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We note that the heading of the plane has been redirected in light of the cross wind. Let us find the new heading of the plane. We use the tangent function as follows

tan ( θ ) = 30 880 size 12{"tan" \( θ \) = { {"30"} over {"880"} } } {}

Applying the inverse tangent leaves us with

θ = tan 1 ( 0 . 341 ) = 1 . 953 0 size 12{θ="tan" rSup { size 8{ - 1} } \( 0 "." "341" \) =1 "." "953" rSup { size 8{0} } } {}

So we conclude the heading of the airplane will be directed 1.953 0 from its initial heading.

Boat and trailer on an inclined ramp

Let us now consider another example that will allow us to apply 2-D vectors. Suppose that a force of 750 pounds is required to pull a boat and trailer up a ramp that is inclined at an angle of 20 0 from the horizon. This situation is depicted in Figure 3 (a). Under the assumption of no friction, what is the combined weight of the boat and trailer?

Depiction of a boat on a landing ramp along with a trigonometric description.

Let us now use the figure to interpret the vectors shown in Figure 3. The vector BA size 12{ widevec { ital "BA"} } {} represents the combined weight of the boat and the trailer. This is the quantity that we need to find. The vector BC size 12{ widevec { ital "BC"} } {} represents the force against the ramp. The vector AC size 12{ widevec { ital "AC"} } {} which is parallel to the ramp represents the force applied to the boat and trailer. From the problem description we know that it has a magnitude of 750 pounds. We use the fact that this is a right triangle to simplify our efforts.

We can apply the definition of the sine function to obtain

sin ( 20 0 ) = AC BA size 12{"sin" \( "20" rSup { size 8{0} } \) = { { lline widevec { ital "AC"} rline } over { lline widevec { ital "BA"} rline } } } {}

Substitution leads to the equation

sin ( 20 0 ) = 750 BA size 12{"sin" \( "20" rSup { size 8{0} } \) = { {"750"} over { lline widevec { ital "BA"} rline } } } {}

We can now solve for the weight of the boat and trailer

BA = 750 / 0 . 342 = 2, 192 lbs size 12{ lline widevec { ital "BA"} rline ="750"/0 "." "342"=2,"192"` ital "lbs"} {}

Cable tension for a hanging weight

A 200-pound weight is suspended from a ceiling. The weight is supported by two cables. One cable makes a 20 0 angle away from the vertical and the other a 30 0 angle as shown in Figure 4. Find the tension in each of the support cables.

Mass suspended from a beam with two cables.

Statics is the field of engineering that is used to solve problems of this sort. Because the object does not move, it is said to be static. Another way to look at this, is that the object is at equilibrium. At equilibrium, the forces acting on an object must balance. Otherwise, the object would indeed move. To better analyze the situation, engineers often make use of what is known as a free body diagram. Figure 5 shows the free body diagram related to our problem.

Free body diagram of mass suspended from a beam with two cables.

The free body diagram shows the three forces that act on the object. The Figure also shows the x-y coordinate system employed in this solution. The tension in cable 1 can be resolved into its x and y components and written in vector notation as

T 1 = T 1 sin ( 20 0 ) x ˆ + T 1 cos ( 20 0 ) y ˆ size 12{T rSub { size 8{1} } = - lline T rSub { size 8{1} } rline "sin" \( "20" rSup { size 8{0} } \) ` { hat {x}}+ lline T rSub { size 8{1} } rline "cos" \( "20" rSup { size 8{0} } \) ` { hat {y}}} {}

Similarly, T 2 can be written as

T 2 = T 2 sin ( 30 0 ) x ˆ + T 2 cos ( 30 0 ) y ˆ size 12{T rSub { size 8{2} } = lline T rSub { size 8{2} } rline "sin" \( "30" rSup { size 8{0} } \) ` { hat {x}}+ lline T rSub { size 8{2} } rline "cos" \( "30" rSup { size 8{0} } \) ` { hat {y}}} {}

The weight ( W ) is expressed as

W = 200 y ˆ size 12{W= - "200"` { hat {y}}} {}

The sum of the forces acting on the object must be 0. Thus

T 1 + T 2 = W size 12{T rSub { size 8{1} } +T rSub { size 8{2} } =W} {}

or

T 1 sin ( 20 0 ) x ˆ + T 1 cos ( 20 0 ) y ˆ + T 2 sin ( 30 0 ) x ˆ + T 2 cos ( 30 0 ) y ˆ 200 y ˆ = 0 size 12{ - lline T rSub { size 8{1} } rline "sin" \( "20" rSup { size 8{0} } \) ` { hat {x}}+ lline T rSub { size 8{1} } rline "cos" \( "20" rSup { size 8{0} } \) ` { hat {y}}+ lline T rSub { size 8{2} } rline "sin" \( "30" rSup { size 8{0} } \) ` { hat {x}}+ lline T rSub { size 8{2} } rline "cos" \( "30" rSup { size 8{0} } \) ` { hat {y}} - "200"` { hat {y}}=0} {}

