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Simultaneous Equations

Module by: Scott Starks. E-mail the author

Summary: This module is part of a collection of modules intended for preengineering students enrolled in MATH 1508 (PreCalculus) at the University of Texas at El Paso.

Simultaneous Equations

Introduction

The applications of simultaneous equations in the modeling, analysis and design of engineering systems are numerous. This module presents several applications drawn from engineering that illustrate uses of simultaneous equations.

Unmanned Air Vehicle (Robotics)

Many applications in engineering involve the quantities speed, distance and time. In some applications, one is presented with two situations. In the two situations, the speed of an object differs. The following is an example involving an Unmanned Air Vehicle (UAV) whose speed varies depending upon whether the UAV travels in the same direction of the wind as opposed to the situation where the UAV travels in the opposite direction in opposition to the wind. Simultaneous equations can be used to solve problems relating to situations such as this.

Example 1: In a closed course surveillance flight, the downwind leg of length18.6 miles is completed in 2.0 hours. The upwind leg which is also 18.6 miles in length is completed in 3.5 hours. Find both the speed of the UAV and the speed of the wind.

To address such a problem it is critical to begin with the definition of variables. We will let V be the airspeed of the UAV measured in miles/hr. The wind speed expressed in miles/hr will be represented by the variable W.

We know that the distance that an object travels is equal the product of its speed and time. When the UAV travels downwind, its speed will be equal to the sum of its airspeed and the windspeed. When the UAV travels upwind, its total speed will equal to the difference of its airspeed minus the windspeed. The distance that the UAV travels is the same (18.6 miles) whether it travels downwind or upwind.

We can use this information to establish two equations

18 . 6 = 2 . 0 × ( V + W ) 18 . 6 = 2 . 0 × ( V + W ) size 12{"18" "." 6=2 "." 0` times \( V+W \) } {}
(1)
18 . 6 = 3 . 5 × ( V W ) 18 . 6 = 3 . 5 × ( V W ) size 12{"18" "." 6=3 "." 5 times \( V - W \) } {}
(2)

From equation (1) we obtain

18 . 6 = 2 . 0 × ( V + W ) 18 . 6 = 2 . 0 × ( V + W ) size 12{"18" "." 6=2 "." 0 times \( V+W \) } {}
(3)

which can can yield an expression for the variable V

V = 9 . 3 W V = 9 . 3 W size 12{V=9 "." 3 - W} {}
(4)

Next, we can substitute this expression for V into equation (2). By doing so, we will be able to solve for W.

18 . 6 = 3 . 5 × ( ( 9 . 3 W ) W ) 18 . 6 = 3 . 5 × ( ( 9 . 3 W ) W ) size 12{"18" "." 6=3 "." 5 times \( \( 9 "." 3 - W \) - W \) } {}
(5)

Dividing each side of the equation by (3.5) yields

5 . 31 = ( 9 . 3 W ) W 5 . 31 = ( 9 . 3 W ) W size 12{5 "." "31"= \( 9 "." 3 - W \) - W} {}
(6)

This equation can be readily solved for the variable W as follows

3 . 99 = 2W W = 1 . 995 miles / hr 3 . 99 = 2W W = 1 . 995 miles / hr alignl { stack { size 12{ - 3 "." "99"= - 2W} {} # size 12{W=1 "." "995"` ital "miles"/ ital "hr"} {} } } {}
(7)

Next, we substitute this value for W into equation (4) to solve for V.

V = 9 . 3 1 . 995 = 7 . 305 miles / hr V = 9 . 3 1 . 995 = 7 . 305 miles / hr size 12{V=9 "." 3 - 1 "." "995"=7 "." "305"` ital "miles"/ ital "hr"} {}
(8)

Thus we conclude that the UAV airspeed is 7.305 miles/hr and the windspeed is 1.995 miles/hr.

Analysis of a Torsion Beam

Torque is a term that is used to describe the tendency of a force to rotate an object about an axis, a fulcrum or a pivot. Just as a force is a push or a pull, a torque can be thought of as a twist. Loosely speaking, torque is a measure of the turning force on an object such as a bolt. For example, pushing or pulling the handle of a wrench connected to a nut or bolt produces a torque (turning force) that loosens or tightens the nut or bolt.

Consider the torsion beam shown in Figure 1.

Figure 1: Diagram of a torsion beam.
Figure 1 (Picture 6.jpg)

Figure 1. Torsion beam.

In this figure, we observe a situation in which forces are applied at each end of the beam. We also note that each force is applied at a right angle to the beam. Under such conditions, torque can be calculated as the product of the force and the length of the beam from its end to its support point atop the pivot point of the beam.

