Dynamics of Mechanical Systems 2009 Part 5 pps

Then, from the definition of equivalent force systems, W and T are: 6.8.8 and 6.8.9 where k is a downward-directed unit vector as in Figure 6.8.4 and M is the total mass of particles of

182 Dynamics of Mechanical Systems

By eliminating S A and S B between these three equations, we have:

(6.7.13)

Now, suppose that A is represented by a single force, say FA, passing through some

common point C of A and B (or A and B extended) together with a couple with torque

TA Similarly, let B be represented by a single force FB passing through C together with

a couple with torque TB Then, because A and B taken together form a zero system (Eq.(6.7.3)), the resultant of A and B and the moment of A and B about C must be zero.

That is,

(6.7.14)and

(6.7.15)

Equations (6.7.13), (6.7.14), and (6.7.15), or the equivalent wording, represent the law of action and reaction.

Consider a particle P with mass m (or, alternatively, a point P with associated mass m) as

depicted in Figure 6.8.1 Let O be an arbitrary reference point, and let p be a position

vector locating P relative to O The first moment of P relative to O, φ P/O, is defined as:

Specifically, if G is the mass center and if r i locates pi relative to G, as in Figure 6.8.3, then the first moment of S relative to G may be expressed as:

(6.8.3)

S A+S B= 0ˆ

φS G

i i i

N m

=

10

From Figure 6.8.3, we have:

(6.8.4)Hence, by substituting into Eq (6.8.3), we obtain:

N

i i G i

N

i N i

N

i N i

1 1

184 Dynamics of Mechanical Systems

Equation (6.8.6) demonstrates the existence of G by providing an algorithm for its location.

We may think of a body as though it were composed of particles, just as a sandstone iscomposed of particles of sand Then, the sums in Eqs (6.8.2) through (6.8.7) become verylarge, and in the limit they may be replaced by integrals

For homogeneous bodies, the mass is uniformly distributed throughout the region, orgeometric figure, occupied by the body The mass center location is then solely determined

by the shape of the figure of the body The mass center position is then said to be at the

centroid of the geometric figure of the body The centroid location for common and simple

geometric figures may be determined by routine integration The results of such tions are listed in figurative form in Appendix I As the name implies, a centroid is at theintuitive center or middle of a figure

integra-As an illustration of these concepts, consider the gravitational forces acting on a body

B with an arbitrary shape Let B be composed of N particles Pi having masses mi (i = 1,…, N), and let G be the mass center of B, as depicted in Figure 6.8.4.

Let the set of all the gravitational forces acting on B be replaced by a single force W passing through G together with a couple with torque T Then, from the definition of

equivalent force systems, W and T are:

(6.8.8)

and

(6.8.9)

where k is a downward-directed unit vector as in Figure 6.8.4 and M is the total mass of

particles of B The last equality of Eq (6.8.9) follows from the definition of the mass center

in Eq (6.8.3)

Equations (6.8.7) and (6.8.8) show how dramatic the reduction in forces can be throughthe use of equivalent force systems

Inertia forces arise due to acceleration of particles and their masses Specifically, the inertiaforce on a particle is proportional to both the mass of the particle and the acceleration of

N

0

the particle However, the inertia force is directed opposite to the acceleration Thus, if P

is a particle with mass m and with acceleration a in an inertial reference frame R, then the

inertia force F* exerted on P is:

(6.9.1)

An inertial reference frame is defined as a reference frame in which Newton’s laws of

motion are valid From elementary physics, we recall that from Newton’s laws we havethe expression:

(6.9.2)

where F represents the resultant of all applied forces on a particle P having mass m and

acceleration a By comparing Eqs (6.9.1) and (6.9.2), we have:

(6.9.3)

Equation (6.9.3) is often referred to as d’Alembert’s principle That is, the sum of the

applied and inertia forces on a particle is zero Equation (6.9.3) thus also presents analgorithm or procedure for the analysis of dynamic systems as though they were staticsystems

For rigid bodies, considered as sets of particles, the inertia force system is somewhatmore involved than for that of a single particle due to the large number of particles making

up a rigid body; however, we can accommodate the resulting large number of inertiaforces by using equivalent force systems, as discussed in Section 6.5 To do this, consider

the representation of a rigid body as a set of particles as depicted in Figure 6.9.1 As B moves in an inertial frame R, the particles of B will be accelerated and thus experience inertia forces; hence, the inertia force system exerted on B will be made up of the inertia forces on the particles of B The inertia force exerted on a typical particle Pi of B is:

(6.9.4)

where m i is the mass of P i and A i is the acceleration of P i in R.

