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Force -- Vector Solutions for Coplanar Forces Concurrent at a Point

Module by: R.G. (Dick) Baldwin. E-mail the author

Summary: This module explains the law of sines and the law of cosines. It also explains vector solutions for coplanar forces that are concurrent at a point in a format that is accessible to blind students.

Preface

General

This module is part of a collection of modules designed to make physics concepts accessible to blind students.

See http://cnx.org/content/col11294/latest/ for the main page of the collection and http://cnx.org/content/col11294/latest/#cnx_sidebar_column for the table of contents for the collection.

The collection is intended to supplement but not to replace the textbook in an introductory course in high school or college physics.

This module explains the law of sines and the law of cosines. It also explains vector solutions for coplanar forces that are concurrent at a point in a format that is accessible to blind students.

Prerequisites

In addition to an Internet connection and a browser, you will need the following tools (as a minimum) to work through the exercises in these modules:

The minimum prerequisites for understanding the material in these modules include:

Viewing tip

I recommend that you open another copy of this document in a separate browser window and use the following links to easily find and view the figures and listings while you are reading about them.

Figures

  • Figure 1 . Mirror image from the file named Phy1100a1.svg.
  • Figure 2 . Non-mirror-image version of the image from the file named Phy1100a1.svg.
  • Figure 3 . Key-value pairs for the image in the file named Phy1100a1.svg.
  • Figure 4 . Mirror image contained in the file named Phy1100b1.svg.
  • Figure 5 . Non-mirror-image version of the image from the file named Phy1100b1.svg.
  • Figure 6 . Key-value pairs for the image in the file named Phy1100b1.svg.
  • Figure 7 . Mirror image from the file named Phy1100c1.svg.
  • Figure 8 . Non-mirror-image version of the image from the file named Phy1100c1.svg.
  • Figure 9 . Key-value pairs for the image in the file named Phy1100c1.svg.
  • Figure 10 . Mirror image from the file named Phy1100d1.svg.
  • Figure 11 . Non-mirror-image version of the image from the file named Phy1100d1.svg.
  • Figure 12 . Key-value pairs for the image in the file named Phy1100d1.svg.
  • Figure 13 . Mirror image from the file named Phy1100e1.svg.
  • Figure 14 . Non-mirror-image version of the image from the file named Phy1100e1.svg.
  • Figure 15 . Key-value pairs for the image in the file named Phy1100e1.svg.
  • Figure 16 . Screen output for Listing #1.
  • Figure 17 . Changes to Listing #1.
  • Figure 18 . Screen output for modified Listing #1.

Listings

Supplemental material

I recommend that you also study the other lessons in my extensive collection of online programming tutorials. You will find a consolidated index at www.DickBaldwin.com .

General background information

Creation of tactile graphics

The module titled Manual Creation of Tactile Graphics at http://cnx.org/content/m38546/latest/ explained how to create tactile graphics from svg files that I will provide.

If you are going to have an assistant create tactile graphics for this module, you will need to download the file named Phy1100.zip , which contains the svg files for this module. Extract the svg files from the zip file and provide them to your assistant.

Also, if you are going to use tactile graphics, it probably won't be necessary for you to perform the graph board exercises. However, you should still walk through the graph board exercises in your mind because I will often embed important physics concepts in the instructions for doing the graph board exercises.

In each case where I am providing an svg file for the creation of tactile graphics, I will identify the name of the appropriate svg file and display an image of the contents of the file for the benefit of your assistant. As explained at http://cnx.org/content/m38546/latest/ , those images will be mirror images of the actual images so that your assistant can emboss the image from the back of the paper and you can explore it from the front.

I will also display a non-mirror-image version of the image so that your assistant can easily read the text in the image.

Also in those cases, I will provide a table of key-value pairs that explain how the Braille keys in the image relate to text or objects in the image.

The law of sines

Much of the background material needed for this module was presented in the earlier module titled Introduction to Statics, Equilibrium, and Forces . However, beginning with this module, you will need some information on trigonometry that was not presented in the earlier module titled Brief Trigonometry Tutorial . That item is known as the law of sines (see http://en.wikipedia.org/wiki/Law_of_sines ).

Proving identities

When I was a student in a trigonometry course many years ago, students were required to "prove identities" which was a process very similar to proving theorems in plane geometry. That activity included doing such things as proving that the law of sines is true.

I don't know if mathematics instructors still require students to prove identities or not. In any event, we will simply accept such mathematical truths in this collection of modules and won't worry about proving that they are true.

We will work through some exercises in this module to show how the law of sines can be used to advantage.