The force (F1) contributes a counter-clockwise torque (TCCW) on the beam. The value of this torque can be calculated

T CCW = F 1 × R 1 T CCW = F 1 × R 1 size 12{T rSub { size 8{ ital "CCW"} } =F rSub { size 8{1} } times R rSub { size 8{1} } } {}
(9)

The force (F2) contributes a clockwise torque (TCW) on the beam. The value of this torque can be calculated as

T CW = F 2 × R 2 T CW = F 2 × R 2 size 12{T rSub { size 8{ ital "CW"} } =F rSub { size 8{2} } times R rSub { size 8{2} } } {}
(10)

Suppose that the beam is not in motion. Engineers would say that the beam is static or that the beam system is in equillibrium. A necessary condition for equillibrium of the system is that the two torques be equal, that is

T CW = T CCW T CW = T CCW size 12{T rSub { size 8{ ital "CW"} } =T rSub { size 8{ ital "CCW"} } } {}
(11)

We now will apply what we know about simultaneous equations to solve a problem involving a torsion beam.

Example 2: Assume that a a beam of length 56 inches is supported by a fulcrum. Force is applied at each end of the beam. Assume that the force (F1) on the left hand side of the beam is 24 lbs while the force on the right hand side of the beam (F2) is 32 lbs. If the beam is static, then what are the values of R1 and R2?

We know from that the clockwise and counter-clockwise torques must balance as a condition for equilibrium. This leads to the first equation that we will use to solve this problem

24 R 1 = 32 R 2 24 R 1 = 32 R 2 size 12{"24"`R rSub { size 8{1} } ="32"`R rSub { size 8{2} } } {}
(12)

which can be written as

24 R 1 32 R 2 = 0 24 R 1 32 R 2 = 0 size 12{"24"`R rSub { size 8{1} } - "32"`R rSub { size 8{2} } =0} {}
(13)

We also know that the overall length of the beam is 56 inches. Therefore,

R 1 + R 2 = 56 R 1 + R 2 = 56 size 12{R rSub { size 8{1} } +R rSub { size 8{2} } ="56"} {}
(14)

which can be solved for R1 as

R 1 = 56 R 2 R 1 = 56 R 2 size 12{R rSub { size 8{1} } ="56" - R rSub { size 8{2} } } {}
(15)

We may substitute the expression for R1 in equation () to obtain

24 ( 56 R 2 ) 32 R 2 = 0 24 ( 56 R 2 ) 32 R 2 = 0 size 12{"24"` \( "56" - R rSub { size 8{2} } \) - "32"`R rSub { size 8{2} } =0} {}
(16)

Multiplication and rearrangement yield the equation

1, 344 = 56 R 2 1, 344 = 56 R 2 size 12{1,"344"="56"`R rSub { size 8{2} } } {}
(17)

which can be solved for R2.

R 2 = 24 in R 2 = 24 in size 12{R rSub { size 8{2} } ="24"` ital "in"} {}
(18)

The value for R1 can be easily found as

R 1 = 56 R 2 = 56 24 = 32 inches R 1 = 56 R 2 = 56 24 = 32 inches size 12{R rSub { size 8{1} } ="56" - R rSub { size 8{2} } ="56" - "24"="32"~ ital "inches"} {}
(19)

Parallel Processing

Parallel processing is a term drawn from Compute Science that involves the simultaneous use of more than one central processing unit (CPU) to execute a program. Ideally, parallel processing makes programs run faster because there are more than one processors in use to execute the program. Though most computers have just one CPU, newer computer architectures that feature several CPU’s are becoming the norm more prevalent.

Example 3: Suppose that we are interested in implementing a computer algorithm on two processors. Let us call these processors A and B. The steps involved in the accomplishment of the algorithm are divided among the two processors. In all there are 17,000,000 computations which will need to be performed on the two processors. Working in parallel, the two processors accomplish the algorithm. In doing so, the processor A is in service for 3 seconds while the processor B is in services for 2 seconds.

Suppose that a second algorithm is to be accomplished by processors A and B working in parallel. In all, the second algorithm requires the execution of 15,500,000 computations. In executing the second algorithm, 2 seconds of computing time are required of processor A while 3 seconds of computing time are required of processor B.

What are the processor rates for processor A and processor B?

Suppose that we define the processing rates (measured in computations/second) for processors A and B to be represented by the variables x and y respectively. With this definition of variables, the number of computations that processor A can perform in 3 seconds is the product (3 x). Likewise the number of computations that processor B can perform in 2 seconds is the product (2 y). The total number of computations that must be performed to complete the first algorithm is 17,000,000. Therefore, we obtain the first of two simultaneous equations

3x + 2y = 17 × 10 6 3x + 2y = 17 × 10 6 size 12{3x+2y="17" times "10" rSup { size 8{6} } } {}
(20)

Let us now consider the implementation of the second algorithm. The number of computations that processor A can make in 2 seconds is equal to the product (2 x). Similarly, the number of computations that processor B can make in 3 seconds is the product (3 y). We obtain the second simultaneous equation as

