify that vector ˆv obtained in Example 3.7.3 is indeed a unit vector by computing its magnitude. If the convoy in Example 2.6.1 was moving across a desert flatland—that is, if the third component of its velocity was zero—what is the unit vector of its direction of motion? Which geographic direction does it represent?


This page titled 3.7: Algebra of Vectors Examples is shared under a CC BY license and was authored, remixed, and/or curated by OpenStax.

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