1. Forging new generations of engineers
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Forging new generations of engineers

2. A lesson on understanding and evaluating moments of forces
My name is Wendy, and I’d like to welcome you to this lesson on moments.

3. In This Lesson What is a moment? How are moments calculated?
How are moments evaluated in dynamics problems?
How are moments evaluated in static equilibrium problems?
How do moments affect unconstrained objects?
In this lesson I will define the term moment ** and introduce the terminology associated with moments. We will look at the formula ** for moments and the units used in the calculations. We will solve some dynamics ** problems involving moments. Then we will solve some static equilibrium ** problems. Finally, we will preview how moments affect objects ** such as airplanes that are completely unconstrained.

4. What Is a Moment?
The moment or torque of a force is a measure of the tendency of the force to rotate the body upon which it acts about an axis.
FORCE
Recall that a force is a push or a pull on an object as a result of an interaction with another object. Depending on where the force is applied and whether the object is confined, the force may cause no motion, or purely translational motion, or it may result in a twisting or spinning motion. The moment ** of a force, which is also referred to as torque, is a measure of the force’s tendency to rotate the object about an axis or point. ** This wrench is confined by the head of the hex bolt, with its axis being the centerline of the threaded shaft of the bolt. The force applied to drive the bolt produces a measurable moment, ** and the wrench rotates about the axis of the bolt.

5. Terminology = F FORCE pivot = D distance lever arm
A review of the terminology associated with moments is necessary at this point. As we mentioned before, a moment is produced by a force **. For these problems, force will be denoted by the letter “F” **. A moment is measured with respect to a pivot point or axis **. The pivot is analogous to the fulcrum of a lever. Finally, a moment depends on the perpendicular length or distance ** from the force to the pivot. The distance is denoted by the letter “D” **. Notice I stated “perpendicular” distance. In this diagram the force is already perpendicular to the wrench, which is essentially a lever arm **. In cases when the force is not perpendicular, some trigonometry must be applied. We will show an example of that later in this lesson.
= D

6. Moment M M = F x D Formula for Moment = F FORCE pivot = D distance
Given the force F and the perpendicular distance of the force from the pivot D, the moment of the force, ** which will be denoted by the letter M, ** is equal to the force times the distance **. Note that the moment is directly proportional to both the force and the distance: if you increase either the force or the distance of the force from the pivot, then you will increase the moment of the force.
= D
Moment
M = F x D M

7. Units for Moments Force Distance Moment English Customary Pound (lb)
Foot (ft)
lb-ft SI
Newton (N)
Meter (m)
N-m
The unit of measure for the moment of a force is derived from its components. In the English customary system, the unit is the pound-foot. In the International System, or SI, moment is measured in Newton-meters.

8. + The Right Hand Rule counterclockwise
Moment is a vector quantity, meaning that it has both magnitude and direction or sense. The convention for the direction of the moment of a force uses what is called the “right hand rule”. Using your right hand, you can curl your fingers in the direction of the rotation produced by the force **. The thumb points in either the positive direction—which is upwards or towards you, or the negative direction—which is down or away from you. If the force produces a counterclockwise ** rotation, then the sense of the moment vector is positive.** A clockwise rotation produces a negative moment.

9. THUMB POINTS TOWARD YOU
Right Hand Rule
FORCE
The problems we will be solving involve 2-dimensional systems. The right hand rule is a 3-dimensional concept and requires a little practice. The force on this wrench ** is spinning the bolt in a counter-clockwise direction. Place your right hand ** so that your fingers curl counter-clockwise. Notice ** that your thumb points toward you. This direction corresponds with the positive-z ** direction in 3-dimensional space and is considered a positive moment.
THUMB POINTS TOWARD YOU
POSITIVE

10. THUMB POINTS AWAY FROM YOU
Right Hand Rule
THUMB POINTS AWAY FROM YOU
FORCE
This wrench ** is spinning the bolt in the clockwise direction. Place your right hand ** so that your fingers curl clockwise. Notice ** that your thumb points away from you. This direction corresponds with the negative-z ** direction in 3-dimensional space and is considered a negative moment.
We will be applying these conventions in the following problems.
NEGATIVE