2x + 3y = 15 . 5 × 10 6 2x + 3y = 15 . 5 × 10 6 size 12{2x+3y="15" "." 5 times "10" rSup { size 8{6} } } {}
(21)

Let us use the first equation to solve for x in terms of y

3x = 2y + 17 × 10 6 3x = 2y + 17 × 10 6 size 12{3x= - 2y+"17" times "10" rSup { size 8{6} } } {}
(22)

which results in the expression

x = 17 × 10 6 2y 3 x = 17 × 10 6 2y 3 size 12{x= { {"17" times "10" rSup { size 8{6} } - 2y} over {3} } } {}
(23)

Next, we substitute this expression for x into equation ()

2 17 × 10 6 2y 3 + 3y = 15 . 5 × 10 6 2 17 × 10 6 2y 3 + 3y = 15 . 5 × 10 6 size 12{2 left ( { {"17" times "10" rSup { size 8{6} } - 2y} over {3} } right )+3y="15" "." 5 times "10" rSup { size 8{6} } } {}
(24)

This equation can be simplified as follows

17 × 10 6 2y + 3 2 ( 3y ) = 3 2 15 . 5 × 10 6 17 × 10 6 2y + 3 2 ( 3y ) = 3 2 15 . 5 × 10 6 size 12{"17" times "10" rSup { size 8{6} } - 2y+ { {3} over {2} } \( 3y \) = { {3} over {2} } left ("15" "." 5 times "10" rSup { size 8{6} } right )} {}
(25)
17 × 10 6 2y + 4 . 5y = 23 . 25 × 10 6 17 × 10 6 2y + 4 . 5y = 23 . 25 × 10 6 size 12{"17" times "10" rSup { size 8{6} } - 2y+4 "." 5y="23" "." "25" times "10" rSup { size 8{6} } } {}
(26)
2 . 5y = 6 . 25 × 10 6 2 . 5y = 6 . 25 × 10 6 size 12{2 "." 5y=6 "." "25" times "10" rSup { size 8{6} } } {}
(27)

This leads to the solution

y = 2 . 5 × 10 6 computations / s y = 2 . 5 × 10 6 computations / s size 12{y=2 "." 5 times "10" rSup { size 8{6} } ~ ital "computations"/s} {}
(28)

We now may solve for x through the use of equation ()

x = 17 × 10 6 2 ( 2 . 5 × 10 6 ) 3 x = 17 × 10 6 2 ( 2 . 5 × 10 6 ) 3 size 12{x= { {"17" times "10" rSup { size 8{6} } - 2` \( 2 "." 5 times "10" rSup { size 8{6} } \) } over {3} } } {}
(29)

which lead to the result

x = 12 × 10 6 3 = 4 × 10 6 x = 12 × 10 6 3 = 4 × 10 6 size 12{x= { {"12" times "10" rSup { size 8{6} } } over {3} } =4 times "10" rSup { size 8{6} } } {}
(30)

Thus we conclude that processor A can perform 4 million computations per second while processor B can perform 2.5 million computations per second.

Alloy Compositions

Metallurgical engineering involves the study of the physical and chemical behavior of metallic elements and their mixtures, which are called alloys. It is also involves the technology of metals. That is, metallurgical engineering encompasses the way in which the science of metals is applied to produce compounds for practical use.

The following example presents how simultaneous equations can be applied in the study of alloys.

Example 4: Let us consider two alloys comprised of different percentages of copper and zinco. Suppose that the first alloy is made up of a mixture of 70% copper and 30% zinc. The second alloy is comprised of 40% copper and 60% zinc.

By melting the two alloys and re-combining their constituents into one 300 gram casting, the resulting alloy is 60% copper and 40% zinc.

Find the amount of copper and zinc in each of the original alloys.

We begin our solution by establishing the variables A and B as being equal to the mass (in grams) of the first alloy and the second alloy respectively. We know that the final casting has a mass of 300 grams, so our first simultaneous equation is

A + B = 300 A + B = 300 size 12{A+B="300"} {}
(31)

Because 60% of the final casting is copper, the amount of copper in the final casting is the product 0.60 (300 grams) or 180 grams.

We know that 70% of the first alloy is copper and that 40% of the second alloy is copper. Thus we may write the second simultaneous equation as

0 . 7 A + 0 . 4 B = 180 0 . 7 A + 0 . 4 B = 180 size 12{0 "." 7`A+0 "." 4`B="180"} {}
(32)

The variable A can be expressed as (300 – B) and substituted into the equation

0 . 7 A + 0 . 4 B = 180 0 . 7 A + 0 . 4 B = 180 size 12{0 "." 7`A+0 "." 4`B="180"} {}
(33)

to yield

0 . 7 ( 300 B ) + 0 . 4B = 180 0 . 7 ( 300 B ) + 0 . 4B = 180 size 12{0 "." 7` \( "300" - B \) +0 "." 4B="180"} {}
(34)

This equation can be simplified as

210 0 . 7B + B = 180 210 0 . 7B + B = 180 size 12{"210" - 0 "." 7B+B="180"} {}
(35)

Combining terms yields

0 . 3B = 30 0 . 3B = 30 size 12{ - 0 "." 3B= - "30"} {}
(36)

The solution for the variable B is 100. So we find that the mass of the second alloy is 100 grams. Because the sum of the masses of the two alloys is 300 grams, the mass of the first allow is 200 grams.

