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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.

Components along the same axis, say the x -axis, are vectors along the same line and, thus, can be added to one another like ordinary numbers. The same is true for components along the y -axis. (For example, a 9-block eastward walk could be taken in two legs, the first 3 blocks east and the second 6 blocks east, for a total of 9, because they are along the same direction.) So resolving vectors into components along common axes makes it easier to add them. Now that the components of R size 12{R} {} are known, its magnitude and direction can be found.

Questions & Answers

Is the force attractive or repulsive between the hot and neutral lines hung from power poles? Why?
Jack Reply
what's electromagnetic induction
Chinaza Reply
electromagnetic induction is a process in which conductor is put in a particular position and magnetic field keeps varying.
Lukman
wow great
Salaudeen
what is mutual induction?
je
mutual induction can be define as the current flowing in one coil that induces a voltage in an adjacent coil.
Johnson
how to undergo polarization
Ajayi Reply
show that a particle moving under the influence of an attractive force mu/y³ towards the axis x. show that if it be projected from the point (0,k) with the component velocities U and V parallel to the axis of x and y, it will not strike the axis of x unless u>v²k² and distance uk²/√u-vk as origin
Gabriel Reply
show that a particle moving under the influence of an attractive force mu/y^3 towards the axis x. show that if it be projected from the point (0,k) with the component velocities U and V parallel to the axis of x and y, it will not strike the axis of x unless u>v^2k^2 and distance uk^2/√u-k as origin
Gabriel Reply
No idea.... Are you even sure this question exist?
Mavis
I can't even understand the question
Ademiye
yes it was an assignment question "^"represent raise to power pls
Gabriel
mu/y³ u>v²k² uk²/√u-vk please help me out
Gabriel
An engineer builds two simple pendula. Both are suspended from small wires secured to the ceiling of a room. Each pendulum hovers 2 cm above the floor. Pendulum 1 has a bob with a mass of 10kg . Pendulum 2 has a bob with a mass of 100 kg . Describe how the motion of the pendula will differ if the bobs are both displaced by 12º .
Imtiaz Reply
no ideas
Augstine
if u at an angle of 12 degrees their period will be same so as their velocity, that means they both move simultaneously since both both hovers at same length meaning they have the same length
Ademiye
Modern cars are made of materials that make them collapsible upon collision. Explain using physics concept (Force and impulse), how these car designs help with the safety of passengers.
Isaac Reply
calculate the force due to surface tension required to support a column liquid in a capillary tube 5mm. If the capillary tube is dipped into a beaker of water
Mildred Reply
find the time required for a train Half a Kilometre long to cross a bridge almost kilometre long racing at 100km/h
Ademiye
method of polarization
Ajayi
What is atomic number?
Makperr Reply
The number of protons in the nucleus of an atom
Deborah
type of thermodynamics
Yinka Reply
oxygen gas contained in a ccylinder of volume has a temp of 300k and pressure 2.5×10Nm
Taheer Reply
why the satellite does not drop to the earth explain
Emmanuel Reply
what is a matter
Yinka
what is matter
Yinka
what is matter
Yinka
what is a matter
Yinka
I want the nuclear physics conversation
Mohamed
because space is a vacuum and anything outside the earth 🌎 can not come back without an act of force applied to it to leave the vacuum and fall down to the earth with a maximum force length of 30kcm per second
Clara
at t=0second,aparticles moving in x-y plain with aconstant acceleration has avelocity of initial velocity =(3i-2j)m/s and is at the origion.at t=3second the particle's velocity is final velocity=(9i+7j)then how to find the acceleration?
Yoni Reply
how about the formula like v^2=u^2+2as
Bayuo
a=v-u/t
Doreen
what is physics
Yinka
why is there a maximum distance at which the image can exist behind a convex mirror
Alfred Reply
The ball of a simple pendulum take 0.255 to swing from its equilibrium position to one extreme. Calculate it period.
Abubakr Reply
The Ball of a simple pendulum take 0.255 to swing from its equilibrium position to one extreme. calculate its period
Abubakr
why is there a maximum distance at which the image can exist behind a convex mirror
Alfred
amplitude=0 .255s period=4×.255=1.02 sec period is one complete cycle
MUKHTAR
Practice Key Terms 1

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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