chapter 7 moments - purdue university · to hold the wrench as far away from the bolt as possible....

Chapter 7: Moments 7-1

Chapter 7

Moments

“Chop your own wood, and it will warm you twice.”

∼ Henry Ford

7.1 Overview

In contrast to the dot product which takes two vectors and forms a scalar, the cross producttakes two vectors and produces another vector. The cross product has a number of differentapplications, but in solid mechanics we use the cross product primarily to calculate moments.Most students have been exposed to moments in one form or another. The idea of a wrenchis a good one because it’s clear that when you are loosening or tightening a bolt, you wantto hold the wrench as far away from the bolt as possible. Intuitively we understand that thefarther away you apply the force, the less force you need to twist the bolt. Later on we willsee how moments explain the mechanical disadvantage of your musculoskeletal system, whyhelicopters are hard to control, and the limitations of levee systems.

7-2 Chapter 7: Moments

Before we go too far, let us specifically define what we mean by a moment due to a force.While we usually think of a force as something pushes or pulls, it is also important toremember that forces can cause things to rotate. So the moment, ~M due to a force is definedas,

~M = ~r × ~F , (7.1)

where ~r is the position vector from the point you are interested in to a point on the lineof action of the force, ~F is the force.

You must remember that

1. Moments are vectors - they have direction and magnitude!

2. The position vector must point from the point you are interested in to a point on theline of action of your force.

FBIC

θ

d

L

b

Figure 7.1: Here is an example of an arm holding a weight. The weight causes a momentabout the elbow as does the force in the biceps muscle. Interestingly, they will cause momentsin opposite directions. In order to calculate the moment due to the weight about the elbow,you must write out the positive vector from the elbow to the weight, ~rEW , and executethe cross product with the vector that represents the force due to the weight. Similarly, tocalculate the moment due to the biceps muscle, you have to write out the position vectorfrom the elbow to ANY point on the line of action of the muscle force, ~rEB and cross it withthe force due to the biceps muscle.


After completing this chapter, the student will be able to:

1. Calculate a cross product in two and three dimensions (take two vectors and create athird vector),

2. Quickly calculate moments in two and three dimensions (of all the concepts in thischapter, this will require the most practice), and

3. Be able to mathematically describe equivalent force-moment systems.

7.2 Calculating Cross Products Quickly and Accurately

Calculating moments quickly and accurately is important for most of what we will do inthis course. For many students this is one of the first topics that requires considerable studyand repetition. For now, let’s calculate some cross products and build up our intuition formoments.

Since the cross product results in a vector, we have to consider both its magnitude anddirection. The magnitude of ~a × ~b is defined as the magnitude of ~a multiplied by themagnitude of ~b and the sine of the angle between,

|~a×~b| = (a)(b)sin(θ). (7.2)

The direction of the resultant vector is perpendicular to both ~a and ~b. So if you make aplane with the two vectors, you know that the resultant will be perpendicular. The questionis whether it points above the plane or below it. In order to figure this out, we use the “righthand rule” which requires us to place our hand over the first vector, in this case ~a with ourpalm exposed to the second vector. If you curl your fingers toward the second vector, yourthumb will be pointing in the direction of the resultant.

If that was not easy to visualize, you are in luck. The basis vectors, ~i, ~j, and ~k come tothe rescue again. If we apply our definition to every unit vector combination, we have ninepossibilities. It should be clear that ~i×~i = ~j ×~j = ~k × ~k = ~0 because the angle between avector and itself is always 0 and sin(0o) is 0. For the other combinations we have,


~i×~j =~j ×~i =~i× ~k =~k ×~i =~j × ~k =~k ×~j =

Now, if we want to take the cross product of two arbitrary vectors, there are three waysto do it. Actually there are probably more than three ways, but we will only discuss threeways here. For the first two, we will utilize the calculated cross products in equation 7.3 tohelp us. First, is the brute force method. We start by writing out the components of bothvectors,

~a×~b = (ax~i+ ay~j + az~k)× (bx~i+ by~j + bz~k), (7.3)

we can then look at all the possible combinations and evaluate them using them using theexpressions in equation 7.3,

~a×~b = (ax~i× bx~i) + (ax~i× by~j) + (ax~i× bz~k) + (ay~j × bx~i) + (ay~j × by~j)+(ay~j × bz~k) + (az~k × bx~i) + (az~k × by~j) + (az~k × bz~k).