11. Moment Calculations ¯ Wrench FORCE F = 20 lb D = 9 in. M = -(F x D)
***Use the right hand rule to determine positive and negative.
D = 9 in. = .75 ft
M = -(20 lb x .75 ft)
M = -15 lb-ft
(15 lb-ft clockwise)
D = 9 in.
This first set of moment problems involves dynamics, which studies how bodies and systems move in reaction to forces acting upon them. In a system which is not fully constrained, moments result in rotational motion. We see this when we open a book or a door, brush our teeth, click on a computer mouse, drive a car—nearly every motion we make involves some rotation. The wrench, loosening or tightening threaded fasteners, is one of the most universal tools, because it translates force so efficiently into rotational motion.
In this first problem, a wrench is used to tighten a bolt with a 20 pound ** force applied 9 ** inches from the pivot point. First write down the formula: Moment M ** equals the negative of the product of the force F and the distance D. The negative sign is applied to the product in this case because I used the right hand rule ** to determine the direction of the moment. Next, ensure that the units are compatible with the units used in moment problems. In this case, the distance is expressed in inches, whereas the customary units for distance is feet. The distance of 9 inches ** is equivalent to .75 feet. Now substitute the values ** for force and distance into the formula. The moment ** resulting from the 20 pound force is negative 15 pound-feet, or 15 pound feet in the clockwise direction.

12. Moment Calculations ¯ Longer Wrench FORCE F = 20 lb D = 1 ft
M = -(F x D)
M = -(20 lb x 1 ft)
M = -20 lb-ft
D = 1 ft
Let’s look at the effect of using a longer wrench. The same force of 20 pounds is applied to the end of the wrench that is 12 inches or 1 foot ** long. We solve the problem in the same manner. First, write the formula. Again, the moment is clockwise or negative. Substitute the force and the distance values **, and solve **. The magnitude of the moment produced by the same force of 20 pounds has increased from 15 pound-feet to 20 pound-feet, just by increasing the length of the wrench. In fact, it’s a proportional increase: the length increased by one third, and the moment increased by one third. The lesson is clear: choose the largest wrench available for the job!

13. Moment Calculations ¯ L-shaped Wrench FORCE F = 20 lb
D = 3 in. = .25 ft
M = -(F x D)
M = -(20 lb x .25 ft)
M = -5 lb-ft
3 in.
This wrench has an interesting L-shape. The wrench designer thought that this might be a more comfortable way to grip the wrench to drive the bolt. The force is the same twenty pounds, applied to the end of the L. For moment problems we must use the perpendicular distance ** of the force from the pivot point. In this case, the perpendicular distance is 3 inches ** or .25 feet **. The calculation follows the same process. Write down the formula **, taking into account that the moment is clockwise, or negative. Substitute the values for force and distance **. The calculated moment ** for this unusual wrench is only 5 pound-feet in a clockwise direction. By adding the L to the wrench, the ability of the wrench to torque the bolt was drastically reduced. Not such a good idea after all.

14. Moment Calculations ¯ FORCE Offset Wrench F = 20 lb
D = 8 in in. = 1.5 ft
M = -(F x D)
M = -(20 lb x 1.5 ft)
M = -30 lb-ft
8 in.
Tight or crowded situations sometimes require that an offset wrench be used to access a bolt. How much will the offset affect the moment produced by the force? The geometry of this offset wrench ** places an 8 inch right angle segment 8 inches from the pivot point, and the wrench handle projects out another 10 inches after the offset. The 20 pound force is at the end of the 10 inch segment. Remember, in moment calculations, we use the perpendicular distance ** of the force from the pivot point. In this case, the perpendicular distance is the sum of the horizontal dimensions: 8 inches plus 10 inches. This converts ** to 1.5 feet. The process is the same: Formula **, substitute **, and solve. ** The total moment produced by the force ** is 30 pound-feet clockwise. Most important to note in this problem is that the 8 inch offset had no effect on the moment. A straight 18 inch wrench will produce the same moment as this offset wrench.
8 in.
10 in.

