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Step 3. To get the magnitude R size 12{R } {} of the resultant, use the Pythagorean theorem:

R = R x 2 + R y 2 . size 12{R= sqrt {R rSub { size 8{x} } rSup { size 8{2} } +R rSub { size 8{y} } rSup { size 8{2} } } "."} {}

Step 4. To get the direction of the resultant:

θ = tan 1 ( R y / R x ) . size 12{θ="tan" rSup { size 8{ - 1} } \( R rSub { size 8{y} } /R rSub { size 8{x} } \) "."} {}

The following example illustrates this technique for adding vectors using perpendicular components.

Adding vectors using analytical methods

Add the vector A size 12{A} {} to the vector B size 12{B} {} shown in [link] , using perpendicular components along the x - and y -axes. The x - and y -axes are along the east–west and north–south directions, respectively. Vector A size 12{A} {} represents the first leg of a walk in which a person walks 53 . 0 m size 12{"53" "." "0 m"} {} in a direction 20 . 0 º size 12{"20" "." 0º } {} north of east. Vector B size 12{B} {} represents the second leg, a displacement of 34 . 0 m size 12{"34" "." "0 m"} {} in a direction 63 . 0 º size 12{"63" "." 0º } {} north of east.

Two vectors A and B are shown. The tail of the vector A is at origin. Both the vectors are in the first quadrant. Vector A is of magnitude fifty three units and is inclined at an angle of twenty degrees to the horizontal. From the head of the vector A another vector B of magnitude 34 units is drawn and is inclined at angle sixty three degrees with the horizontal. The resultant of two vectors is drawn from the tail of the vector A to the head of the vector B.
Vector A size 12{A} {} has magnitude 53 . 0 m size 12{"53" "." "0 m"} {} and direction 20 . 0 º size 12{"20" "." 0 { size 12{ circ } } } {} north of the x -axis. Vector B size 12{B} {} has magnitude 34 . 0 m size 12{"34" "." "0 m"} {} and direction 63 . 0 º size 12{"63" "." 0° } {} north of the x -axis. You can use analytical methods to determine the magnitude and direction of R size 12{R} {} .

Strategy

The components of A size 12{A} {} and B size 12{B} {} along the x - and y -axes represent walking due east and due north to get to the same ending point. Once found, they are combined to produce the resultant.

Solution

Following the method outlined above, we first find the components of A size 12{A} {} and B size 12{B} {} along the x - and y -axes. Note that A = 53.0 m size 12{"A" "=" "53.0 m"} {} , θ A = 20.0º size 12{"θ" "subA" "=" "20.0°" } {} , B = 34.0 m size 12{"B" "=" "34.0" "m"} {} , and θ B = 63.0º size 12{θ rSub { size 8{B} } } {} . We find the x -components by using A x = A cos θ size 12{A rSub { size 8{x} } =A"cos"θ} {} , which gives

A x = A cos θ A = ( 53. 0 m ) ( cos 20.0º ) = ( 53. 0 m ) ( 0 .940 ) = 49. 8 m alignl { stack { size 12{A rSub { size 8{x} } =A"cos"θ rSub { size 8{A} } = \( "53" "." 0" m" \) \( "cos""20" "." 0 { size 12{ circ } } \) } {} #" "= \( "53" "." 0" m" \) \( 0 "." "940" \) ="49" "." 8" m" {} } } {}

and

B x = B cos θ B = ( 34 . 0 m ) ( cos 63.0º ) = ( 34 . 0 m ) ( 0 . 454 ) = 15 . 4 m . alignl { stack { size 12{B rSub { size 8{x} } =B"cos"θ rSub { size 8{B} } = \( "34" "." 0" m" \) \( "cos""63" "." 0 { size 12{ circ } } \) } {} #" "= \( "34" "." 0" m" \) \( 0 "." "454" \) ="15" "." 4" m" {} } } {}

