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By: ConnexionsAs a part of collection: "Accessible Physics Concepts for Blind Students"

"Blind students should not be excluded from physics courses because of inaccessible textbooks. The modules in this collection present physics concepts in a format that blind students can read […]"

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# Force -- Center of Gravity

Module by: Richard Baldwin. E-mail the author

Summary: This module explains force and center of gravity 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 force and center of gravity 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 while you are reading about them.

#### Figures

• Figure 1 . Mirror image from the file named Phy1120a1.svg.
• Figure 2 . Non-mirror-image version of the image from the file named Phy1120a1.svg.
• Figure 3 . Key-value pairs for the image in the file named Phy1120a1.svg.
• Figure 4 . Mirror image from the file named Phy1120b1.svg.
• Figure 5 . Non-mirror-image version of the image from the file named Phy1120b1.svg.
• Figure 6 . Key-value pairs for the image in the file named Phy1120b1.svg.

### 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

The gravitational constant

The gravitational constant, G, is a constant that is used in the calculation of the gravitational attraction between objects with mass. The value of G is approximately:

G = 6.67 * 10^(-11) *m^3*kg^(-1)*s^(-2), or

G = 6.67 * 10^(-11) N*(m/kg)^2

According to the law of universal gravitation, the attractive force (F) between two bodies is proportional to the product of their masses (m1 and m2), and inversely proportional to the square of the distance (r) between them:

F = G*m1*m2/r^2

Newton's law of gravitation

Newton's law of gravitation proposes that every small element of matter, m, attracts every other small element of matter, m' with a force that is proportional to the product of the masses and inversely proportional to the square of the distance between them, so that at the small element level,

F = G*m*m'/r^2

As a result, we tend to fall toward the earth unless we are supported by upward forces, such as the floor that we are standing on.

An important simplification

It would be very difficult to deal with the attraction of gravity if we were required to deal with every object (including the earth) in terms of the many small elements of mass that make up the object.

Newton simplified that process by proving mathematically that the earth regarded as a sphere attracts a body as if the whole mass of the earth were concentrated at its center.

Almost parallel forces

The attraction of the earth on the elements of our body produces forces on those elements that are almost parallel. Those forces lie along lines that all end at a point that is almost 4000 miles beneath the surface of the earth.

At that distance, each pair of force lines forms the sides of a triangle with extremely long sides, an extremely small angle, and an extremely narrow base. As a practical matter, therefore, we can assume that those forces are parallel insofar as estimating the effect that the sum of all of those elemental forces have on our bodies.

The center of gravity C.G.

It can be proved that:

The resultant of the weights of all the elements of a rigid body passes through a certain fixed point commonly called its center of gravity, C.G., regardless of the orientation of that body relative to the earth.

For purposes of computations involving statics, the mass of a rigid body may be considered as if concentrated at it C.G., and its weight may be considered to center at this same point.

(Recall that the weight of an object is the force required to cause the mass of that object to fall toward the center of the earth with an acceleration of approximately 32.2 ft/s^2 or 9.81 m/s^2.)

A plumb-bob or plummet

A plumb-bob or a plummet is a weight, usually with a pointed tip on the bottom, that is suspended from a string and used as a vertical reference line, or plumb-line.

When the string is attached to an object that is fixed relative to the surface of the earth and the plummet is allowed to hang down, it will eventually (when it stops swinging) point directly at the center of mass of the earth.

Common uses for a plummet

The plummet is most commonly used in construction projects to ensure that constructions are "plumb", or vertical.

It is also used in land surveying to set a surveying instrument exactly over a fixed survey marker, or to transcribe positions onto the ground for placing a marker.

Experimental verification of the C.G.

I'm aware that as a blind student, you probably can't perform this experiment, but hopefully you can imagine it.

Make a hanger

Use a plummet to draw a vertical line on a flat, smooth, vertical surface (such as a wall). Drive a finishing nail (a nail with a small head) into the surface on the vertical line allowing a small portion of the nail to stick out. Cause the nail to be as horizontal as practical.

