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Then A = 10.3 size 12{A} {} blocks and θ = 29.1º size 12{"29.1º"} , so that

A x = A cos θ = ( 10.3 blocks ) ( cos 29.1º ) = 9.0 blocks size 12{}
A y = A sin θ = ( 10.3 blocks ) ( sin 29.1º ) = 5.0 blocks . size 12{""}

Calculating a resultant vector

If the perpendicular components A x size 12{A rSub { size 8{x} } } {} and A y size 12{A rSub { size 8{y} } } {} of a vector A size 12{A} {} are known, then A size 12{A} {} can also be found analytically. To find the magnitude A size 12{A} {} and direction θ size 12{θ} {} of a vector from its perpendicular components A x size 12{A rSub { size 8{x} } } {} and A y size 12{A rSub { size 8{y} } } {} , we use the following relationships:

A = A x 2 + A y 2 size 12{A= sqrt {A rSub { size 8{x} rSup { size 8{2} } } +A rSub { size 8{y} rSup { size 8{2} } } } } {}
θ = tan 1 ( A y / A x ) . size 12{θ="tan" rSup { size 8{ - 1} } \( A rSub { size 8{y} } /A rSub { size 8{x} } \) } {}
Vector A is shown with its horizontal and vertical components A sub x and A sub y respectively. The magnitude of vector A is equal to the square root of A sub x squared plus A sub y squared. The angle theta of the vector A with the x axis is equal to inverse tangent of A sub y over A sub x
The magnitude and direction of the resultant vector can be determined once the horizontal and vertical components A x size 12{A rSub { size 8{x} } } {} and A y size 12{A rSub { size 8{y} } } {} have been determined.

Note that the equation A = A x 2 + A y 2 size 12{A= sqrt {A rSub { size 8{x} rSup { size 8{2} } } +A rSub { size 8{y} rSup { size 8{2} } } } } {} is just the Pythagorean theorem relating the legs of a right triangle to the length of the hypotenuse. For example, if A x size 12{A rSub { size 8{x} } } {} and A y size 12{A rSub { size 8{y} } } {} are 9 and 5 blocks, respectively, then A = 9 2 +5 2 =10 . 3 size 12{A= sqrt {9 rSup { size 8{2} } "+5" rSup { size 8{2} } } "=10" "." 3} {} blocks, again consistent with the example of the person walking in a city. Finally, the direction is θ = tan –1 ( 5/9 ) =29.1º size 12{θ="tan" rSup { size 8{–1} } \( "5/9" \) "=29" "." 1 rSup { size 8{o} } } {} , as before.

Determining vectors and vector components with analytical methods

Equations A x = A cos θ size 12{A rSub { size 8{x} } =A"cos"θ} {} and A y = A sin θ size 12{A rSub { size 8{y} } =A"sin"θ} {} are used to find the perpendicular components of a vector—that is, to go from A size 12{A} {} and θ size 12{θ} {} to A x size 12{A rSub { size 8{x} } } {} and A y size 12{A rSub { size 8{y} } } {} . Equations A = A x 2 + A y 2 size 12{A= sqrt {A rSub { size 8{x} rSup { size 8{2} } } +A rSub { size 8{y} rSup { size 8{2} } } } } {} and θ = tan –1 ( A y / A x ) are used to find a vector from its perpendicular components—that is, to go from A x and A y to A and θ . Both processes are crucial to analytical methods of vector addition and subtraction.

Adding vectors using analytical methods

To see how to add vectors using perpendicular components, consider [link] , in which the vectors A size 12{A} {} and B size 12{B} {} are added to produce the resultant R size 12{R} {} .

Two vectors A and B are shown. The tail of vector B is at the head of vector A and the tail of the vector A is at origin. Both the vectors are in the first quadrant. The resultant R of these two vectors extending from the tail of vector A to the head of vector B is also shown.
Vectors A size 12{A} {} and B size 12{B} {} are two legs of a walk, and R size 12{R} {} is the resultant or total displacement. You can use analytical methods to determine the magnitude and direction of R size 12{R} {} .

If A and B represent two legs of a walk (two displacements), then R is the total displacement. The person taking the walk ends up at the tip of R . There are many ways to arrive at the same point. In particular, the person could have walked first in the x -direction and then in the y -direction. Those paths are the x - and y -components of the resultant, R x and R y size 12{R rSub { size 8{y} } } {} . If we know R x and R y size 12{R rSub { size 8{y} } } {} , we can find R and θ using the equations A = A x 2 + A y 2 and θ = tan –1 ( A y / A x ) size 12{θ="tan" rSup { size 8{–1} } \( A rSub { size 8{y} } /A rSub { size 8{x} } \) } {} . When you use the analytical method of vector addition, you can determine the components or the magnitude and direction of a vector.