Beyond the right triangle

Previous modules using trigonometry have dealt almost exclusively with right triangles. Beginning in this module, we will start using trigonometry to solve problems that involve triangles that are not right triangles.

Tactile graphics

The svg file that is required to create tactile graphics for this exercise is named Phy1100a1.svg. You should have downloaded that file earlier.

Figure 1 shows the mirror image that is contained in that file for the benefit of your assistant who will create the tactile graphic for this exercise.

Figure 1: Mirror image from the file named Phy1100a1.svg.
Mirror image from the file named Phy1100a1.svg.
Missing image

Figure 2 shows a non-mirror-image version of the same image.

Figure 2: Non-mirror-image version of the image from the file named Phy1100a1.svg.
Non-mirror-image version of the image from the file named Phy1100a1.svg.
Missing image.

Key-value pairs

Figure 3 shows the key-value pairs that go with the image in the file named Phy1100a1.svg.

Figure 3: Key-value pairs for the image in the file named Phy1100a1.svg.
Key-value pairs for the image in the file named Phy1100a1.svg.

m: Laws of sines and cosines
n: C = 40 degrees
o: a = 10
p: B = 80 degrees
q: A = 60 degrees
r: c = 7.4
s: File: Phy1100a1.svg
t: b = 11.4

Create a triangle

Please use your graph board to create a triangle with the dimensions, angles, and labels specified below.

Use your Braille labeler to label the sides of the triangle a, b, and c. Label the angle opposite from side a as A, the angle opposite side b as B, and the angle opposite side c as C.

Adjust your triangle to make side a equal to 10 units, angle B equal to 80 degrees, and angle C equal to 40 degrees. Using the above relationship, we also know that angle A is equal to 60 degrees because the angles in a triangle must sum to 180 degrees.

Using the law of sines

The law of sines can be used to compute the remaining sides of a triangle when two angles and a side are known. Of course, if you know two of the angles of a triangle, you also know the third angle because the sum of the angles in a triangle is always 180 degrees.

Use the Google calculator

All of the arithmetic and trigonometric requirements in this section can be satisfied using the Google calculator. Go to Google and enter the following in the search box:

sine 80 degrees

Google should produce a line of output text that reads:

sine(80 degrees) = 0.984807753

Record this value for future use.

Get and record two more values

Go to Google again and enter sine 40 degrees. The output should be:

sine(40 degrees) = 0.64278761

Doing the same thing again for 60 degrees produces:

sine(60 degrees) = 0.866025404

The law of sines formula

Given the names of the sides and angles that you have applied to your triangle, the law of sines says that the following is true:

(a/sin A) = (b/sin B) = (c/sin C)

In general, the law of sines states that the ratio of the length of any side to the sine of the opposite angle is equal to the ratio of any other side to the sine of its opposite angle.

Compute the length of side b

Substituting the known values in the above formula gives us:

(10/0.866) = (b/0.985) = (c/0.643)

Combining the first and second ratios and rearranging terms gives us:

b = 10 * 0.985/0.866 = 11.4

(You can also use the search box at Google to perform arithmetic such as 10 * 0.985/0.866.)

Compute the length of side c

Combining the first and third ratios and rearranging terms gives us:

c = 10*0.643/.866 = 7.42

Hopefully, you can confirm these values (approximately) by making measurements on the triangle on your graph board.

Another use for the law of sines

The law of sines can also be used when two sides and one of the non-enclosed angles is known to find the values of the other side and the values of the other two angles.

Let's give that scenario a test drive

Pretend that all we know about the triangle is the following:

  • Side a = 10
  • Side b = 11.3741339
  • Angle B = 80 degrees
  • sin B = 0.984807753

We need to solve for angles A and C as well as side c.

Substitution of known values

Recall that the general formula for the law of sines is:

(a/sin A) = (b/sin B) = (c/sin C)

Substitution of the known values into the formula gives:

(10/sin A) = (11.3741339/0.984807753) = (c/sin C)

Rearranging terms

Using the first two ratios and rearranging terms gives us:

sin A = 10*0.984807753/11.3741339 = 0.865830983

A = arcsine(0.865830983) = 1.04680884 radians = 60 degrees

(Note that I used an extreme number of decimal digits to avoid rounding errors and cause the final answer to come out almost exactly correct.)

Compute the value of angle c

Now that we know the values of angles A and B, we also know that

Angle C = 180 - 80 - 60 = 40 degrees.

Compute the length of side c

Using the second and third ratios and rearranging terms, we can solve for side c as:

c = sin C * (11.3741339/0.984807753)

= 0.64278761 * (11.3741339/0.984807753)

= 7.42393866

Thus, we have determined that:

  • Angle A = 60 degrees
  • Angle C = 40 degrees
  • Side c = 7.42

These values all match what we know to be true from the first example.