Reducing these cross products term-by-term yields,

~a×~b = ~0 + axby~k − axbz~j − aybx~k +~0 + aybz~i+ azbx~j − azby~i+~0, (7.4)

and collecting terms yields,

~a×~b = (aybz − azby)~i+ (azbx − axbz)~j + (axby − aybx)~k. (7.5)

The second way to calculate cross products of vectors in three dimensions is to use thedeterminant method,

~a×~b =

∣∣∣∣∣∣~i ~j ~kax ay azbx by bz

∣∣∣∣∣∣ , (7.6)

which can be broken up in the following way,


~a×~b =

∣∣∣∣∣∣~i ~j ~kax ay azbx by bz

∣∣∣∣∣∣ =~i

∣∣∣∣ay azby bz

∣∣∣∣−~j ∣∣∣∣ax azbx bz

∣∣∣∣+ ~k

∣∣∣∣ax aybx by

∣∣∣∣ . (7.7)

This leads directly to the same expression found in equation 7.5. Calculating cross productscan be tedious, but it is a very important thing to be able to do quickly and accurately.For three dimensional vectors, we recommend one of these two methods. It does not reallymatter which one you choose, but you should test them out, time yourself, and make surethat you can always write them out quickly and accurately .

The third method is generally the most useful for two dimensional problems, where thevectors all lie in a plane (one component in each vector is always 0). You can still use thefirst two methods described above, but the third method, finding a perpendicular distance,is sometimes quicker. Consider the example shown below which illustrates the muscle forceexerted by the biceps brachii when carrying a weight in the hand. If you place a coordinatesystem at the center of the elbow, you can calculate the moment that the biceps muscleexerts by drawing a position vector from the elbow center to the attachment point of themuscle (using the distance, d) and calculating it as described above. Alternatively, one canmeasure the perpendicular distance from the center of the elbow to the line of action ofthe force, b. In this case, the magnitude of the moment will be bFBIC . How you label thedirection depends on how you choose your coordinate system.

FBIC

θ

d

L

b


Example: Gymnastics

The figure below shows a gymnast, executing an iron cross. This maneuver is an excellentexample of an athlete in static equilibrium who is subjecting his shoulders to extraordinaryforces and moments. Calculating these reactions will allow us to minimize injuries anddevelop training regimens that better prepare elite athletes for international competition.

Figure 7.2: An iron cross. The gymnast’s arms make an angle of approximately 15o withthe horizontal and the ropes make an angle of 83o with the horizontal.

First, find the force in the cables holding the gymnast up. Treat him as a particle with aweight of 154 lbs. and use the equations of equilibrium to find the tension in the cables.

In order to build our intuition about moments, we will compute the moment that therings generate about the gymnast’s shoulder. First, determine the position vector from theshoulder to any point on the line of action of the force, then write out the force in vectorform and execute the cross product. You may assume that the distance from the shoulderto the ring is approximately 26 inches.

How does this moment compare to the moment generated about your shoulder whileholding a 10 lbs. weight as far from your body as possible?

Why are Olympic level gymnasts rarely taller than 5 ft. 6 inches?


Example: Robot Arms

Shown below is the European Robot Arm developed for use with the Space Shuttle Programand the International Space Station.

We will use a simplified model of a planar robotic arm to begin with.

F

θ1

θ3

A

B

θ2

C

D

Figure 7.3: A simplified model of a planar robotic arm. You may assume that each segmenthas length 1.5 m, and the applied force is 250 N. θ1 = 35oandθ2 = θ3 = 20o

Calculate the moment that the force, F , exerts about the point A.

What is the worst case scenario for the moment that F can exert about point A?


Example: 3D Bent Bar

Calculate the moment that the force, F , exerts about point A.

The distance, a is 3 m, b = 1.5m, c = 2m, and d = 5m. You may assume that the force,F , is 900 N.

A

B

C

F

a

d

c

b

z

y

x


Example: Tower Crane

Given: The Tower crane revisited once more. This tower crane has the same dimensionsas the one you encountered previously. You may assume that a = 22m, b = 65m, c = 10m,and d = 5m. You may use the force you calculated previously for the tension in the cable.

Find:

1. The moment that the cable force pulling on the load exerts about the base of the crane,C.

a

b

c

d

A

B

Cx

z

y


7.3 Force Couples

The methods described above will always work for calculating moments, but there are timeswhen something special happens. If you look at Fig. 7.4 below, you see a very simple object;a rectangle with two forces applied to it. As shown, the forces are equal and opposite sothey should cause no acceleration of the object’s center of mass. They do have an effect,however, in that they will cause the object to rotate.

c

a d

F

F

O

A

B

a

x

y

Figure 7.4: A free body diagram of a simple box with two equal and opposite forces appliedto it.

Briefly, calculate the moment caused by the two forces about the center of the object,

∑~M/O =

and do it again for the lower left corner,

∑~M/A =

and finally, try it one more time about the upper right corner,

∑~M/B =

where it should be noted that you should obtain the same answer for each of the aforemen-tioned calculations, (Fd)(−~k) = −Fd~k.