15. Moment Calculations Wheel and Axle + D = r = 50 cm = .5 m M = F x D
***Use the right hand rule to determine positive and negative.
M = 100 N x .5 m
M = 50 N-m +
The wheel and axle, another simple machine, provides another opportunity for evaluating moments. This ship’s wheel rotates counterclockwise as the result of a 100 N force applied as shown. Notice in this problem we are using metric units. The radius of this ship’s wheel is 50 cm. Recall that the metric units for moment problems are Newtons and meters. The radius of the wheel is also the perpendicular distance of the force from the pivot point. The 50 cm distance ** converts to .5 meters. The solution process once more includes first writing the formula. For this wheel, the right hand rule ** indicates that the moment is positive. Substitute the force and distance values **, and then solve the equation. ** The 100 N force produces a 50 N-m moment.
F = 100 N

16. Moment Calculations Wheel and Axle Fy = Fsin50° = (100 N)(.766)
r = 50 cm
Fy = Fsin50° = (100 N)(.766)
Fy = 76.6 N
D = r = 50 cm = .5 m
M = Fy x D
M = 76.6 N x .5 m
M = 38.3 N-m Fx
A casual sailor might apply the 100 N force ** at an angle other than tangent to the wheel. How will this affect the moment? The force we see in this problem is 50 degrees ** from the horizontal. The wheel radius is still 50 cm. The force shown can be broken down into its x ** and y components. Since the x component is in line with the pivot point, its perpendicular distance from the pivot point is zero. Therefore the magnitude of its moment is zero. The y component of the force vector has a perpendicular distance from the pivot point which is equal to the radius of the wheel. It is this y component of the force that we will use for the moment calculation. Using trigonometry, the y force is the side opposite the 50 degree angle. ** Thus Fy equals the force times the sine of 50 degrees.** Solving, Fy ** equals 76.6 N. Again, the distance is .5 meters. ** The formula is M ** equals Fy times D, substitutions are made, ** and the moment of the force applied at an angle of 50 degrees is 38.3 N-m. This is significantly less than the 50 N-m provided by a force applied tangentially.
50o Fy
50o
F = 100 N

17. What Is Equilibrium?
The state of a body or physical system at rest or in unaccelerated motion in which the resultant of all forces acting on it is zero. The sum of all moments about any point or axis is zero.
ΣM = 0
M1 + M2 + M = 0
The problems solved so far involved the force causing the object to rotate feely around the pivot. The upcoming problems will examine objects that tend to remain at rest because all of the forces and moments acting on them are balanced, meaning they cancel each other out. These objects are said to be in equilibrium.** Engineers are very interested in objects in equilibrium, as solutions to these types of problems allow engineers to further study the sizes and types of materials suitable for the different applications. In this lesson we will use the fact that in equilibrium the sum of all moments about ANY point is equal to zero. The formula is written sigma M ** equals zero. Sigma is the Greek letter meaning sum of all. Sigma M indicates that all of the moments for the system being evaluated are added together, taking into account the positive or negative sense of each moment. The sum of all of the moments equals zero.

18. Moment Calculations See-Saw
The see-saw is familiar to most people who have spent time at a playground. See-saws work very well when two bodies of the same size want to gently go up and down—two bodies of the same size balance each other, causing the sum of the moments to equal zero. However, when one body is considerably larger than the other, ** sitting at the end of the see-saw will cause a definitely unbalanced situation. Riders are not able to have an enjoyable see-saw ride when this happens.

