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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.
Practice Key Terms 1

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Source:  OpenStax, Physics 110 at une. OpenStax CNX. Aug 29, 2013 Download for free at http://legacy.cnx.org/content/col11566/1.1
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