Using the procedures of Section 6.5, we can replace this system of many forces by a

single force F* passing through an arbitrary point, together with a couple having a torque

186 Dynamics of Mechanical Systems

T* It is generally convenient to let F* pass through the mass center G of the body Then,

F* and T* are:

(6.9.5)

and:

(6.9.6)

where ri locates P i relative to G.

Using Eq (4.9.6), we see that because both P i and G are fixed on B, a i may be expressed as:

(6.9.7)

where αααα and ωωωω are the angular acceleration and angular velocity of B in R Hence, by

substituting into Eq (6.9.5), F* becomes:

(6.9.8)

or

(6.9.9)

where M is the total mass of B and where the last two terms of Eq (6.9.8) are zero because

G is the mass center of B (see Eq (6.8.3)).

Similarly, by substituting for ai in Eq (6.9.6), T* becomes:

(6.9.10)

F*= (− a)

=

∑ m i i i

i i

N

1 1

(6.9.11)

where the first term of Eq (6.9.10) is zero because G is the mass center, and the last term

is obtained from its counterpart in the previous line by using the identity:

(6.9.12)

To see this, simply expand the triple products of Eq (6.9.10) using the identity Specifically,

(6.9.13)and

(6.9.14)

where the first terms are zero because ri and ωω are perpendicular to ωωωω × ri Comparing

Eqs (6.9.13) and (6.9.14), we see the results are the same That is,

(6.9.15)

Neither of the terms of Eq (6.9.11) is in a form convenient for computation or analysis;however, the terms have similar forms Moreover, we can express these forms in terms of

the inertia dyadic of the body as discussed in the next chapter This, in turn, will enable us

to express T* in terms of the moments and products of inertia of B for its mass center G.

Section 6.2 Forces and Moments

P6.2.1: A force F with magnitude 12 lb acts along a line L passing through points P and

Q as in Figure P6.2.1 Let the coordinates of P and Q relative to a Cartesian reference frame

be as shown, measured in feet Express F in terms of unit vectors nx, ny, and nz, which are

parallel to the X-, Y-, and Z-axes.

188 Dynamics of Mechanical Systems

P6.2.2: See Problem P6.2.1 Find the moment of F about points O, A, and B of Figure P6.2.1.

Express the results in terms of nx, ny, and nz.

P6.2.3: A force F with magnitude 52 N acts along diagonal OA of a box with dimensions

as represented in Figure P6.2.3 Express F in terms of the unit vectors n1, n2, and n3, which

are parallel to the edges of the box as shown

P6 2.4: See Problem P6.2.3 Find the moment of F about the corners A, B, C, D, E, G, and

H of the box Express the results in terms of n1, n2, and n3

Section 6.3 Systems of Forces

P6 3.1: Consider the force system exerted on the box as represented in Figure P6.3.1 The

force system consists of ten forces with lines of action and magnitudes as listed in Table

P6.3.1 Let the dimensions of the box be 12 m, 4 m, and 3 m, as shown

a Express the forces F1,…, F10 in terms of the unit vectors n1, n2, and n3 shown in

Figure P6.3.1

b Find the resultant of the force system expressed in terms of n1, n2, and n3

c Find the moment of the force system about O (expressed in terms of n1, n2,

P6.3.2: See Problem 6.3.1, and (a) find the moment of the force system about points C and

H, and (b) verify Eq (6.3.6) for these results That is, show that:

where R is the resultant of the force system.

P6 3.3: A cube with 2-ft sides has forces exerted upon it as shown in Figure P6.3.3 The

magnitudes and lines of action of these forces are listed in Table P6.3.3

4m 12m

190 Dynamics of Mechanical Systems

a Determine the resultant R of this system of forces.

b Find the moments of the force system about points C, H, E, and G.

