1. RIGID BODIES: EQUIVALENT SYSTEM OF FORCES
CHAPTER 3
RIGID BODIES: EQUIVALENT SYSTEM OF FORCES

2. 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

3. In this Chapter you will learn
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.

4. 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.

5. Examples of external Forces
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
FBD F W R1 R2

6. 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.

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

8. Principle of transmissibility
Conditions of motion are unaffected. F and F’ are equivalent F F’ = W W R1 R1 R2 R2

9. 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.
V=PxQ Q  P

10. Vector product
The magnitude of V is also equal to the area of the parallelogram that has P and Q for sides.
V=PxQ A1 V
V=PxQ’ A2 Q Q’ P
V=P x Q=P x Q’

11. 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)

12. 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 y j i k x z
(+):CCW j k i
(-):CW

13. 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

14. Vx=(PyQz-PzQy) Vy=(PzQx-PxQz) Vz=(PxQy-PyQx)
The right hand members represent the expansion of a determinant.
i j k
Px Py Pz
Qx Qy Qz
i j k i j
Px Py Pz Px Py
Qx Qy Qz Qx Qy V=
(-)
(+)

15. 3.6 Moment of a Force around a point
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. M0 F O r A d 
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.

16. Units of Moment
SI: Nm
US Units: lb ft or lb in

17. 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’

18. 3.7 Varignon’s Theorem r x (F1+F2+……)= r x F1+ r x F2+…
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+…

19. 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

20. Rectangular components of the moment of a force
i j k i j
x 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)

21. 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.

22. 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

23. 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= 0
i i= 1, j j= 1, k k= 1
Therefore:
P Q =
PxQx + PyQy + PzQz

24. Special Case
P=Q
P P =
Px Px + PyPy + PzPz
=P2

25. Applications 1) To determine the angle between two vectors
P Q cos  =PxQx + PyQy + PzQz
Solving for cos …
cos =
PxQx + PyQy + PzQz
P Q

26. 2nd Application: Projection of a vector in a given axis
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) L
axis y A Q P  O x
Consider Q directed along the axis OL: z
P Q= POL Q
POL= P Q Q
P Q= P Q cos 
= POL Q
POL

27. P Q Q
POL= 
POL= P , POL= Px x+Py y+Pz z

28. 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)
(+) If the vectors are read ccw order or circular permutation
(-) cw direction P S Q S Q P

29. 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:
Sx Sy Sz
Px Py Pz
Qx Qy Qz
S (P x Q)=
Application of triple product

30. 3.11 Moment of a Force around a given axis
Given a force, a position vector and a moment : L
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 M0 F
POL= P , 
A(x,y,z) r
MOL= M0  O
(r x F)
Mixed Triple Product
MOL=  (r x F)

31. In determinant form: x y z x y z Fx Fy Fz MOL=
Moment MOL: Measures the tendency of F to impart to the rigid body rotation about a fixed axis OL
x y z
x y z
Fx Fy Fz
MOL=
Where x , y , z are direction cosines of OL
x, y, z are coordinates of point of application of F
Fx, Fy, Fz are components of F
What is the difference between MOL and M0?

32. 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 L
rA/B=rA-rB  B F A O
Determine the projection on the axis BL of MB of F about B.
MBL= MB BL
= BL(rA/B x F)

33. Moment MBL x y z xA/B yA/B zA/B Fx Fy Fz MBL= L rA/B=rA-rB  B F A
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.

34. 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

35. 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

36. Being rA and rB the position vectors of the points of application of F and -F.M A B  d
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 F
M= moment of a couple.
It’s perpendicular to the plane containing the 2 forces.
Its magnitude is
Its sense is defined by the right hand rule
M=r F sin = F d

37. 3.13 Equivalent Couples
Two couples that have the same moment M are equivalents.
Equivalent systems =
4 inch
4 inch
30 lb
4 inch
4 inch
20 lb
30 lb
6 inch
6 inch
20 lb

38. 3.14 Addition of Couples Given two systems of couples: M1=r x F1
F1 and -F1
F2 and -F2
M1=r x F1
M2=r x F2
Resultant Moment M=M1+M2

39. 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 -F F y M
“couple vector” x O z

40. Couple vector: is a free vector (point of application can be moved)
can be resolved into components Mx, My, Mz y M
“couple vector” x O z

41. 3.16 Resolution of a given force into a force at O and a couple
Consider F
We’d rather have a force acting at O
Mo=r x F
add 2 forces at O M0 F F F F r = = A O O O r
- F
Force couple system

42. 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 M0 F F F F r = = A O O O r
- F
Force couple system

43. To move F from O to O’, it’s necessary to add a couple vector.
If F is moved from A to O’ M0 F F A A F r r r = = A O r’ O r’ O s’ O’ s’ O’ M0
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)

44. Sample problem 3.6
Determine the components of a single couple equivalent to the couples shown in the figure (page 113).

45. 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. M1
MoR M2 F2 F1 R = F3 M3
R= F, MoR= Mo= (r x F)

46. THE END…..
…….FOR CHAPTER 3