Similarly, the y -components are found using A y = A sin θ A size 12{A rSub { size 8{y} } =A"sin"θ rSub { size 8{A} } } {} :

A y = A sin θ A = ( 53 . 0 m ) ( sin 20.0º ) = ( 53 . 0 m ) ( 0 . 342 ) = 18 . 1 m alignl { stack { size 12{A rSub { size 8{y} } =A"sin"θ rSub { size 8{A} } = \( "53" "." 0" m" \) \( "sin""20" "." 0 { size 12{ circ } } \) } {} #" "= \( "53" "." 0" m" \) \( 0 "." "342" \) ="18" "." 1" m" {} } } {}

and

B y = B sin θ B = ( 34 . 0 m ) ( sin 63 . 0 º ) = ( 34 . 0 m ) ( 0 . 891 ) = 30 . 3 m . alignl { stack { size 12{B rSub { size 8{y} } =B"sin"θ rSub { size 8{B} } = \( "34" "." 0" m" \) \( "sin""63" "." 0 { size 12{ circ } } \) } {} #" "= \( "34" "." 0" m" \) \( 0 "." "891" \) ="30" "." 3" m" "." {} } } {}

The x - and y -components of the resultant are thus

R x = A x + B x = 49 . 8 m + 15 . 4 m = 65 . 2 m size 12{R rSub { size 8{x} } =A rSub { size 8{x} } +B rSub { size 8{x} } ="49" "." 8" m"+"15" "." 4" m"="65" "." 2" m"} {}

and

R y = A y + B y = 18 . 1 m + 30 . 3 m = 48 . 4 m . size 12{R rSub { size 8{y} } =A rSub { size 8{y} } +B rSub { size 8{y} } ="18" "." 1" m"+"30" "." 3" m"="48" "." 4" m."} {}

Now we can find the magnitude of the resultant by using the Pythagorean theorem:

R = R x 2 + R y 2 = ( 65 . 2 ) 2 + ( 48 . 4 ) 2 m size 12{R= sqrt {R rSub { size 8{x} } rSup { size 8{2} } +R rSub { size 8{y} } rSup { size 8{2} } } = sqrt { \( "65" "." 2 \) rSup { size 8{2} } + \( "48" "." 4 \) rSup { size 8{2} } } " m"} {}

so that

R = 81.2 m. size 12{R ="81.2" "m."} {}

Finally, we find the direction of the resultant:

θ = tan 1 ( R y / R x ) =+ tan 1 ( 48 . 4 / 65 . 2 ) . size 12{θ="tan" rSup { size 8{ - 1} } \( R rSub { size 8{y} } /R rSub { size 8{x} } \) "=+""tan" rSup { size 8{ - 1} } \( "48" "." 4/"65" "." 2 \) "."} {}

Thus,

θ = tan 1 ( 0 . 742 ) = 36 . 6 º . size 12{θ="tan" rSup { size 8{ - 1} } \( 0 "." "742" \) ="36" "." 6 { size 12{ circ } } "."} {}
The addition of two vectors A and B is shown. Vector A is of magnitude fifty three units and is inclined at an angle of twenty degrees to the horizontal. Vector B is of magnitude thirty four units and is inclined at angle sixty three degrees to the horizontal. The components of vector A are shown as dotted vectors A X is equal to forty nine point eight meter along x axis and A Y is equal to eighteen point one meter along Y axis. The components of vector B are also shown as dotted vectors B X is equal to fifteen point four meter and B Y is equal to thirty point three meter. The horizontal component of the resultant R X is equal to A X plus B X is equal to sixty five point two meter. The vertical component of the resultant R Y is equal to A Y plus B Y is equal to forty eight point four meter. The magnitude of the resultant of two vectors is eighty one point two meters. The direction of the resultant R is in thirty six point six degree from the vector A in anticlockwise direction.
Using analytical methods, we see that the magnitude of R size 12{R} {} is 81 . 2 m size 12{"81" "." "2 m"} {} and its direction is 36 . size 12{"36" "." 6°} {} north of east.