Express the results in terms of the unit vectors n1, n2, and n3 shown in Figure P6.3.3

6.4 Special Force Systems: Zero Force Systems and Couples

P6 4.1: In Figure P6.4.1, the box is subjected to forces as shown The magnitudes and directions of the forces are listed in Table P6.4.1 Determine the magnitudes of forces F1,

F3, F4, F5, F6, and F10 so that the system of forces on the box is a zero system

P6 4.2: Three cables support a 1500-lb load as depicted in Figure P6.4.2 By considering

the connecting joint O of the cables to the load to be in static equilibrium and by recalling

that forces in cables are directed along the cable, determine the forces in each of the cables

The location of the fixed ends of the cables (A, B, and C) are determined by their x, y, z

coordinates (measured in feet) as shown in Figure P6.4.2

P6 4.3: Repeat Problem P6.4.2 if the load is 2000 lb.

P6.4.4: Repeat Problem P6.4.3 if the coordinates of the cable supports (A, B, and C) are

given in meters instead of feet

P6 4.5: Show that the force system exerted on the box shown in Figure P6.4.5 is a couple.

The magnitudes and directions of the forces are listed in Table P6.4.5

B(-2,2,10) A(6,-3,8)

192 Dynamics of Mechanical Systems

P6 4.6: See Problem P6.4.5 Find the torque of the couple.

P6 4.7: See Problems P6.4.5 and P6.4.6 Find the moment of the force system about points

O, A, D, and G.

P6 4.8: Show that the magnitude of the torque of a simple couple (two equal-magnitude

but oppositely directed forces) is simply the product of the magnitude of one of the forcesmultiplied by the distance between the parallel lines of action of the forces Show furtherthat the orientation of the torque is perpendicular to the plane of the forces, with sense

determined by the right-hand rule.

P6 4.9: Show that a set of simple couples is also a couple Show further that the torque of

this composite couple is then simply the resultant (sum) of the torques of the simplecouples

6.5 Equivalent Force Systems

P6 5.1: See Problem P6.3.1 Consider again the force system exerted on the box of Problem

P6.3.1 as shown in Figure P6.5.1, where the magnitudes and lines of action of the forcesare listed in Table P6.5.1 Suppose this force system is to be replaced by an equivalent

force system consisting of a single force F passing through O together with a couple having

torque T Find F and T (Express the results in terms of the unit vectors n1, n2, and n3 ofFigure P6.5.1.)

P6.5.2: Repeat Problem P6.5.1 with the single force F passing through G Compare the

magnitudes of the respective couple torques

P6.5.3: Consider a homogeneous rectangular block with dimensions a, b, and c as

repre-sented in Figure P6.5.3 Let the gravitational forces acting on the block be replaced by an

equivalent force system consisting of a single force W If ρ is the uniform mass density of

the block, find the magnitude and line of action of W.

P6 5.4: See Problem P6.5.3 Suppose the gravitational force system on the block is replaced

by an equivalent force system consisting of a single force W passing through O together

with a couple with torque T Find F and T Express the results in terms of the unit vectors

n1, n2, and n3 shown in Figure P6.5.3

P6 6.1: See Problems P6.5.1 and P6.3.1 For the force system of Problems P6.5.1 and P6.3.1,

find the point Q* for which there is a minimum moment of the force system Specifically,

find the position vector OQ* locating Q* relative to O Express the result in terms of the

unit vectors n1, n2, and n3 shown in Figure P6.5.1

P6 6.2: See Problem P6.6.1 Find a wrench that is equivalent to the force system of Problems

P6.3.1, P6.5.1, and P6.6.1 Express the results in terms of the unit vectors of Figure P6.5.1

6.7 Physical Forces: Applied (Active) Forces

P6.7.1: Suppose the human arm is modeled by three bodies B1, B2, and B3 representing theupper arm, lower arm, and hand, as shown in Figure P6.7.1A Suppose further that the

194 Dynamics of Mechanical Systems

gravitational forces acting on these bodies are represented by equivalent gravity (weight)

force systems consisting of single vertical forces W1, W2, and W3 as in Figure P6.7.1B.Finally, suppose we want to find an equivalent gravity force system for the entire arm

consisting of a single force W passing through the shoulder joint O together with a couple

with torque M Find W and M Express the results in terms of the angles θ1, θ2, and θ3;

the distances r1, r2, r3, 1, 2, and 3; the force magnitudes W1, W2, and W3; and the unit

vectors nx, ny, and nz shown in Figures P6.7.1A and B

P6 7.2: See Problem P6.7.1 Table P6.7.2 provides numerical values for the geometric

quantities and weights of the arm model of Figures P6.7.1A and B Using these values,

determine the magnitudes of W and M.