The law of cosines

In addition to the law of sines, there is also a law of cosines. The law of cosines is useful for computing the third side of a triangle when two sides and their enclosed angle are known. The law is also useful in computing the angles of a triangle when all three sides are known.

Using the same names for the sides and the angles, there are three formulas for the law of cosines :

  • c^2 = a^2 + b^2 - 2*a*b*cos C
  • a^2 = b^2 + c^2 - 2*b*c*cos A
  • b^2 = a^2 + c^2 - 2*a*c*cos B

Compute the length of side c

Using the known values from the earlier example, we can compute the length of side c by entering the following expression into the Google search box:

sqrt(10^2 + 11.3^2 - 2*10*11.3*cos(40*pi/180))

This produces an answer of 7.39, which is close enough to the known value for the length of side c.

Compute angle C in degrees

Similarly, using the known values from the earlier example, we can compute the angle C knowing the lengths of the three sides. Enter the following expression into the Google search box:

arccos((10^2 + 11.3^2 - 7.39^2)/(2*10*11.3))*180/pi

This produces an answer of 40.02 which is close enough to the known value for the angle C in degrees.

An exercise for the student

I will leave it as an exercise for the student to work through the algebra to determine how I used the known values and the first formula given above to construct the expressions that I entered into the Google search box.

Discussion and sample code

Now that we have expanded our knowledge of trigonometry, let's examine some solutions to problems involving force using vectors.

All of the exercises that we will work through in this module involve forces in a two-dimensional plane (coplanar) that meet at a point (concurrent at a point). This excludes forces that create moments or torques, which will be the topic of future modules, and also excludes forces that may come from any direction in a third dimension.

Using the graph board

You will be making heavy use of your graph board in this module. In order to make the instructions less awkward, I am going to begin using drawing syntax when talking about the graph board. For example, when I say draw a line from 3,5 to 10,15, that will mean for you to insert a pin at coordinates x=3 and y=5, to insert a second pin at x=10 and y=15, and to connect the two pins with a rubber band, a pipe cleaner, a wire, a piece of yarn, or whatever works for you.

Parallelogram of forces

Force is a vector. It has both magnitude and direction. You learned in an earlier module that you can resolve vectors two at a time using a method that involves parallelograms.

Stated more formally, any two forces meeting at a point can be replaced by a single resultant force, which is represented by the diagonal of a parallelogram whose two adjacent sides represent the two forces.

Tactile graphics

The file named Phy1100b1.svg contains an image from which your assistant can create a tactile graphic to illustrate the scenario described below .

Figure 4 shows the mirror image that is contained in that file for the benefit of your assistant who will create the tactile graphic for this exercise.

Figure 4: Mirror image contained in the file named Phy1100b1.svg.
Mirror image contained in the file named Phy1100b1.svg.
Missing image

Figure 5 shows a non-mirror-image version of the same image.

Figure 5: Non-mirror-image version of the image from the file named Phy1100b1.svg.
Non-mirror-image version of the image from the file named Phy1100b1.svg.
Missing image

Figure 6 shows the key-value pairs that go with the image in the file named Phy1100b1.svg.

Figure 6: Key-value pairs for the image in the file named Phy1100b1.svg.
Key-value pairs for the image in the file named Phy1100b1.svg.
m: Schematic diagram of the scenario
n: pipe 
o: cord under tension 
p: slack cord 
q: cord under tension 
r: cord under tension 
s: 1.02 kg mass 
t: File: Phy1100b1.svg
u: small connecting ring

The scenario

Please draw a schematic diagram of this scenario on your graph board.

A strong pipe extends from left to right in the work area supported on both ends. You can get your hands around the pipe to attach things to it.

Light, strong, flexible cords

You tie two light, strong, flexible cords to the pipe approximately one meter apart. (the distance isn't critical). We will ignore the weight of the cords. You thread the ends of the two cords through a small ring that is just large enough for you to tie four cords onto the ring. (The only purpose of the ring is to provide a convenient way to attach cords together at a single point.)

Adjust the position of the ring

You adjust the position of the ring on each of the cords so that the cord on the left forms a 60-degree south of east angle with the pipe and the cord on the right forms a 45 degree south of west angle with the pipe. Then you tie both cords to the ring at that point.

A triangle

When you pull straight down on the ring, the two cords and the pipe form a triangle. The interior angle at the upper-left vertex is 60 degrees, the interior angle at the upper-right vertex is 45 degrees, and because the sum of the angles in a triangle is 180 degrees, the interior angle at the bottom vertex is 75 degrees.