Step 1. Identify the x- and y-axes that will be used in the problem. Then, find the components of each vector to be added along the chosen perpendicular axes . Use the equations A x = A cos θ size 12{A rSub { size 8{x} } =A"cos"θ} {} and A y = A sin θ size 12{A rSub { size 8{y} } =A"sin"θ} {} to find the components. In [link] , these components are A x size 12{A rSub { size 8{x} } } {} , A y size 12{A rSub { size 8{y} } } {} , B x size 12{B rSub { size 8{x} } } {} , and B y size 12{B rSub { size 8{y} } } {} . The angles that vectors A size 12{A} {} and B size 12{B} {} make with the x -axis are θ A size 12{θ rSub { size 8{A} } } {} and θ B size 12{θ rSub { size 8{B} } } {} , respectively.

Two vectors A and B are shown. The tail of the vector B is at the head of vector A and the tail of the vector A is at origin. Both the vectors are in the first quadrant. The resultant R of these two vectors extending from the tail of vector A to the head of vector B is also shown. The horizontal and vertical components of the vectors A and B are shown with the help of dotted lines. The vectors labeled as A sub x and A sub y are the components of vector A, and B sub x and B sub y as the components of vector B..
To add vectors A size 12{A} {} and B size 12{B} {} , first determine the horizontal and vertical components of each vector. These are the dotted vectors A x size 12{A rSub { size 8{x} } } {} , A y size 12{A rSub { size 8{y} } } {} , B x size 12{B rSub { size 8{x} } } {} and B y size 12{B rSub { size 8{y} } } {} shown in the image.

Step 2. Find the components of the resultant along each axis by adding the components of the individual vectors along that axis . That is, as shown in [link] ,

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} } } {}
Two vectors A and B are shown. The tail of vector B is at the head of vector A and the tail of the vector A is at origin. Both the vectors are in the first quadrant. The resultant R of these two vectors extending from the tail of vector A to the head of vector B is also shown. The vectors A and B are resolved into the horizontal and vertical components shown as dotted lines parallel to x axis and y axis respectively. The horizontal components of vector A and vector B are labeled as A sub x and B sub x and the horizontal component of the resultant R is labeled at R sub x and is equal to A sub x plus B sub x. The vertical components of vector A and vector B are labeled as A sub y and B sub y and the vertical components of the resultant R is labeled as R sub y is equal to A sub y plus B sub y.
The magnitude of the vectors A x size 12{A rSub { size 8{x} } } {} and B x size 12{B rSub { size 8{x} } } {} add to give the magnitude R x size 12{R rSub { size 8{x} } } {} of the resultant vector in the horizontal direction. Similarly, the magnitudes of the vectors A y size 12{A rSub { size 8{y} } } {} and B y size 12{B rSub { size 8{y} } } {} add to give the magnitude R y size 12{R rSub { size 8{y} } } {} of the resultant vector in the vertical direction.

Questions & Answers

If a prism is fully imersed in water then the ray of light will normally dispersed or their is any difference?
Anurag Reply
the same behavior thru the prism out or in water bud abbot
Ju
If this will experimented with a hollow(vaccum) prism in water then what will be result ?
Anurag
What was the previous far point of a patient who had laser correction that reduced the power of her eye by 7.00 D, producing a normal distant vision power of 50.0 D for her?
Jaydie Reply
What is the far point of a person whose eyes have a relaxed power of 50.5 D?
Jaydie
What is the far point of a person whose eyes have a relaxed power of 50.5 D?
Jaydie
A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?
Jaydie
29/20 ? maybes
Ju
In what ways does physics affect the society both positively or negatively
Princewill Reply
how can I read physics...am finding it difficult to understand...pls help
rerry Reply
try to read several books on phy don't just rely one. some authors explain better than other.
Ju
And don't forget to check out YouTube videos on the subject. Videos offer a different visual way to learn easier.
Ju
hope that helps
Ju
I have a exam on 12 february
David Reply
what is velocity
Jiti
the speed of something in a given direction.
Ju
what is a magnitude in physics
Jiti Reply
Propose a force standard different from the example of a stretched spring discussed in the text. Your standard must be capable of producing the same force repeatedly.
Giovani Reply
What is meant by dielectric charge?
It's Reply
what happens to the size of charge if the dielectric is changed?
Brhanu Reply
omega= omega not +alpha t derivation
Provakar Reply
u have to derivate it respected to time ...and as w is the angular velocity uu will relace it with "thita × time""
Abrar
do to be peaceful with any body
Brhanu Reply
the angle subtended at the center of sphere of radius r in steradian is equal to 4 pi how?
Saeed Reply
if for diatonic gas Cv =5R/2 then gamma is equal to 7/5 how?
Saeed
define variable velocity
Ali Reply
displacement in easy way.
Mubashir Reply
binding energy per nucleon
Poonam Reply
Practice Key Terms 1

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