Discussion

This example illustrates the addition of vectors using perpendicular components. Vector subtraction using perpendicular components is very similar—it is just the addition of a negative vector.

Subtraction of vectors is accomplished by the addition of a negative vector. That is, A B A + ( –B ) size 12{A – B equiv A+ \( - B \) } {} . Thus, the method for the subtraction of vectors using perpendicular components is identical to that for addition . The components of –B are the negatives of the components of B size 12{B} {} . The x - and y -components of the resultant A B = R size 12{A- bold "B = R"} {} are thus

R x = A x + ( B x ) size 12{R rSub { size 8{x} } =A rSub { size 8{x} } +-B rSub { size 8{x} } } {}

and

R y = A y + ( B y ) size 12{R rSub { size 8{y} } =A rSub { size 8{y} } +-B rSub { size 8{y} } } {}

and the rest of the method outlined above is identical to that for addition. (See [link] .)

Analyzing vectors using perpendicular components is very useful in many areas of physics, because perpendicular quantities are often independent of one another. The next module, Projectile Motion , is one of many in which using perpendicular components helps make the picture clear and simplifies the physics.

In this figure, the subtraction of two vectors A and B is shown. A red colored vector A is inclined at an angle theta A to the positive of x axis. From the head of vector A a blue vector negative B is drawn. Vector B is in west of south direction. The resultant of the vector A and vector negative B is shown as a black vector R from the tail of vector A to the head of vector negative B. The resultant R is inclined to x axis at an angle theta below the x axis. The components of the vectors are also shown along the coordinate axes as dotted lines of their respective colors.
The subtraction of the two vectors shown in [link] . The components of –B size 12{B} {} are the negatives of the components of B size 12{B} {} . The method of subtraction is the same as that for addition.

Summary

  • The analytical method of vector addition and subtraction involves using the Pythagorean theorem and trigonometric identities to determine the magnitude and direction of a resultant vector.
  • The steps to add vectors A size 12{A} {} and B size 12{B} {} using the analytical method are as follows:

    Step 1: Determine the coordinate system for the vectors. Then, determine the horizontal and vertical components of each vector using the equations

    A x = A cos θ B x = B cos θ alignl { stack { size 12{A rSub { size 8{x} } =A"cos"θ} {} #B rSub { size 8{x} } =B"cos"θ {} } } {}

    and

    A y = A sin θ B y = B sin θ . alignl { stack { size 12{A rSub { size 8{y} } =A" sin"θ} {} #B=B suby " sin "θ {} } } {}

    Step 2: Add the horizontal and vertical components of each vector to determine the components R x size 12{R rSub { size 8{x} } } {} and R y size 12{R rSub { size 8{y} } } {} of the resultant vector, R size 12{R} {} :

    R x = A x + B x size 12{R rSub { size 8{x} } =A rSub { size 8{x} } +B rSub { size 8{x} } } {}

    and

    R y = A y + B y . size 12{R rSub { size 8{y} } =A rSub { size 8{y} } +B rSub { size 8{y} } } {}

    Step 3: Use the Pythagorean theorem to determine the magnitude, R size 12{R} {} , of the resultant vector R size 12{R} {} :

    R = R x 2 + R y 2 . size 12{R= sqrt {R rSub { size 8{x} } rSup { size 8{2} } +R rSub { size 8{y} } rSup { size 8{2} } } } {}

    Step 4: Use a trigonometric identity to determine the direction, θ size 12{θ} {} , of R size 12{R} {} :

    θ = tan 1 ( R y / R x ) . size 12{θ="tan" rSup { size 8{ - 1} } \( R rSub { size 8{y} } /R rSub { size 8{x} } \) } {}

Conceptual questions

Suppose you add two vectors A size 12{A} {} and B size 12{B} {} . What relative direction between them produces the resultant with the greatest magnitude? What is the maximum magnitude? What relative direction between them produces the resultant with the smallest magnitude? What is the minimum magnitude?