Attach a mass to the ring

You tie a short cord to the bottom of the ring and tie a 1.02 kg mass to the cord. (A 1.02 kg mass has a weight of approximately 10 newtons.). Then you allow the two cords that are tied to the pipe to support the mass, which pulls straight down with a force of 10 newtons.

The structure of the system

Now, let's think about the structure of this system. It consists of three cords emanating out from a small ring with each cord exerting a force on the ring.

Compression or tension?

To begin with, a flexible cord cannot support a compressive force. If you push on one end of a cord in an attempt to push the object attached to the other end, the cord will simply bend. Therefore, the cords in this scenario cannot possibly exert a pushing force on the ring.

However, a cord can support a tension or pulling force. In other words, any or all of the cords can support a pulling force on the ring.

Nothing else is touching the ring

The ring is not in contact with anything other than the three cords at this point in time, and the ring is in equilibrium. By that I mean that the ring has zero velocity and no acceleration.

(We are also assuming that the ring is, for all practical purposes, a point and it is not experiencing a moment or a torque. Once again, we'll get into moments and torques in a future module.)

Thee pulling forces

So, we know that the ring is being subjected to three pulling forces. A pulling force of 10 newtons is pulling straight down toward the center of the earth. The forces associated with the two upper cords must be along the lengths of the cords. Otherwise, the cords and the ring would be moving.

Tactile graphics

The file named Phy1100c1.svg contains a force vector diagram for this scenario.

Figure 7 shows the mirror image that is contained in that file for the benefit of your assistant who will create the tactile graphic for this exercise.

Figure 7: Mirror image from the file named Phy1100c1.svg.
Mirror image from the file named Phy1100c1.svg.
Missing image.

Figure 8 shows a non-mirror-image version of the same image.

Figure 8: Non-mirror-image version of the image from the file named Phy1100c1.svg.
Non-mirror-image version of the image from the file named Phy1100c1.svg.
Missing image.

Figure 9 shows the key-value pairs that go with the image in the file named Phy1100c1.svg.

Figure 9: Key-value pairs for the image in the file named Phy1100c1.svg.
Key-value pairs for the image in the file named Phy1100c1.svg.
m: Force vector diagram.
n: Parallelogram lines
o: 7.3 newtons 
p: Vector C 
q: Resultant 
r: 10 newtons 
s: Vector B 
t: 5.3 newtons 
u: 60 degrees 
v: 45 degrees 
w: 10 newtons 
x: Vector A 
y: File: Phy1100c1.svg 

Directions of the forces

The fact that the cords can only support tension tells us that one of the two forces pulling in an upward direction is actually pulling at an angle of 45 degrees north of east(the direction of the cord). The other force is pulling at angle of 60 degrees north of west.

What is the tension in each cord?

So, three interesting questions are:

  • What is the tension in each of the cords.
  • How much of the 10-newton load is being carried by the cord on the right.
  • How much of the load is being carried by the cord on the left?

We can determine the answers to those questions using the parallelogram rule. At least we can find the tension in each of the cords and determine if those tensions are sufficient to carry the 10-newton load.

The parallelogram rule

As I told you earlier, any two forces meeting at a point (we're considering the ring to be a point) can be replaced by a single resultant force, which is represented by the diagonal of a parallelogram whose two adjacent sides represent the two forces.

If we can figure out what the resultant force is, we can figure out the answer to the above questions regarding the tensions and the load being borne by each of the upper cords.

We should be able to figure out what the resultant force is with a thought experiment based on our life experiences.

Tie a fourth cord to the ring

Pretend that you tie another cord to the top of the ring and loop it over the pipe somewhere between the tie points of the other two cords. (This cord is indicated by the object labeled "slack cord" in the image in the file named Phy1100b1.svg and also in the image in Figure 5 .)

Then pull very slowly on that fourth cord until both of the other cords go slack. At that point, the fourth cord, (which is somewhat analogous to the resultant of the forces in the other two cords) has assumed the entire 10-newton load.

Is the ring in equilibrium?

Did the ring and the mass move sideways when the other two cords went slack. If so, the equilibrium of the system was disrupted. Relax the tension on the fourth cord, move it slightly sideways, and repeat the experiment. Continue this process until you find a location where the ring doesn't move sideways when the fourth cord assumes the total load.