P6 7.3: See Problems P6.7.1 and P6.7.2 Suppose the equivalent force system is to be a

wrench, where the couple torque M is a minimum Locate a point G on the line of action

of the equivalent force Find the magnitudes of the equivalent wrench force and minimummoment

P6 7.4: Three springs are connected in series as in Figure P6.7.4 Find the elongation for

the applied forces The spring moduli and force magnitudes are listed in Table P6.7.4

P6 7.5: See Problem P6.7.4 Let the springs of Problem P6.7.4 be arranged in parallel as in

Figure P6.7.5 Find the elongation δ for the applied forces Assume that the springs aresufficiently close (or even coaxial) so that the rotation of the attachment plates can beignored

P6 7.6: A block is resting on an incline plane as shown in Figure P6.7.6 Let µ be thecoefficient of friction between the block and the plane Find the inclination angle θ of theincline where the block is on the verge of sliding down the plane

P6 7.7: See Problem P6.7.6 Let the inclination angle θ be small Let the drag factor (f) be defined as an effective coefficient of friction that accounts for the small slope Find f in

terms of µ and θ What would be the value of f if the block is sliding up the plane?

TABLE P6.7.2

Numerical Values for the Geometric Parameters and Weights

of Figures P6.7.1A and B

Physical Data for the System of Figure P6.7.4

Force F Spring Moduli 12 lb (lb/in.) 50 N (N/mm)

P6.7.8: A homogeneous block is pushed to the right by a force P as in Figure P6.7.8 If

the friction coefficient between the block and plane is µ, find the maximum elevation

h above the surface where the force can be applied so that the block will slide and not tip.

P6.7.9: Figure P6.7.9 shows a schematic representation of a short-shoe drum brake When

a force F is applied to the brake lever, the friction between the brake shoe and the drum

creates a braking force and braking moment on the drum Let µ be the coefficient of friction

between the shoe and the drum If the force F is known, determine the braking moment

M on the drum if the drum is rotating: (a) clockwise, and (b) counterclockwise Express

M in terms of µ, F, and the dimensions a, b, c, d, and r shown in Figure P6.7.9.

P6 7.10: See Problem P6.7.9 Find a relation between µ and the dimensions of Figure P6.7.9

so that the brake is self-locking when the drum is rotating counterclockwise (Self-locking means that a negligible force F is required to brake the drum.)

196 Dynamics of Mechanical Systems

Section 6.8 Mass Center

P6.8.1: Let a system of 10 particles P i (i = 1,…, 10) have masses m i and coordinates (x i , y i , z i)relative to a Cartesian coordinate system as in Table P6.8.1 Find the coordinates –x, – y, –z

of the mass center of this set of particles if the m i are expressed in kilograms and the x i,

y i , z i are in meters How does the result change if the m i are expressed in slug and the x i,

y i , z i in feet?

P6 8.2: See Problem P6.8.1 From the definition of mass center as expressed in Eq (6.8.3)

show that the coordinates of the mass center may be obtained by the simple expressions:

where

P6 8.3: Use the definition of mass center as expressed in Eq (6.8.3) to show that the mass

center of a system of bodies may be obtained by: (a) letting each body Bi be represented

by a particle Gi located at the mass center of the body and having the mass mi of the body;

and (b) by then locating the mass center of this set of particles

P6 8.4: See Problem P6.8.3 Consider a thin, uniform-density, sheet-metal panel with a

circular hole as in Figure P6.8.4 Let the center O of the hole be on the diagonal BC 13 in from corner B Locate the mass center relative to corner A (that is, distance from A in the

AB direction and distance from A in the AC direction).

N

i i i

31 in

17 in

OB = 13 in.