19. Moment Calculations ¯ See-Saw + ΣM = 0 M1 + (–M2) = 0
***Use the right hand rule to determine positive and negative.
M1 = M2
F1 x D1 = F2 x D2
25 lb x 4 ft = 40 lb x D2
100 lb-ft = 40 lb x D2
See-Saw
F2 = 40 lb
F1 = 25 lb +
By moving the larger body closer to the pivot point, this can cause the moments on the two sides to balance each other, bringing the system into a state of equilibrium. The question is where should the larger body sit? We can use the formula for equilibrium of moments to solve this problem. Body one provides a downward force F1 ** of 25 pounds due to its weight. Body two provides a downward force F2 ** of 40 pounds. F1 is located 4 feet ** from the pivot point, while F2 is seated at some position closer than 4 feet from the pivot. The formula is sigma M equals zero **. Substitute for sigma M all of the moments for this system acting about the pivot **. M1 is positive because it is rotating the see-saw in a counter-clockwise direction. I check this using the right hand rule: ** as I curl my fingers around in the direction of the rotation caused by F1, my thumb is pointing toward me in a positive z direction. Likewise, M2 is negative as it tends to rotate the see-saw in a clockwise direction. I confirm that my right thumb points away from me when I curl my fingers in the direction of rotation. I can rearrange the terms to read that M1 equals M2 **. I then substitute the moment formula: ** M1 is replaced by F1 times D1, and M2 is replaced by F2 times D2. I then substitute the actual numerical values. ** Substitute 25 pounds for F1, 4 feet for D1, and 40 pounds for F2. D2 remains the unknown for which we are solving. Moment M1 is calculated to be 100 ** pound-feet. We can then divide both sides ** by 40 pounds to isolate the D2 term. The 40 pounds in the right numerator ** cancels the 40 pounds in the right denominator, and the pounds unit in the left numerator cancels the pounds unit in the left denominator. The final calculation, 100 divided by 40 ft, results in D2 equal to 2.5 feet.
When I get to the end of a solution, I take a step back and see if my result makes sense. By doing this, many times I’ve caught myself with a decimal place error or some other trivial mistake. Does 2.5 feet make sense? Yes, it does, because we initially assumed that the heavier body would be seated closer to the pivot.
40 lb lb
D1 = 4 ft
D2 = ? ft
2.5 ft = D2

20. Moment Calculations Loaded Beam C 10 ft 10 ft A B
Select the pivot location A. Solve for RB.
Loaded Beam
ΣM = 0
MB + (–MC) = 0
MB = MC
RB x DAB = FC x DAC
RB x 10 ft = 35 lb x 3 ft
RB x 10 ft = 105 lb-ft
DAB = 10 ft
DAC= 3 ft C
10 ft ft
This problem involves moment calculations applied to a static beam that does not rotate. The beam is perched upon simple supports at both ends, labeled A and B. ** An object is placed on the beam ** and labeled C. The object has a weight of 35 pounds, which exerts force FC ** on the beam. For this problem the beam itself is considered massless. The length of the beam, DAB ** is 10 feet, and the object is located at distance DAC ** which is 3 feet from A. Completing the free body diagram of the beam, replace the simple supports with their reaction forces, ** RA and RB. We will use the equilibrium and moment formulas to solve for the unknown reaction forces, RA and RB.
For all moment problems, we get to select the pivot point. By selecting the pivot to be located at one of the unknown forces, we reduce the number of variables in our equation. Let’s select the pivot to be at A **. The moment of RA is zero since RA acts through the pivot. We can now solve for RB.
The procedure we follow for solving this problem is much like the last problem. After writing the formulas, **substitute the known values, DAB, FC, and DAC ** and solve the moment on the right side of the equation. ** Divide each side of the equation by 10 ft ** to isolate the variable RB. Cancel out 10 ft on the left side of the equation ** and the feet units on the right side. The reaction force at B is 10.5 pounds **.
Does this make sense? I would expect RB to be sharing the 35 pounds with RA, and I would also expect it to have less than half the load, since the object is closer to A. So, yes, this makes sense.
In this simple problem, we can see that RA and RB acting upward support the 35 pounds acting downward. Since the sum of the forces in the y direction equals zero in equilibrium, we can solve for RA ** and find that it equals 24.5 pounds. A B
RB = 10.5 lb
RA + RB = 35 lb
RA = 35 lb – 10.5 lb = 24.5 lb
FC = 35 lb RA RB

21. Moment Calculations Truss FB = 500 lb B RAX A C D RAY Fc = 600 lb RDY
Replace the pinned and rolling supports with reaction forces.
12 ft
RAX A
It is common for engineers to calculate the loads on a truss as the first step to determining material types and sizes that will support expected loads. Notice the support on the left is a pinned support, which allows no movement in either the x or y directions. The support on the right is a rolling support, which allows movement in the x direction, but not in the y direction. This is really important as truss structures, like bridges, undergo expansion and contraction with temperature changes.
This truss is loaded with 600 pounds downward ** at point C, called FC, and a horizontal force ** of 500 pounds, at point B called FB. To complete the free body diagram of the truss, replace the pinned support with reaction forces RAX and RAY, and the rolling support with reaction force RDY.
Now we are ready to apply the equilibrium and moment formulas.
24 ft C
8 ft D
DAC = 24 ft
DCD = 8 ft
DCB = 12 ft
DAD = 32 ft
RAY
Fc = 600 lb
RDY