Only one location will suffice

You can probably agree, from prior experience on the swing set at the playground as a child, that the only location at which the fourth cord can assume the total load without the ring moving sideways is when the fourth cord passes over the pipe directly above the ring and the mass. Therefore, we know that the resultant of the forces being exerted by the original two upper cords must be on a vertical line directly above the ring and the mass.

The diagonal

We also know on the basis of the above rule that the length of the resultant force is proportional to the length of the diagonal of a parallelogram whose two adjacent sides represent the two forces that are being resolved.

The resultant force

The resultant force is indicated by a heavy dashed vector pointing upward in the image in Phy1100c1.svg. Two sides of the parallelogram are represented by lighter dashed lines. The other two sides of the parallelogram are represented by heavy solid vectors, one at 45 degrees relative to the horizontal and the other at 60 degrees relative to the horizontal.

The length of the resultant force vector

Let's think a bit about the required length of the resultant force vector. The above experiment resolves the force system into only two forces acting on the ring. (The original two upward force vectors are eliminated from the picture when those two cords go slack.) One force is directed up and the other force is directed down.

The vector sum must be zero for equilibrium

We also know that the vector sum of all the forces acting on a point must be zero in in order for the point to be in equilibrium. Since there are the only two forces acting on the ring, the only way that they can sum to zero is for them to be equal in magnitude and opposite in direction from one another.

Equal and opposite

The force vector pulling down on the ring has a magnitude of 10 newtons. We know because we tied a ten-newton load to the bottom of the ring. Therefore, the magnitude of the resultant vector pulling up on the ring must also be 10 newtons. That is the required length of the diagonal of the parallelogram discussed earlier.

The graphical solution

Draw the force vector pointing straight up from the ring with a length of ten units. (You can use whatever drawing scale is convenient for your drawing.) That is the diagonal of a parallelogram. Two force vectors lie along the lines of the two cords. Those two force vectors form the adjacent sides of a parallelogram, but we don't yet know their lengths.

Complete the parallelogram

The lengths of the force vectors that lie along the lines of the cords are equal to the sides of the parallelogram. That provides the answer to the original question regarding the tension in each of the cords. The tension in each cord is equal to the length of the side of the parallelogram using the same scale that was used to draw the 10-newton diagonal length.

Drawing the parallelogram might be difficult

Drawing the parallelogram is not particularly easy even for a sighted person without special drawing tools. Here is one way that I have found to do it.

Draw a horizontal line (parallel to the pipe) that just touches the tip of the force vector that represents the diagonal. Then use your protractor to draw lines that begin at the tip of the force vector and emanate downward on both sides of the diagonal at angles of 60 degrees south of east and 45 degrees south of west. Be sure to do it in the correct order so that you end up with a parallelogram.

Mark the spots

Mark the spots where those lines cross the two upper cords. Those spots are the tips of the force vectors in the cords.

My answer

When I did it graphically (using pencil, paper, protractor, and ruler), I got the length of the force vector pointing 45 degrees north of east to have a length of 5.3 newtons. I got the length of the force vector pointing north of west to be 7.3 newtons. These are the values on the upper force vectors in the image in Phy1100c1.svg.

Numeric sum exceeds 10 newtons

You may have noticed that the numeric sum of the magnitudes of these two force vectors exceeds the load of 10 newtons. How can that be?

The answer is that when force vectors that are supporting a vertical load go off in a direction other than vertical, only the vertical component of the force vector counts insofar as supporting the load is concerned.

The vertical components of the two force vectors are:

  • 5.3*sin(45 degrees) = 3.74
  • 7.3*sin(60 degrees) = 6.32

The sum of the vertical components is:

6.32 + 3.74 = 10.06

which is close enough to the downward force being exerted on the ring by the mass.

Your answer

Were you able to get reasonably close to my answers? If so, congratulations. If not, don't worry too much about it. The most important thing is for you to understand the process and not to develop expertise in constructing accurate drawings using pushpins and rubber bands. However, I do believe that constructing vector diagrams will help you gain a better understanding of the process.

Later I will show you how to solve the same .problem mathematically which is probably the better choice for a blind student.

Triangle of forces

Now let's take another look at the same problem from a somewhat different graphical viewpoint. To recap, we have a point in a plane (the ring) which is subjected to the following three vector forces and the ring is in equilibrium. (All angles are given relative to the positive horizontal axis).

  • Vector A -- A force of 10 newtons at an angle of -90 degrees.
  • Vector B -- A force of an unknown magnitude at an angle of 45 degrees.
  • Vector C -- A force of an unknown magnitude at an angle of 120 degrees.

We know that in order for the ring to be in equilibrium, the sum of the vectors for these three forces must add to zero. What are the magnitudes of the forces that vector B and vector C exert on the ring?

