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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

Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
Kate Reply
To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.
Muhammed
What is the direction and net electric force on q_{1}= 5µC located at (0,4)r due to charges q_{2}=7mu located at (0,0)m and q_{3}=3\mu C located at (4,0)m?
Kate Reply
what is the change in momentum of a body?
Eunice Reply
what is a capacitor?
Raymond Reply
Capacitor is a separation of opposite charges using an insulator of very small dimension between them. Capacitor is used for allowing an AC (alternating current) to pass while a DC (direct current) is blocked.
Gautam
A motor travelling at 72km/m on sighting a stop sign applying the breaks such that under constant deaccelerate in the meters of 50 metres what is the magnitude of the accelerate
Maria Reply
please solve
Sharon
8m/s²
Aishat
What is Thermodynamics
Muordit
velocity can be 72 km/h in question. 72 km/h=20 m/s, v^2=2.a.x , 20^2=2.a.50, a=4 m/s^2.
Mehmet
A boat travels due east at a speed of 40meter per seconds across a river flowing due south at 30meter per seconds. what is the resultant speed of the boat
Saheed Reply
50 m/s due south east
Someone
which has a higher temperature, 1cup of boiling water or 1teapot of boiling water which can transfer more heat 1cup of boiling water or 1 teapot of boiling water explain your . answer
Ramon Reply
I believe temperature being an intensive property does not change for any amount of boiling water whereas heat being an extensive property changes with amount/size of the system.
Someone
Scratch that
Someone
temperature for any amount of water to boil at ntp is 100⁰C (it is a state function and and intensive property) and it depends both will give same amount of heat because the surface available for heat transfer is greater in case of the kettle as well as the heat stored in it but if you talk.....
Someone
about the amount of heat stored in the system then in that case since the mass of water in the kettle is greater so more energy is required to raise the temperature b/c more molecules of water are present in the kettle
Someone
definitely of physics
Haryormhidey Reply
how many start and codon
Esrael Reply
what is field
Felix Reply
physics, biology and chemistry this is my Field
ALIYU
field is a region of space under the influence of some physical properties
Collete
what is ogarnic chemistry
WISDOM Reply
determine the slope giving that 3y+ 2x-14=0
WISDOM
Another formula for Acceleration
Belty Reply
a=v/t. a=f/m a
IHUMA
innocent
Adah
pratica A on solution of hydro chloric acid,B is a solution containing 0.5000 mole ofsodium chlorid per dm³,put A in the burret and titrate 20.00 or 25.00cm³ portion of B using melting orange as the indicator. record the deside of your burret tabulate the burret reading and calculate the average volume of acid used?
Nassze Reply
how do lnternal energy measures
Esrael
Two bodies attract each other electrically. Do they both have to be charged? Answer the same question if the bodies repel one another.
JALLAH Reply
No. According to Isac Newtons law. this two bodies maybe you and the wall beside you. Attracting depends on the mass och each body and distance between them.
Dlovan
Are you really asking if two bodies have to be charged to be influenced by Coulombs Law?
Robert
like charges repel while unlike charges atttact
Raymond
What is specific heat capacity
Destiny Reply
Specific heat capacity is a measure of the amount of energy required to raise the temperature of a substance by one degree Celsius (or Kelvin). It is measured in Joules per kilogram per degree Celsius (J/kg°C).
AI-Robot
specific heat capacity is the amount of energy needed to raise the temperature of a substance by one degree Celsius or kelvin
ROKEEB
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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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