22. Moment Calculations Truss FB = 500 lb B RAX A C D RAY Fc = 600 lb RDY
Select the pivot at A. Solve for RDY.
Truss
ΣM = 0
MD – MB – MC = 0
MD = MB + MC
RDY x DAD = (FB x DCB) + (FC x DAC)
RDY x 32 ft = (500 lb x 12 ft) (600 lb x 24 ft)
RDY x 32 ft = 6000 lb-ft lb-ft
RDY x 32 ft = lb-ft
FB = 500 lb B
12 ft
12 ft
RAX A
Select the pivot to be located at A. This way, two of the five forces act through the pivot and produce no moment. Add the three other moments, ** noting that MD is a positive moment while MB and MC are both negative moments. Rearranging the terms ** eliminates the negative values. We can see the perpendicular distances of the forces at C and at D from the pivot are 24 feet and 32 feet respectively. But what is the perpendicular distance of the force at B from the pivot? I can extend the line of FB back to construct the perpendicular. ** Notice that the perpendicular distance from the pivot to FB is equal to the length of member CB, so we will call it DCB. Now for each of the moments, substitute the moment formula ** force times distance. Substitute the values for our known quantities ** and solve. ** Isolate RDY by dividing both sides by 32 feet and cancelling units. The result for RDY is pounds. Does the answer make sense? The force up at D is greater than the force down at C. That means the support at D is carrying the entire 600 pound load at C, and then some. That extra force is due to the rotation produced by the 500 pound force at B. The force at B is pulling up on the support at A and pushing down on the support at D through the truss members. Evaluating horizontal loads at the tops of structures relates to studying how the force of strong winds affect the design of the base supports of a bridge or building.
24 ft C
8 ft D
32 ft ft
DAC = 24 ft
DCD = 8 ft
DCB = 12 ft
DAD = 32 ft
RDY = lb
RAY
Fc = 600 lb
RDY

23. Moments on An Airplane RUDDER Yaw ELEVATORS Pitch AILERON Roll
We have examined moments causing rotation about a point or axis and moments that do not cause rotation but do affect static equilibrium. Finally, we will briefly discuss how moments affect an object that is not constrained in any way—specifically how moments act on an airplane.
Aircraft have several features called control surfaces, which are movable flaps located on the wings and tail. These surfaces have the job of controlling moments caused by the force of air on the wings and tail of the airplane. By controlling the moments of these forces, pilots can steer and maneuver aircraft in all directions. The motion caused by these moments—the roll, pitch, and yaw—rotates around three mutually perpendicular axes that meet at the center of gravity. As you can imagine, the interplay of all of the forces and moments is quite complex and is a field of study in itself.

24. Translation and Rotation
Moments on an Airplane
PATH WITH MOMENTS
Translation and Rotation
PATH WITHOUT MOMENTS
Pure Translation
Moments affect the flight of aircraft ** by adding rotational components to the straight-ahead translation, providing the ability to dive, bank, roll, and to perform other complex maneuvers.

25. References Halpern, A.M. (1988). Schaum’s 3000 solved problems in
physics. New York, NY: McGraw-Hill.
NASA. (n.d.). The beginners' guide to aeronautics. Retrieved
June 11, 2008, from
12/airplane/
Nave, C.R. (2005). HyperPhysics. Retrieved June 12, 2008,
from
National Institute of Standards and Technology. (2000). The
NIST reference on constants, units and uncertainty.
Retrieved June 11, 2008, from
Moments are present in most physical systems, from transportation systems like aircraft, to construction systems like bridges, to biological systems like skeletal structures. Calculating moments and the effect of moments is a key skill in many aspects of mechanical, civil, and bioengineering.

26. Credits Writer: Wendy DeMane Content Editor: Wes Terrell
Production Work: CJ Amarosa
We would like to thank you for joining us for this lesson on moments. We hope it was helpful and your Virtual Academy learning experience enjoyable.