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0.14 Two-dimensional vectors  (Page 2/2)

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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

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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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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