Photovoltaic Arrays

A photovoltaic system is a system which uses one or more solar panels to convert sunlight into electricity. It consists of multiple components, including the photovoltaic modules, mechanical and electrical connections and mountings and means of regulating and/or modifying the electrical output.

Due to the low voltage of an individual solar cell (typically on the order of 0.5V), several cells are wired in series in the manufacture of a "laminate". The laminate is assembled into a protective weatherproof enclosure, thus making a photovoltaic module or solar panel. Modules may then be strung together into a photovoltaic array. The electricity generated can be either stored, used directly (island/standalone plant)or fed into a large electricity grid powered by central generation plants (grid-connected/grid-tied plant) or combined with one or many domestic electricity generators to feed into a small grid (hybrid plant)

A photovoltaic array (or solar array) is a linked collection of solar panels. The power that one module can produce is seldom enough to meet requirements of a home or a business, so the modules are linked together to form an array. Most arrays use a device known as an inverter to convert the DC power produced by the modules into alternating current that can be used to power lights, household appliances, motors, and other loads. The modules in a photovoltaic array are usually first connected in series to obtain the desired voltage. Once this is accomplished, the individual strings are then connected in parallel to allow the system to produce more current.

Example 6: A photvoltaic array is to be installed. It is a requirement that the length of the array be 8 inches longer than its width. A second requirement states that the total area of the array be 180 in2.

Find the length (L) and width (W) of the array.

We may use the first requirement to construct the first simultaneous equation

L = W + 8 L = W + 8 size 12{L=W+8} {}
(37)

The second requirement contributes a second simultaneous equation

L W = 180 L W = 180 size 12{L`W="180"} {}
(38)

We may substitute the expression for L given by equation (18) into equation (19)

W 2 + 8W = 180 W 2 + 8W = 180 size 12{W rSup { size 8{2} } +8W="180"} {}
(39)

or equivalently

W 2 + 8W 180 = 0 W 2 + 8W 180 = 0 size 12{W rSup { size 8{2} } +8W - "180"=0} {}
(40)

This quadratic equation can be factored as

( W + 18 ) ( W 10 ) = 0 ( W + 18 ) ( W 10 ) = 0 size 12{ \( W+"18" \) ` \( W - "10" \) =0} {}
(41)

The two roots for W are 10 in and -18 in. Because it is physically impossible to have a negative value for width, we ignore the second root and conclude that the width of the photovoltaic array is 10 in. We use this value to solve for the length (L) of the array

L = W + 8 = 10 + 8 in = 18 in L = W + 8 = 10 + 8 in = 18 in size 12{L=W+8="10"+8` ital "in"="18"` ital "in"} {}
(42)

Therefore, the dimensions of the photovoltaic array are 18 x 10 in.

Exercises

  1. A rowing team competes in a tournament that has two legs. The first leg covers 15 miles in the direction of the current. The second leg covers the same 15 miles, but against the current. It takes 25 minutes to complete the first leg and 30 minutes to complete the second leg. Assuming that there is no current present, what would be the speed of the boat? What is the speed of the current?
  2. A beam of length 3 meters is supported by a fulcrum. Force is applied at each end of the beam. Assume that the force (F1) on the left hand side of the beam is 6 N while the force on the right hand side of the beam (F2) is 7 N. If the beam is static, then what are the lengths between the each end of the beam and its fulcrum?
  3. Suppose that we are interested in implementing a computer algorithm on three processors. Let us call these processors A and. The steps involved in the accomplishment of the algorithm are divided among the three processors. In all there are 36,000,000 computations which will need to be performed on the three processors. In performing the overall task, processor A is in service for 2 seconds while the processor B is in services for 4 seconds. A second algorithm is implemented in parallel on processors A and B. In it, a total of 23,000,000 computations are required. In performing the task, processor A is in service for 1 second and processor B is in service for 1.5 seconds. Find the processing rate (computations/s) for each processor.
  4. One alloy is made up of a concentration of 25% copper and 75% manganese. A second alloy consists of a concentration of 50% copper and 50% manganese. The two alloys are melted and re-combed. The resulting mixture weighs 450 gram and consists of 35% copper and 65% zinc. Find the weights of the original alloys.

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