Drawing tail to tip

You learned in an earlier module that you can add vectors graphically by drawing them tail to tip. Once you have done that, the resultant vector is a vector that extends from the tail of the first vector to the tip of the last vector.

Tactile graphics

The file named Phy1100d1.svg contains the image that is required to create a tactile graphic for this scenario.

Figure 10 shows the mirror image that is contained in that file for the benefit of your assistant who will create the tactile graphic for this exercise.

Figure 10: Mirror image from the file named Phy1100d1.svg.
Mirror image from the file named Phy1100d1.svg.
Missing image.

Figure 11 shows a non-mirror-image version of the same image.

Figure 11: Non-mirror-image version of the image from the file named Phy1100d1.svg.
Non-mirror-image version of the image from the file named Phy1100d1.svg.
Missing image

Figure 12 shows the key-value pairs that go with the image in the file named Phy1100d1.svg.

Figure 12: Key-value pairs for the image in the file named Phy1100d1.svg.
Key-value pairs for the image in the file named Phy1100d1.svg.
m: Tip-to-tail force vector diagram
n: Vector C
o: 7.3 newtons
p: 10 newtons
q: Vector A
r: 120 degrees
s: 45 degrees
t: File: Phy1100d1.svg
u: Vector B
v: 5.3 newtons

A closed path for equilibrium

For this case where the sum must be zero, the resultant vector must have a zero length. This means that the tip of the last vector must touch the tail of the first vector. In other words, the connected vectors must describe a closed path.

In the earlier modules, you were given the magnitude and direction of several vectors and you arranged them tail to tip in order to find the resultant vector. However, for this problem, you know the direction and magnitude of one vector but you only know the directions for the other two vectors. You need to find their magnitudes.

Knowing that the three vectors must form a closed path makes a graphical solution to this problem possible.

Draw a vector diagram

Begin by drawing vector A with a length of 10 newtons at an angle of -90 degrees, using whatever plotting scale works for you. The tail for this vector will be at the top of the vertical line and the tip will be at the bottom.

Add vector B to the diagram

Go to the tip of vector A and draw vector B with an unspecified length at an angle of 45 degrees. The tail of vector B will touch the tip of vector A and the tip of vector B will be somewhere up and to the right of the tip of vector A.

Add vector C to the diagram

Go to the tail of vector A and draw vector C with an unspecified length at an angle of 120 degrees. The tip of vector C will touch the tail of vector A and the tail of vector C will be somewhere down and to the right of the tail of vector A.

The lines should cross

If you did everything right, the lines that describe vector B and vector C should cross. Use a pushpin to mark the place where they cross. This point will be the tip of vector B and will be the tail of vector C.

Measure the lengths of vector B and vector C

At this point, you can measure the lengths of the lines that describe vector B and vector C. Using the same plotting scale that you used for the 10-newton vector A, you can determine the magnitudes of each of those vectors.

Since this is the same problem as before, the magnitudes for those two vectors should be the same as before. Vector B should be about 5.3 newtons and vector C should be about 7.3 newtons.

Was this easier than the parallelogram?

I suspect that you probably found the physical construction of the triangle solution to be easier than the physical construction of the parallelogram solution. I further suspect that the trigonometric solution, which I will explain in the next section, will be even easier for blind students.

Trigonometric solution to the triangle of forces

Let's pretend that you don't have a graphic solution for this problem. However, you do have a rough sketch of what the triangle looks like on your graph board. Use your Braille labeler to label the three vectors in the sketch A, B, and C. Then label the angles opposite each side as a, b, and c respectively.

Here are the facts as we know them at this point:

  • Vector A = 10 newtons.
  • Angle a = 105 degrees.
  • Sine 105 degrees = 0.966
  • Angle b = 30 degrees
  • Sine 30 degrees = 0..5
  • Angle c = 45 degrees
  • Sine c = 0.707

The law of sines

We will solve this problem using the law of sines that I explained earlier. Given the names that we have applied to the sides and the angles for this triangle, the law of sines can be written as follows:

(A/sin a) = (B/sin b) = (C/sin c)

(Note that upper and lower-case letters were switched relative to the law of sines given near the beginning of this module.)

Solve for the length of side B

Using the first two ratios and rearranging terms gives

B = A*sin b/sin a = 5.18

Solve for the length of side C

Using the first and third ratio and rearranging terms gives

C = A*sin c/sin a = 7.32

The answers

These are the answers that we are looking for:

  • B = 5.18 newtons
  • C = 7.32 newtons

Using components to analyze for equilibrium

Tent pole exercise

Please sketch this scenario on your graph board. Then we will solve the problem using a script and components.

