CHAPTER 3

RIGID BODIES: EQUIVALENT SYSTEM OF FORCES

Objectives• 3.1 Introduction • 3.2 External and Internal Forces• 3.3 Principle of Transmissibility• 3.4 Vector Product of two vectors• 3.5 Vector Product expressed in terms of rectangular components• 3.6 Moment of a Force about a point• 3.7 Varignon’s Theorem• 3.8 Rectangular components of the moment of the force• 3.9 Scalar Product of 2 vectors• 3.10 Mixed triple product of 2 vectors• 3.11 Moment of a force about a given axis• 3.12 Moment of a couple• 3.13 and 3.14 Equivalent and Addition of Couples• 3.15 Couples represented by vectors• 3.16 Resolution of a given force into a force at O and a couple• 3.17 Resolution of a system of forces to one force and one couple

So far, we have reviewed• Fundamental Principles:

– Parallelogram law– Newton’s laws– Principle of transmissibility

• Assumption: A rigid body could be treated as a single particle (not always true)

In this Chapter you will learn• The effect of forces exerted on a rigid body and how to

replace a given system of forces by a simpler equivalent system.

3.2 External and Internal Forces

• External Forces: – responsible for the external behavior of the

rigid bodies.– cause the rigid bodies to move or ensure that

they remain at rest.

• Internal Forces:– hold together the particles or parts forming the

rigid body.

Examples of external Forces

FBD

W

R1 R2

W: weight of the truck. Point of application center of gravity.

R1 and R2: reactions by the ground.

F: force of exertion, point of application truck’s bumper. It causes translation

F

3.3 Principle of transmissibility, equivalent forces

• Forces acting on a particle: vectors with a well defined point of application, called bound or fixed vectors.

• Forces acting on a rigid body: vectors whose point of application of the force doesn’t matter, as long as the line of action remains unchanged, called sliding vectors.

Principle of transmissibility

=F

F’

F and F’ have the same effect if their magnitude , direction and line of action are the same. Based on experimental evidence.

Principle of transmissibility

W

R1 R2

Conditions of motion are unaffected. F and F’ are equivalent

W

=

R1 R2

F F’

3.4 Vector product of 2 vectors• Vector product of 2 vectors is defined as a vector V,

which satisfies the following:– 1) Line of action of V is perpendicular to the plane

containing P and Q.

– 2) The magnitude of V is:

V=P Q sin – 3) The direction of V is obtained by the right hand rule.

Q

P

V=PxQ

Vector product

• The magnitude of V is also equal to the area of the parallelogram that has P and Q for sides.

Q

P

VV=PxQ

Q’

A1

V=PxQ’

V=P x Q=P x Q’

A2

Other properties of vector product:

• Commutative:

• Distributive:

• Associative Property:

NO Q x P P x Q

YES P x (Q1+Q2) = P x Q1 +P x Q2

NO(P x Q) x S P x (Q x S)

Vector expressed in terms of rectangular components

• Vector product of any 2 units vectors: i, j, k

• i x j= k

• j x i= -k

• j x k= i

• k x j= -i

• k x i= j

• i x k= -j

• i x i= 0, j x j= 0, k x k= 0

x

y

z

ij

k

i

j

k

(+):CCW

(-):CW

• We can express the vector product of two given vectors P and Q in terms of rectangular components:

• V=P x Q=(Pxi+Pyj+Pzk) x (Qxi+Qyj+Qzk)

• Using distributive property and the products of unit vectors:

• V=(PyQz-PzQy)i + (PzQx-PxQz)j+(PxQy-

-PyQx)k

• Vx=(PyQz-PzQy)

• Vy=(PzQx-PxQz)

• Vz=(PxQy-PyQx)

The right hand members represent the expansion of a determinant.

V=i j kPx Py Pz

Qx Qy Qz

i j k i jPx Py Pz Px Py Qx Qy Qz Qx Qy

(+)(-)

3.6 Moment of a Force around a point

A

F

r

M0

O

d

•The effect of the force F on the rigid body depends on its:•Magnitude•Direction•Point of application (A)

•Position of A: represented by r•r and F define a plane.•M0= r Fsin = Fd , where d= perpendicular distance from O to the line of action of F.Moment of F around O: M0= r x F, M0 is perpendicular to the

plane containing O and F The sense of F is defined by the right hand rule.

Units of Moment

• SI: Nm

• US Units: lb ft or lb in

3.8 Rectangular Components of the moment of a force

• We restate the principle of transmissibility as:

• Two forces F and F’ are equivalent if, only if, :– they are equal (=magnitude, = direction)– And have equal moment about a given point O

– F=F’ M0=M0’

3.7 Varignon’s Theorem

Moment of the resultant Sum of the moments

of several forces = of each force around

the same point O

r x (F1+F2+……)= r x F1+ r x F2+…

3.8 Rectangular Components of the moment of a Force

• We can simplify the calculation of the moment of a force by resolving the force and the position vector into components.

r = x i +y j +z k

F=Fx i +Fy j +Fz k

M0= r x F

Rectangular components of the moment of a force

i j k i jx y z x y Fx Fy Fz Fx Fy

(+)(-)

M0=(y Fizz-z Fy) i+ (z Fx-x Fz) j+ (xFy- y Fx) k

Mx=(y Fz-z Fy) My=(z Fx-x Fz) Mz=(xFy- y Fx)

Sample Problem 3.1

• A vertical force of 100 lb is applied to the end of the rod bar which is attached to a shaft at O. Determine:– A) Moment of a 100 lb force about O– B) The horizontal force applied at A which

creates the same moment about O.– C)The smallest force at A which creates the

same moment about O.