Problems&Exercises

Find the following for path C in [link] : (a) the total distance traveled and (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.

A map of city is shown. The houses are in form of square blocks of side one hundred and twenty meter each. Four paths A B C and D are shown in different colors. The path c shown as blue extends to one block towards north, then five blocks towards east and then two blocks towards south then one block towards west and one block towards north and finally three blocks towards west. It is asked to find out the total distance traveled the magnitude and the direction of the displacement from start to finish for path C.
The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.

(a) 1.56 km

(b) 120 m east

Find the following for path D in [link] : (a) the total distance traveled and (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.

Find the north and east components of the displacement from San Francisco to Sacramento shown in [link] .

A map of northern California with a circle with a radius of one hundred twenty three kilometers centered on San Francisco. Sacramento lies on the circumference of this circle in a direction forty-five degrees north of east from San Francisco.

North-component 87.0 km, east-component 87.0 km

Solve the following problem using analytical techniques: Suppose you walk 18.0 m straight west and then 25.0 m straight north. 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 size 12{A} {} and B size 12{B} {} , as in [link] , then this problem asks you to find their sum R = A + B size 12{R=A+B} {} .)

In the given figure displacement of a person is shown. First movement of the person is shown as vector A from origin along negative x axis. He then turns to his right. His movement is now shown as a vertical vector in north direction. The displacement vector R is also shown. In the question you are asked to find the displacement of the person from the start to finish.
The two displacements A size 12{A} {} and B size 12{B} {} add to give a total displacement R size 12{R} {} having magnitude R size 12{R} {} and direction θ size 12{θ} {} .

Note that you can also solve this graphically. Discuss why the analytical technique for solving this problem is potentially more accurate than the graphical technique.

Repeat [link] using analytical techniques, but reverse the order of the two legs of the walk and show that you get the same final result. (This problem shows that adding them in reverse order gives the same result—that is, B + A = A + B .) Discuss how taking another path to reach the same point might help to overcome an obstacle blocking you other path.

30.8 m, 35.8 west of north

A farmer wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A , size 12{A,} {} B , size 12{B,} {} and C size 12{C} {} in [link] , and then correctly calculates the length and orientation of the fourth side D size 12{D} {} . What is his result?

A quadrilateral with sides A, B, C, and D. A begins at the end of D and is 4 point seven zero kilometers  at an angle of 7 point 5 degrees south of west. B begins at the end of A and is 2 point four eight kilometers in a direction sixteen degrees west of north. C begins at the end of B and is 3 point zero 2 kilometers in a direction nineteen degrees north of west. D begins at the end of C and runs distance and direction that must be calculated

Suppose a pilot flies 40 . 0 km size 12{"40" "." "0 km"} {} in a direction 60º size 12{"60"°} {} north of east and then flies 30 . 0 km size 12{"30" "." "0 km"} {} in a direction 15º size 12{"15"°} {} north of east as shown in [link] . Find her total distance R size 12{R} {} from the starting point and the direction θ size 12{θ} {} of the straight-line path to the final position. Discuss qualitatively how this flight would be altered by a wind from the north and how the effect of the wind would depend on both wind speed and the speed of the plane relative to the air mass.

A triangle  defined by vectors A, B, and R. A begins at the origin and run forty kilometers in a direction sixty degrees north of east. B begins at the end of A and runs thirty kilometers in a direction fifteen degrees north of east. R is the resultant vector and runs from the origin (the beginning of A) to the end of B for a distance and in a direction theta that need to be calculated.
Practice Key Terms 1

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Source:  OpenStax, 2d kinematics. OpenStax CNX. Sep 04, 2015 Download for free at http://legacy.cnx.org/content/col11879/1.3
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