A bird's eye view of the top of a tent pole shows guy wires extending outward and exerting radial 10-newton forces on a tent pole at the following angles :

  • 0 degrees
  • 45 degrees
  • 90 degrees
  • 135 degrees
  • 180 degrees
  • 225 degrees
  • 270 degrees

Tactile graphics

The image in the file named Phy1100e1.svg contains seven vectors of equal length emanating out from a common point at the angles given above.

Figure 13 shows the mirror image that is contained in that file for the benefit of your assistant who will create the tactile graphic for this exercise.

Figure 13: Mirror image from the file named Phy1100e1.svg.
Mirror image from the file named Phy1100e1.svg.
Missing image

Figure 14 shows a non-mirror-image version of the same image.

Figure 14: Non-mirror-image version of the image from the file named Phy1100e1.svg.
Non-mirror-image version of the image from the file named Phy1100e1.svg.
Missing image

Figure 15 shows the key-value pairs that go with the image in the file named Phy1100e1.svg.

Figure 15: Key-value pairs for the image in the file named Phy1100e1.svg.
Key-value pairs for the image in the file named Phy1100e1.svg.
m: Tent pole exercise
n: File: Phy1100e1.svg

Equilibrium

Is the tent pole in equilibrium? If not, why not? What new force would need to be added to cause it to be in equilibrium?

You can probably solve this exercise in your head by inspection of the force vectors because they are all of the same length and they emanate outward at regularly spaced angles around the pole.

Could be more complex

Note, however, that the scenario could be much more complicated involving force vectors with different lengths arranged at seemingly random angles around the pole.

The important point is, the same general scheme would be used to solve the more complicated problem as is used to solve this simple exercise. However, the code in the script for the more complicated version would probably be somewhat more messy.

The scheme for the solution

The solution is to compute the sum of the force vectors to determine if the resultant vector is zero. If it is not zero, the tent pole would not be in equilibrium. It would be necessary to add a new force vector with the same magnitude as the resultant vector but in a direction that is the exact opposite of the direction of the resultant vector to cause the tent pole to be in equilibrium.

Create a script

Please copy the code from Listing 1 into an html file and open the file in your browser.

Listing 1: Tent pole exercise.

<!---------------- File JavaScript01.html --------------------->
<html><body>
<script language="JavaScript1.3">

document.write("Start Script </br></br>");

//The purpose of this function is to receive the adjacent
// and opposite side values for a right triangle and to
// return the angle in degrees in the correct quadrant.
function getAngle(x,y){
  if((x == 0) && (y == 0)){
    //Angle is indeterminate. Just return zero.
    return 0;
  }else if((x == 0) && (y > 0)){
    //Avoid divide by zero denominator.
    return 90;
  }else if((x == 0) && (y < 0)){
    //Avoid divide by zero denominator.
    return -90;
  }else if((x < 0) && (y >= 0)){
    //Correct to second quadrant
    return Math.atan(y/x)*180/Math.PI + 180;
  }else if((x < 0) && (y <= 0)){
    //Correct to third quadrant
    return Math.atan(y/x)*180/Math.PI + 180;
  }else{
    //First and fourth quadrants. No correction required.
    return Math.atan(y/x)*180/Math.PI;
  }//end else
}//end function getAngle



var resultX = 0;//magnitude in newtons
var resultY = 0;//magnitude in newtons
var ang = 0;//degrees
var mag = 10;//newtons

while(ang <= 270){
  resultX = resultX + mag*Math.cos(ang*Math.PI/180);
  resultY = resultY + mag*Math.sin(ang*Math.PI/180);
  ang = ang + 45;
}//end while loop

var resultAng = getAngle(resultX,resultY);
var resultMag = Math.sqrt(resultX*resultX +
                          resultY*resultY);

document.write("The resultant is:" + "</br>");
document.write("Magnitude = " + resultMag.toFixed(2) +
               "</br>");
document.write("Angle = " + resultAng.toFixed(2) +
               "</br>");

document.write("</br>End Script");

</script>
</body></html>

Screen output

The text shown in Figure 16 should appear in your browser window when you open the html file in your browser.

Figure 16: Screen output for Listing #1.
Screen output for Listing #1.
Start Script

The resultant is:
Magnitude = 10.00
Angle = 135.00

End Script 

Analysis of the output

Figure 16 shows the resultant force when the vector forces shown earlier are added.

The requirement for equilibrium

Two or more coplanar forces concurrent at a point are in equilibrium if the sums of their components, taken along two mutually perpendicular axes are zero.