This is an interesting result because it implies that the moment caused by the two forcesdepends only on their magnitude and the perpendicular distance between them. To reiterate,any time you have equal and opposite forces, each with magnitude, F , separated by a


perpendicular distance, d (more commonly known as a force couple), the magnitude of themoment they generate is FD. In fact, we often go a step further and assert that a forcecouple can be replaced by a concentrated moment, (Fig. 7.5)1. When modeling our objectas a rigid body, it does not actually matter where we put the moment (as evidenced by thefact that it has the same result no matter which point we sum the moments about).

c

a d

F

F

O

A

B

a

c

a d

M=Fd

O

A

B

a

=

Figure 7.5: A free body diagram of a simple box with two equal and opposite forces appliedto it and the equivalent representation of the force couple as a single applied moment.

This is the first example of what we call equivalent systems, such as the two shown in Fig.7.5.

7.4 Equivalent Systems

The concept of equivalent systems is a very powerful one in mechanics. Moreover, introducingit here should pose few difficulties for students because expertise in working with vectors andcalculating moments has already been established. The concept is quite simple. We considerone set of forces and moments and replace them with a second set of forces and momentsthat has the same net effect. The only real difficulty is distinguishing between equivalentsystems and equilibrium. It is a subtle distinction, but for equivalent systems, we just wantto simplify a system of forces and moments. We may not consider all the reactions acting onthe object and the net effect is often different from zero. For equilibrium, we must considerall the forces and moments and they must add up to zero.

In truth, we have already touched on this concept in the last section when we discussedforce couples. There we replaced two equal and opposite forces offset by some distance,d with a single concentrated moment. One thing to keep in mind is that you must draw

1Realistically, any time you see a problem where there is a concentrated moment, it is really a forcecouple. We just use the moment notation to simplify things a little, especially when we know there is a forcecouple, but we do not necessarily know the perpendicular distance between the two forces.


a separate diagram for each system. That is very important because trying to put bothsystems on the same diagram can only lead to confusion and catastrophe.

Let’s begin with an illustration of equivalent systems.


Example: Bent Bar

Consider the example below. System I consists of four forces and a concentrated moment2

and System II consists of a force located at point B and a different concentrated moment.If the force in System II has a magnitude of 300 lbs. and the concentrated moment has amagnitude of 600 ft.*lbs., determine the magnitudes of F1, F2, and F3.

For System I we have,

∑Fx = F − F1 = 200lbs.− F1∑Fy = F3 − F2∑

M/B = (2ft.)F1 + (4ft.)F3 + (2ft.)F2 −M = (2ft.)F1 + (4ft.)F3 − 500ft.lbs.

For System II we have,

2Recall that a concentrated moment is really a force couple.


∑Fx = Rcos(53.13o)∑Fy = Rsin(53.13o)∑

M/B = 600ft.lbs.

In order to make the two systems equivalent, the forces in the x and y direction must bethe same3 and the net moments must be the same. This leads us to,

∑F1 = 20lbs.∑F3 = 265lbs.∑F2 = 25lbs.

3N.B. the system may or may not be in equilibrium


Example: Cantilevered Beam

Here is a cantilevered beam with a system of forces and moments on it. What force andmoment must act at point A in order to produce an equivalent system of forces (not includingthe reactions)? What about points B, C, and D?

The beam has a length, L = 3.5m, a = 0.25m, b = 0.75m, c = 1.5m, and d = 0.5m. Thevalue of F1 = 100N, F2 = 200N, F3 = 300N.

c

a d

F

F

O

A

B

a

1

1

C

F2

F3

x

y

b

D


Example: Wind Turbine

Here is our wind turbine again. Replace the wind forces on the blades with an equivalentforce and concentrated moment at the origin (the center of the actual turbine). We mayassume that the thrust force on each blade (FD) is approximately 35 N (in the −x direction).The lift force (FL) causing each blade to rotate is approximately 200N. You may assume thatthe forces act at the center of each 20 m blade. These forces are all in the y − z plane. Inaddition, d1 = 65 m, d2 = 5 m.

FD

x

y

z

FL

FDFD

FL

FL

2d

1d


Example: Simple Bars

For each bar, replace the forces and moment shown with an equivalent force and momentlocated at the left end of the beam.

12 kN

12 kN

4 m

12 kN*m

1 kN

6 kN12 kN

3 kN

4 kN

2 kN*m

4

3

5

2 kN


Example: Current Turbine - 3D Force Couples

Below is an example of a “wind” turbine modified for use under water. This so-calledcurrent turbine is potentially able to generate large amounts of power while dissipating onlya fraction of the river’s momentum.

Fcurrent

If the drag and lift forces act at the middle of each blade (assume each blade is 5 m long)and are 200 N and 800 N, respectively. If the rotor on the left turns clockwise, calculate thetotal moment about the center of the rotor.

If the rotor centers are 14 m apart, sit approximately 7 m above the floor, and are 1.5 min front of the main support, calculate the total moment about the base of the main supportdue to all the forces acting on the blades.

chapter 7 moments - purdue university · to hold the wrench as far away from the bolt as possible....

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