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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.

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?
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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
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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
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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
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definitely of physics
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what is field
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physics, biology and chemistry this is my Field
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field is a region of space under the influence of some physical properties
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determine the slope giving that 3y+ 2x-14=0
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Another formula for Acceleration
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a=v/t. a=f/m a
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innocent
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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. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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