Make an irregular object

Cut an irregular shape out of a piece of flat cardboard and punch several holes around the perimeter of the shape.

Perform the experiment

Using several holes in succession, including holes on all four sides of the cardboard, hang the cardboard on the nail in such a way that the cardboard is free to swing and settle into a stable orientation.

Once the cardboard stops swinging, mark the point on the bottom of the cardboard that coincides with the vertical line that you drew on the vertical surface.

After you have done this several times, draw lines from each hole to the corresponding mark that you made on the opposite edge of the cardboard. All of these lines should cross at a single point. That point is the C.G. for that particular irregular shape of cardboard.

Modify the experiment

Cut regular shapes of cardboard such as rectangles, triangles, hexagons, octagons, etc, and repeat the experiment with those shapes. Observe the location of the C.G. for those shapes.

Rectangular shapes

You should find that the C.G. for a rectangular piece of cardboard is at the center of the rectangle. This should be the case for all rectangles ranging from squares to long skinny rectangles (although it may difficult to measure for long skinny rectangles.)

Triangular shapes

The C.G. for a triangular shape should be at the intersection of its medians.

In case you have forgotten, the median of a triangle is a line segment from a vertex of the triangle to the midpoint of the side opposite that vertex. Because there are three vertices, there are three medians. No matter what shape the triangle, all three medians intersect at a single point, which is called the centroid of the triangle. That point would also be the C.G. for a triangular-shaped piece of cardboard.

The doughnut hole

Cut a piece of cardboard in the general shape of a doughnut or tire with a hole in the middle and perform the experiment using that shape. Observe that the C.G. can be located in the doughnut hole. The C.G. of an object doesn't have to be in a solid part of the object.

A void in a 3D object

It is also possible for the C.G. to be located in a cavity in a three-dimensional object. For example, the C.G. for a small section of cylindrical pipe that is cut square on both ends would be in the center of the void half way between the ends of the pipe.

## Discussion and computations

### 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 Phy1120.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.

### Equilibrium: stable, unstable, and neutral

When a vertical line through the C.G. of a body falls within the area covered by the supporting base, the body can rest in equilibrium. A body in equilibrium can be further qualified as being either stable, unstable, or neutral.

• Stable equilibrium: If a body at rest receives a small displacement and tends to return to its former position, that body is said to be in stable equilibrium.
• Unstable equilibrium: If that body tends to move further away, then that body is said to be in unstable equilibrium.
• Neutral equilibrium: If it does neither, the body is said to be in neutral equilibrium.

A practical example

Consider the case of a round pencil, sharpened on one end and flat on the other end. If you lay the pencil flat on the table and give it a very small push, it will move and then stop. That pencil is said to be in neutral equilibrium. It doesn't tend to return to its original position, and except for rolling a small distance as a result of inertia, it doesn't tend to move further away.

Balance it on the flat end

If you balance the pencil on its flat end and give it a very small push at the other end, it will tip slightly and then return to its original position. This assumes that you don't push it beyond its tipping point, which we will discuss in more detail later.

The pencil in this configuration is said to be in stable equilibrium. The key to success is that you don't tip it so far that a vertical line through the C.G. moves outside the circle that defines the area of the supporting base of the pencil.

Balance it on its point

Assume that you rub the pencil lead back and forth a few times to slightly flatten the pointed end. If you were able to balance the pencil on that very small flattened surface, and then blow on the pencil very lightly, it would probably tip over and fall flat on the table. The smaller the flattened surface supporting the pencil, the easier it would be to cause it to tip over. The pencil in that state would be said to be in unstable equilibrium.

Stability is important

Stability is very important for many things such as boats and airplanes. Usually, the lower the C.G., the more stable will be the object.

A high-wire artist

For example, a high-wire artist at the circus may carry a long flexible pole, perpendicular to the wire held near the center of the pole at about thigh height. In this case, the supporting surface is the feet on the wire.

The flex in the pole allows the ends of the pole, and conceivably a large portion of the pole to hang lower than the soles of the feet. This has the effect of lowering the C.G. of the combination of the person and the pole considerably. If the C.G. is below the supporting surface, which in this case is the soles of the feet, the combination of the person and the pole will be very stable.