P6 8.5: See Problem P6.8.4 Let the panel of Problem P6.8.4 be augmented by the addition

of: (a) a triangular plate as in Figure P6.8.5A, and (b) by both a triangular plate and asemicircular plate as in Figure P6.8.5B Locate the mass center in each case

P6 8.6: See Problem P6.8.5 Let the semicircular plate of Figure P6.8.5b be bent upward as

depicted in a side view of the panel as in Figure P6.8.6 Locate the mass center parallel to

the main panel relative to corner A as before and in terms of its distance, or elevation,

above the main panel

P6 8.7: See Problems P6.7.1, P6.7.2, and P6.7.3 Consider again the model of the human arm

as depicted in Figures P6.7.1A and B and in Figures P6.8.7A and B Let the physical andgeometrical data for the arm model be as in Table P6.7.2 and as listed again in Table P6.8.7

Determine the location of the mass center of the arm model relative to the shoulder joint O

for the configuration of Table P6.8.7 Compare the results with those of Problem P6.7.3

FIGURE P6.8.5

(A) A thin, uniform-density panel consisting of a triangular plate and a rectangular plate with a circular hole (see Figure P6.8.4) (B) A thin, uniform-density panel consisting of a triangular plate, a semicircular plate, and a rectangular plate with a circular hole (see Figures P6.8.4 and P6.8.5A).

FIGURE P6.8.6

Side view of the panel of Figure

P6.8.5B with bent upward

semicircu-lar plate (see Figure P6.8.5B).

FIGURE P6.8.7

(A) A model of the human arm (B)

Equivalent gravity (weight) forces on

the human arm model.

TABLE P6.8.7

Numerical Values for the Geometric Parameters and Weights of

Figures P6.8.7A and B

In this chapter we review various topics and concepts about inertia Many readers will

be familiar with a majority of these topics; however, some topics, particularly thoseconcerned with three-dimensional aspects of inertia, may not be as well understood, yetthese topics will be of most use to us in our continuing discussion of mechanical systemdynamics In a sense, we have already begun our review with our discussion of masscenters in the previous chapter At the end of the chapter, however, we discovered that

we need additional information to adequately describe the inertia torque of Eq (6.9.11),shown again here:

(7.1.1)

Indeed, the principal motivation for our review of inertia is to obtain simplified sions for this torque Our review will parallel the development in Reference 7.4 with abasis found in References 7.1 to 7.3 We begin with a discussion about second-momentvectors — a topic that will probably be unfamiliar to most readers As we shall see, though,second-moment vectors provide a basis for the development of the more familiar topics,particularly moments and products of inertia

200 Dynamics of Mechanical Systems

Observe that the second moment is somewhat more detailed than the first moment (mp)defined in Eq (6.8.1) The second moment depends upon the square of the distance of P

from O and it also depends upon the direction of the unit vector na

The form of the definition of Eq (7.2.1) is motivated by the form of the terms of theinertia torque of Eq (7.1.1) Indeed, for a set S of particles, representing a rigid body B

(Figure 7.2.2), the second moment is defined as the sum of the second moments of theindividual particles That is,

(7.2.2)

Then, except for the presence of na instead of αααα or ω, the form of Eq (7.2.2) is identical tothe forms of Eq (7.7.1) Hence, by examining the properties of the second moment vector,

we can obtain insight into the properties of the inertia torque We explore these properties

in the following subsections

Consider again a particle P, with mass m, a reference point O, unit vector na, and a second

unit vector nb as in Figure 7.3.1 The moment and product of inertia of P relative to O forthe directions na and nb are defined as the scalar projections of the second moment vector(Eq (7.2.1)) along na and nb Specifically, the moment of inertia of P relative to O for thedirection na is defined as:

(7.3.1)Similarly, the product of inertia of P relative to O for the directions na and nb is defined as:

Iab P O D= Ia P O⋅nb

0593_C07_fm Page 200 Monday, May 6, 2002 2:42 PM

Inertia, Second Moment Vectors, Moments and Products of Inertia, Inertia Dyadics 201

Observe that by substituting for from Eq (7.2.1) that and may be expressed

in the form:

(7.3.3)and

(7.3.4)

Observe further that (p×na)2 may be identified with the square of the distance da from

P to a line passing through O and parallel to na (see Figure 7.3.2) This distance is often

referred to as the radius of gyration of P relative to O for the direction na

Observe also for the product of inertia of Eq (7.3.3) that the unit vectors na and nb may

be interchanged That is,

(7.3.5)