This problem can be simplified by writing an equation for solely the x -component

T 1 sin ( 20 0 ) + T 2 sin ( 30 0 ) = 0 size 12{ - lline T rSub { size 8{1} } rline "sin" \( "20" rSup { size 8{0} } \) `+ lline T rSub { size 8{2} } rline "sin" \( "30" rSup { size 8{0} } \) =0} {}

We can do the same for the y-component

T 1 cos ( 20 0 ) + T 2 cos ( 30 0 ) 200 = 0 size 12{ lline T rSub { size 8{1} } rline "cos" \( "20" rSup { size 8{0} } \) + lline T rSub { size 8{2} } rline "cos" \( "30" rSup { size 8{0} } \) - "200"`=0} {}

We begin by substituting in for the trigonometric function values to yield the following set of equations

0 . 342 T 1 + 0 . 5 T 2 = 0 size 12{ - 0 "." "342"` lline T rSub { size 8{1} } rline +0 "." 5` lline T rSub { size 8{2} } rline =0} {}

and

0 . 940 T 1 + 0 . 866 T 2 = 200 size 12{0 "." "940"` lline T rSub { size 8{1} } rline +0 "." "866"` lline T rSub { size 8{2} } rline ="200"} {}

First, we find

T 2 = 0 . 684 T 1 size 12{ lline T rSub { size 8{2} } rline =0 "." "684"` lline T rSub { size 8{1} } rline } {}

This expression can then be substituted into the other equation

0 . 940 T 1 + 0 . 866 ( 0 . 684 ) T 1 = 200 size 12{0 "." "940"` lline T rSub { size 8{1} } rline +0 "." "866"` \( 0 "." "684" \) ` lline T rSub { size 8{1} } rline ="200"} {}

This leads to the solution

T 1 = 130 . 5 lbs size 12{ lline T rSub { size 8{1} } rline ="130" "." 5` ital "lbs"} {}

Next, we solve for the other variable

T 2 = 0 . 684 T 1 = 0 . 684 ( 130 . 5 ) = 89 . 3 lbs size 12{ lline T rSub { size 8{2} } rline =0 "." "684"` lline T rSub { size 8{1} } rline =0 "." "684"` \( "130" "." 5 \) ="89" "." 3` ital "lbs"} {}

So we conclude that the tension in the cables are 130.5 lbs and 89.3 lbs respectively.

Exercises

  1. A load is of mass 150 kg is situated atop a moving dolly. A force with a magnitude of 15 Newtons is applied at an angle of 30 0 with respect to the horizontal. Resolve the force into its x- and y- components.
  2. A ship travels 300 miles due East, then 700 miles North of due East. Sketch the geometry of the situation. Be sure to include all angles in your sketch. In all, how far does the ship travel on its voyage?
  3. A small airplane travels at a velocity of 320 km/hr at an angle that is 40 0 South of East. The airplane ecounters a wind whose speed is 25 km/hr. (a) If the wind travels in a direction from West to East, what is the resulting speed and direction of the airplane? (b) Repeat for a 25 km/hr wind directed from East to West. (c) Repeat for a 25 km/hr wind directed from North to South. (d) Repeat for a 25 km/hr wind directed from South to North.
  4. A 50 kg mass is suspended by two cables of equal length from a beam. Each cable makes a 450 angle with the horizontal beam. Sketch a free body diagram that represents the situation. Determine the tension in each cable.
  5. A complex number has two components. One component is real while the other is imaginary. A complex number can be represented as a two-dimensional vector using what is known as the complex plane. The complex plane is a special plane whose abscissa is use to specify the real part of the complex number and whose ordinate is used to specify the imaginary part of the complex number. Consider the complex number z = 4 + i 6 size 12{z=4+i`6} {} . This complex number is shown as a vector in the complex plane in the figure below. The symbol M represents the magnitude of the complex number and θ represents the angle or argument of the complex number. (a) Find the magnitude and angle of the complex number z = 4 + i 6 size 12{z=4+i`6} {} . (b) Repeat for the complex number z = 4 i 6 size 12{z=4 - i`6} {} . (c) Repeat for the complex number z = 4 + i 6 size 12{z= - 4+i`6} {} . (d) Repeat for the complex number z = 4 i 6 size 12{z= - 4 - i`6} {} .
Complex number representation in the complex plane.

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
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Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
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what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Math 1508 (laboratory) engineering applications of precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11337/1.3
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