Cargo ships

The people that load cargo ships try to keep the C.G. of the loaded ship low in the hull of the ship. Sometimes they add extra weight called ballast very low in the hull to get the C.G. as low as possible.

Cargo aircraft

The people that load large military cargo aircraft are very careful how they load the cargo in order to control the location of the C.G. of the aircraft to maintain stability of the aircraft. The position of the C.G. relative to the axis of the wings is very important.

Children's toys

When my children were young, they had toys with a round bottom and something like a clown's head on the top. The toys had a heavy weight inside and very close to the bottom. This caused the C.G. to be very low in the toy. When the child pushed the toy over, it would return to an upright position. The toy was very stable.

### An exercise involving the tipping point

Let's see if we can demonstrate stability or the lack thereof with a thought experiment. Begin by drawing two rectangles on your graph board side by side. Make each rectangle 2 units wide and five units tall.

Two boxes

Pretend that these rectangles represent side views of two cardboard boxes with dimensions of 2x2x5 ft. Assume that each box contains identical heavy material at one end and the remainder of the box is filled with very light packing material. Assume that the heavy contents are at the bottom of the left box and at the top of the right box. (The boxes and their contents are just alike, but one is upside down relative to the other.)

Tactile graphics

The file named Phy1120a1.svg contains an image that represents this scenario. The image shows two boxes, as described above , tipped over to their tipping points .

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

Figure 3 shows the key-value pairs that go with the image in the file named Phy1120a1.svg.

Figure 3: Key-value pairs for the image in the file named Phy1120a1.svg.
Key-value pairs for the image in the file named Phy1120a1.svg.
 m: An exercise involving the tipping point n: Tipping point = 45 degrees o: Tipping point = 14 degrees p: Center of gravity q: Center of gravity r: 45 degrees s: 76 degrees t: W u: W v: File: Phy1120a1.svg w: Floor x: Box y: Box

Mark the C.G. of each box

Getting back to your drawing, assume that the C.G. of the left box is one foot up from the bottom and the C.G. of the right box is one foot down from the top (4 feet up from the bottom). In both cases, the C.G. is at the horizontal center of the box. Insert a pin at the location of the C.G. for both boxes and label it C.G. in both cases.

The weight of each box

The downward force that is the weight of each box is on a vertical line that goes through the C.G. and the total weight appears to be concentrated at the C.G. Therefore, when the box is setting flat on the floor, the weight vector for each box is on a line that is 1 ft. from either side of the box.

Vertical weight vectors

Begin at the C.G. for each box and draw a vertical line that extends 7 units down from the C.G. for each box. Label each line as W. These are weight vectors.

Both boxes are in equilibrium

We know from experience that both of these boxes are in equilibrium. We have also just learned that so long as the vertical weight vector that goes through the C.G. also goes through the two-foot wide supporting base of the box, the box will be in stable equilibrium.

What is the tipping point?

Label the bottom-right corner of each box as b. Pretend that you drive a nail immediately to the right of point b and allow it to protrude up from the floor so that when you push on the top-left corner of the box, it will rotate around point b instead of sliding to the right.

How far can we tip each box in a clockwise direction around the bottom right corner before we lose equilibrium?

Tactile graphics

The file named Phy1120b1.svg contains an image that shows the triangles discussed 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 5 shows a non-mirror-image version of the same image.

Figure 6 shows the key-value pairs that go with the image in the file named Phy1120b1.svg.

Figure 6: Key-value pairs for the image in the file named Phy1120b1.svg.
Key-value pairs for the image in the file named Phy1120b1.svg.
 m: An exercise involving the tipping point, part 2 n: K1 = 1.414 o: K2 = 4.123 p: Z1 = 1 q: Z2 = 4 r: R = 1 s: R = 1 t: 45 degrees u: 14 degrees v: Center of gravity w: Center of gravity x: b y: b z: Floor mm: W mn: W mo: File: Phy1120b1.svg

The lines K1 , K2, Z1, Z2, and R

To answer this question, begin by drawing a line from the C.G. to point b for each box. Label this line K1 for the left box and K2 for the right box.

Label the line segment that extends from the C.G. to the bottom of the left box as Z1 and label the line segment that extends from the C.G. to the bottom of the right box as Z2. Label the line segment that is the right half of the bottom of each box as R.

Right triangles

In the left box, the line segments K1, Z1, and R form a right triangle for which the lengths of two line segments are:

• Z1 = 1
• R = 1

In the right box, the line segments K2, Z2, and R form a right triangle for which the lengths of two line segments are:

• Z2 = 4
• R = 1

Compute the lengths of K1 and K2

Using the Pythagorean theorem and the Google calculator, we can compute the following lengths:

K1 = sqrt(1^2 + 1^2) = 1.414

K2 = sqrt(1^2 + 4^2) = 4.123

The sides of the right triangles

Next, we will compute the interior angle for each triangle at the C.G. vertex. Label the interior angle for the left box P1 and label the interior angle for the right box P2.

We know the lengths of all three sides of each triangle. There are a couple of ways that we can use trigonometry to find the interior angles. Considering the triangle from the viewpoint of the interior angle at the C.G. for the left triangle:

• hypotenuse = K1 = 1.414
• base = Z1 = 1
• opposite side = R = 1

For the right triangle,

• hypotenuse = K2 = 4.123
• base = Z2 = 4
• opposite side = R = 1

Compute the interior angles at each C.G. vertex

Using the Google calculator, we can compute the interior angle at the C.G. for each box as follows:

• P1 = arcsine(1/1.414)*180/PI = 45 degrees
• P2 = arcsine(1/4.123)*180/PI = 14 degrees

Tilt each box and observe the angles at the C.G. vertices

Now pretend that you exert a force on the top-left corner of each box causing it to rotate around the lower-right corner at point b. As you do that, the weight vector would rotate around the C.G. in a counter-clockwise direction (always pointing straight down) and the angle between the weight vector and the line K1 or K2 would decrease.

The tipping point

When the angle between the weight vector and the line K1 or K2 goes to zero degrees for either box, the tipping point for that box will have been reached. (See the image in the file named Phy1120a1.svg, or Figure 1 .)

At that exact point, the box will still be in equilibrium, but it will be in unstable equilibrium. Exerting a little more force to the right will cause the box to turn all the way over onto its side under its own weight. Relaxing the force a little when the box is in that position will allow it to rotate back towards its original position.

What are the tipping point angles?

The box on the left would need to rotate slightly more than 45 degrees clockwise around point b before the direction of the weight vector through the C.G. would move to the right of point b, causing the box to continue rotating under its own weight.

However, the box on the right would only need to rotate slightly more than 14 degrees clockwise around point b before the direction of the weight vector through the C.G. would move to the right of point b causing the box to continue rotating under its own weight.

Conclusion

Therefore, the box on the left is more stable than the box on the right. If you want to lessen the likelihood of an object tipping over, cause the C.G. of the object to be near the bottom of the object.

An exercise for the student

I will leave it as an exercise for the student to compute moments about point b to confirm the tipping angle for each box where the torque about point b changes from counter-clockwise to clockwise.

### A real-world example

Let me illustrate this situation with a real-world example that may seem familiar to you. Assume that you put some flowers with long stems in a lightweight plastic vase and set the vase on a table. The C.G. of the vase and the flowers would be relatively high, and it wouldn't take much of a sideways push to cause the vase to turn over.

Now assume that you add water to the vase until it is about half full. This would cause the C.G. of the vase, the water, and the flowers to move down, probably into the bottom half of the vase. This would make it more difficult to tip the vase over. Of course, when it does tip over, it would make a bigger mess than would be the case without the water.

## Do the computations

I encourage you to repeat the computations that I have performed in this module. Experiment with the computations, 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 -- Center of Gravity
• File: Phy1120.htm
• Revised: 07/09/2011
• Keywords:
• physics
• accessible
• blind
• graph board
• protractor
• refreshable Braille display
• JavaScript
• trigonometry
• force
• center of gravity
• gravitational constant

### 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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