Note that no restrictions are placed upon the unit vectors na and nb If, however, na and

nb are perpendicular, or, more generally, if we have three mutually perpendicular unit

vectors, we can obtain additional geometric interpretations of moments and products of

inertia Specifically, consider a particle P with mass m located in a Cartesian reference

frame R as in Figure 7.3.3 Let (x, y, z) be the coordinates of P in R Then, from Eq (7.2.1)

the second moment vectors of P relative to origin O for the directions nx, ny, and nz are:

202 Dynamics of Mechanical Systems

and, similarly,

(7.3.7)and

(7.3.8)

Using the definitions of Eqs (7.3.1) and (7.3.2), we see that the various moments andproducts of inertia are then:

(7.3.9)

Observe that the moments of inertia are always nonnegative or zero, whereas the products

of inertia may be positive, negative, or zero depending upon the position of P.

It is often convenient to normalize the moments and products of inertia by dividing by

the mass m Then, the normalized moment of inertia may be interpreted as a length squared, called the radius of gyration and defined as:

(7.3.10)

(See also Eq (7.3.3).)

Finally, observe that the moments and products of inertia of Eqs (7.3.9) may be niently listed in the matrix form:

I I I

I I I

mxy

mzy

mxz myz

,,,

where the repeated index denotes a sum over the range of the index.

These expressions can be simplified even further by using the concept of a dyadic A

dyadic is the result of a product of vectors employing the usual rules of elementary algebra,

except that the pre- and post-positions of the vectors are maintained That is, if a and b

are vectors expressed as:

(7.4.4)

then the dyadic product of a and b is defined through the operations:

(7.4.5)

where the unit vector products are called dyads The nine dyads form the basis for a general

dyadic (say, D), expressed as:

P O

yx x yy y yz z z

204 Dynamics of Mechanical Systems

(7.4.6)

Observe that a dyadic may be thought of as being a vector whose components are vectors;

hence, dyadics are sometimes called vector–vectors Observe further that the scalar

com-ponents of D (the d ij of Eq (7.4.6)) can be considered as the elements of a 3 × 3 matrix and

as the components of a second-order tensor (see References 7.5, 7.6, and 7.7).

In Eq (7.3.12), we see that the moments and products of inertia may be assembled as

elements of a matrix In Eq (7.4.3) these elements are seen to be I ij (i, j = 1, 2, 3) If these

matrix elements are identified with dyadic components, we obtain the inertia dyadicdefined as:

Equation (7.3.5) shows that the matrix of moments and products of inertia is symmetric

(that is, I ij = I ji) Then, by rearranging the terms of Eq (7.4.7), we see that IP/O may also beexpressed as:

(7.4.11)and

Finally, for systems of particles or for rigid bodies, the inertia dyadic is developed from

the contributions of the individual particles That is, for a system S of N particles we have:

Then, by forming the projection of onto nb we obtain the product of inertia , which

in view of Eq (7.5.3) can be expressed as:

or

(7.5.5)

IS O IP O i

N i

206 Dynamics of Mechanical Systems

Observe that the scalar components a i and b i of na and nb of Eqs (7.5.2) and (7.5.3) may

be identified with transformation matrix components as in Eq (2.11.3) Specifically, let

(j = 1, 2, 3) be a set of mutually perpendicular unit vectors, distinct and noncollinear with

the ni (i = 1, 2, 3) Let the transformation matrix components be defined as:

(7.5.6)Then, in terms of the Sij, Eq (7.5.5) takes the form:

(7.5.7)

Consider once more the definition of the second moment vector of Eq (7.2.1):

(7.6.1)

Observe that just as is directly dependent upon the direction of na, it is also dependent

upon the choice of the reference point O The transformation rules discussed above enable

us to evaluate the second-moment vector and other inertia functions as na changes Theparallel axis theorems discussed in this section will enable us to evaluate the second-

moment vector and other inertia functions as the reference point O changes.

To see this, consider a set S of particles P i (i = 1,…, N) with masses m i as in Figure 7.6.1

Let G be the mass center of S and let O be a reference point Let pG locate G relative to

O, let p i locate P i relative to O, and let r i locate P i relative to G Finally, let n a be an arbitrary

unit vector Then, from Eq (7.2.2), the second moment of S for O for the direction of n a is: