CHAPTER 3

1. 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. 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’ 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 Vis 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. A1 V=PxQ V A2 V=PxQ’ 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 P x (Q1+Q2) = P x Q1 +P x Q2 YES 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. ijk Px Py Pz Qx Qy Qz ijkij 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 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=Fxi +Fyj +Fzk M0= r x F

20. Rectangular components of the moment of a force ijkij x y zx 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) Mz=(xFy- y Fx) My=(z Fx-x Fz)

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: PQ= P Q cos  (scalar) • Satisfies: • Commutative Property: PQ= Q P • Distributive Property: P (Q1 + Q2)= PQ1+PQ2

23. Scalar product of 2 vectors • The scalar product of 2 vectors P and Q can be expressed as: • PQ=(Pxi + Pyj + Pzk) (Qxi + Qyj + Qzk) • And ij= 0, jk= 0, ki= 0 ii= 1, jj= 1, kk= 1 • Therefore: • PQ = PxQx + PyQy + PzQz

24. Special Case • P=Q • PP = 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 z • Consider Q directed along the axis OL: PQ= POL Q POL= PQ Q • PQ= P Q cos  = POL Q POL

27. PQ 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: 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 andM0?

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 L rA/B=rA-rB  B xyz xA/B yA/B zA/B Fx Fy Fz F A MBL= O 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. B r -F M  F rB A rA O Being rA and rB the position vectors of the points of application of F and -F. 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 30 lb 20 lb 6 inch 6 inch 20 lb

38. 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

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 O A 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 O A O O r - F Force couple system

43. 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