The vector sum is not zero

As you can see from Figure 16 , the vector sum of the forces shown earlier is not zero. The resultant vector has a magnitude of 10 newtons and a direction of 135 degrees.

Therefore, the tent pole is not in equilibrium and is subject to falling down. If it falls, it will probably fall in the direction of 135 degrees because there is a resultant force of 10 newtons pulling it in that direction.

Add a 10-newton force at 315 degrees

To put the tent pole in equilibrium, we would need to add one more force of 10 newtons in the opposite direction of the resultant shown in Figure 16 . In particular, one more guy wire is needed that adds a 10-newton pulling force in the direction of 315 degrees.

Analysis of the code

We will deal with causing the tent pole to be in equilibrium shortly. Right now, let's make certain that we understand the code shown in Listing 1 .

I doubt that you will find anything new in this script because I explained the use of components for computing vector sums in an earlier module.

A function named getAngle

Listing 1 begins with the definition of a function named getAngle that I explained in an earlier module. Therefore, I won't explain that code again in this module.

Declare working variables

Following the definition of the getAngle function, Listing 1 declares and initializes several working variables.

Enter a while loop

Then control goes into a while loop that computes the sum of the x-components and the sum of the y-components for each of the forces shown earlier .

Those forces are uniformly spaced around the tent pole with a 10-newton force vector every 45 degrees beginning at 0 and extending to 270 degrees inclusive. That makes it easy to compute and add the components using a while loop.

The code in the while loop begins by computing the components of the force vector at 0 degrees, and then computes and accumulates the x and y components for each of the remaining force vectors with a new force vector every 45 degrees up to and including the force vector at 270 degrees.

Storage locations for the vector sum

When the while loop exits, after computing the sum of the x-components and the sum of the y-components for the seven forces shown earlier , the sum of the x-components is stored in the variable named resultX and the sum of the y-components is stored in the variable named resultY.

Compute the magnitude and angle of the resultant vector

The magnitude of the resultant vector is computed as the square root of the sum of the squares of the x and y components. The getAngle method is called to get the angle of the resultant force vector.

Then those values are displayed in the browser window as shown in Figure 16 .

Now let's fix the problem

We said earlier that we can cause the tent pole to be in equilibrium by adding one additional 10-newton force at an angle of 315 degrees. We can easily accomplish this by making the changes shown in Figure 17 to the code in Listing 1 .

Figure 17: Changes to Listing #1. .
Changes to Listing #1. .

Change the code that reads:

while(ang <= 270){

to read

while(ang <= 315){

This change will cause one more 10-newton force vector at an angle of 315 degrees to be added to the sum of the vectors.

Make the change

Make that change and open the modified html file in your browser. When you do, the text shown in Figure 18 should appear in your browser window.

Figure 18: Screen output for modified Listing #1.
Screen output for modified Listing #1.
Start Script

The resultant is:
Magnitude = 0.00
Angle = 198.43

End Script 

Magnitude of resultant vector is zero

The significant thing about Figure 18 is that the magnitude of the resultant vector is now 0, which satisfies the requirement for equilibrium stated earlier .

Pay no attention to the angle shown in Figure 18 . It is meaningless. It is not possible to compute a valid angle for a vector that has a magnitude of zero.

Run the scripts

I encourage you to run the script and perform the calculations that I have presented in this lesson to confirm that you get the same results. Experiment with the code and the calculations, making changes, and observing the results of your changes. Make certain that you can explain why your changes behave as they do.

Resources

I will publish a module containing consolidated links to resources on my Connexions web page and will update and add to the list as additional modules in this collection are published.

Miscellaneous

This section contains a variety of miscellaneous information.

Note:

Housekeeping material
  • Module name: Force -- Vector Solutions for Coplanar Forces Concurrent at a Point
  • File: Phy1100.htm
  • Revised: 07/06/2011
  • Keywords:
    • physics
    • accessible
    • blind
    • graph board
    • protractor
    • screen reader
    • refreshable Braille display
    • JavaScript
    • trigonometry
    • coplanar forces
    • law of sines
    • law of cosines
    • parallelogram rule
    • triangle rule
    • resultant
    • equilibrium

Note:

Disclaimers:

Financial : Although the Connexions site makes it possible for you to download a PDF file for this module at no charge, and also makes it possible for you to purchase a pre-printed version of the PDF file, you should be aware that some of the HTML elements in this module may not translate well into PDF.

I also want you to know that I receive no financial compensation from the Connexions website even if you purchase the PDF version of the module.

Affiliation : I am a professor of Computer Information Technology at Austin Community College in Austin, TX.

-end-

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