3.9 Scalar product of 2 vectors

• The scalar product, (or dot product) of two vectors is defined as:

P Q= P Q cos (scalar)

• Satisfies: – Commutative Property: P Q= Q P

– Distributive Property: P (Q1 + Q2)= P Q1+P Q2

Scalar product of 2 vectors

• The scalar product of 2 vectors P and Q can be expressed as:

• P Q=(Pxi + Pyj + Pzk) (Qxi + Qyj + Qzk)• And

i j= 0, j k= 0, k i= 0i i= 1, j j= 1, k k= 1

• Therefore:• P Q = PxQx + PyQy + PzQz

Special Case

• P=Q

• P P = Px Px + PyPy + PzPz

=P2

Applications

• 1) To determine the angle between two vectors

• P Q cos =PxQx + PyQy + PzQz

• Solving for cos …

• cos =

PxQx + PyQy + PzQz

P Q

2nd Application: Projection of a vector in a given axis

O

Laxis

P

The projection of P along the axis OL is defined as a scalar:POL=P cos (+) if OA has the

same sense as OL (axis) (-) if OA has the

opposite sense as OL (axis)QA

•Consider Q directed along the axis OL:

POL

= POL Q• P Q= P Q cos P Q= POL QPOL= P Q

Q

x

y

z

P Q Q

POL=

POL= P , POL= Px x+Py y+Pz z

3.10 Mixed triple product of 3 vectors

• Mixed triple product=S (P x Q)

• Geometrically:

• Mixed triple product=Volume of the parallelepiped having S, P and Q for sides

(Scalar expression)

S

P

Q

(+) If the vectors are read ccw order or circular permutation

(-) cw direction

S

P

Q

Mixed triple product in terms of rectangular components

S (P x Q) = Sx(Py Qz-Pz Qy) + Sy(Pz Qx-Px Qz) + Sz(PxQy- -Py Qx)

In compact form:

S (P x Q)=

Sx Sy Sz Px Py Pz

Qx Qy Qz

Application of triple product

3.11 Moment of a Force around a given axis

• Given a force, a position vector and a moment :

F

r

M0

O

L

A(x,y,z)

axis•We define moment MOL of F about OL: projection of the moment M0 onto the axis OL.•Projection of a vector onto an axis

POL= P ,

MOL= M0

(r x F)

MOL= (r x F)

Mixed Triple Product

In determinant form:

MOL=

x y z x y z Fx Fy Fz

Where x , y , z are direction cosines of OLx, y, z are coordinates of point of application of

FFx, Fy, Fz are components of F

Moment MOL:

Measures the tendency of F to impart to the rigid body rotation about a fixed axis OL

What is the difference between MOL and M0?

How is the moment of a force applied at A, about an axis, which does not pass through the origin obtained?

By choosing an arbitrary point B on the axis

FB

A

rA/B=rA-rB

O

L

Determine the projection on the axis BL of MB of F about B.

MBL= MB BL = BL(rA/B x F)

Moment MBL

FB

A

rA/B=rA-rB

O

L

MBL=

x y z xA/B yA/B zA/B Fx Fy Fz

Where xA/B=xA-xB, yA/B= yA-yB, zA/B= zA-zB

The result of the moment of F about the axis L is independent of the choice of the point B on the given axis.

Sample problem 3.5

• A force P acts on a cube of side a. Determine the moment of P:– a) about A– b) about AB– c) about AG

3.12 Moment of a couple

• Couple: Two forces F and -F having the same magnitude, parallel lines of action and opposite sense. The forces tend to make the body on which they act rotate.

-F

F

Being rA and rB the position vectors of the points of application of F and -F.

The sum of the moments of F and -F about O M=rA x F + rB x (-F)= (rA-rB) x F

M = r x FM= moment of a couple. It’s perpendicular to the plane containing the 2 forces. Its magnitude isIts sense is defined by the right hand rule

M=r F sin = F d

rA

FrB

-F

O

r

M

A

B

d

3.13 Equivalent Couples

• Two couples that have the same moment M are equivalents.

20 lb20 lb

6 inch

4 inch

4 inch

6 inch

4 inch

4 inch

30 lb

30 lb=

Equivalent systems

3.14 Addition of Couples

• Given two systems of couples: – F1 and -F1– F2 and -F2

• M1=r x F1

• M2=r x F2

• Resultant Moment M=M1+M2

3.15 Couples can be represented by vectors

• Instead of drawing actual forces:

• Draw an arrow equal in magnitude and direction to the moment M of the couple

M

“couple vector”

-F

F

Ox

y

z

• Couple vector:– is a free vector (point of application can be moved)

– can be resolved into components Mx, My, Mz

M

“couple vector”

Ox

y

z

3.16 Resolution of a given force into a force at O and a couple

• Consider F

F

OAr

•We’d rather have a force acting at O

F

- Fr

F F

O O

M0

• add 2 forces at O Mo=r x F

Force couple system

= =

F

OAr

F

- Fr

F F

O O

M0

Force couple system

= =

Conclusion: Any F acting on a rigid body can be moved to

an arbitrary point O provided that a couple is added whose moment is equal to the moment of F about O

F

OAr

F

O

M0

=

If F is moved from A to O’

A

O’s’

rr’

F

O

M0

A

O’s’

rr’=

•To move F from O to O’, it’s necessary to add a couple vector.

Mo’= r’ x F= (r + s) x F = (r x F) + (s x F)

Mo’ = MO + (s x F)

Sample problem 3.6

• Determine the components of a single couple equivalent to the couples shown in the figure (page 113).

3.17 Resolution of a system of forces to one force and a couple

• Any system of forces can be reduced to an equivalent force-couple system acting at a point O.

F1

F2

F3

M1

M2

M3

R

MoR

=

R= F, MoR= Mo= (r x F)

THE END…..

…….FOR CHAPTER 3