1. 1 The scalar product or dot product between two vectors P and Q is defined as Scalar products: -are commutative, -are distributive, -are not associative, Scalar products with Cartesian unit components, 3.9 SCALAR PRODUCT OF TWO VECTORS

2. 2 Angle between two vectors: Projection of a vector on a given axis: For an axis defined by a unit vector:

3. 3 Two forces F and -F having the same magnitude, parallel lines of action, and opposite sense are said to form a couple. Moment of the couple, The moment vector of the couple is independent of the choice of the origin of the coordinate axes, i.e., it is a free vector that can be applied at any point with the same effect. 3.12 MOMENT OF A COUPLE

4. 4 Two couples will have equal moments if the two couples lie in parallel planes, and the two couples have the same sense or the tendency to cause rotation in the same direction.

5. 5 A couple is formed by any two parallel forces acting on the same plane with equal magnitude and opposite directions. The moment of a couple is a vector whose line of action is perpendicular to the plane of the couple and whose direction can be determined by the right-hand rule. MOMENT OF A COUPLE

6. 6 APPLICATIONS A torque or moment of 12 N · m is required to rotate the wheel. Which one of the two grips of the wheel above will require less force to rotate the wheel?

7. 7 APPLICATIONS (continued) The crossbar lug wrench is being used to loosen a lug net. What is the effect of changing dimensions a, b, or c on the force that must be applied?

8. 8 F -F Couple The forces will result in rotation of the body ≠ 0

9. 9 Couple Moment If F is the magnitude of the forces forming the couple and d is the perpendicular distance between the lines of actions of the forces, then the magnitude of the couple-moment is: M=dF F F A B d d-b b M = dF = (d – b)F + bF

10. 10 MOMENT OF A COUPLE A couple is defined as two parallel forces with the same magnitude but opposite in direction separated by a perpendicular distance d. The moment of a couple is defined as M O = F d (using a scalar analysis) or as M O = r  F (using a vector analysis). Here r is any position vector from the line of action of –F to the line of action of F.

11. 11 Couple Moment The couple-moment is known as a free vector, meaning that it can be moved anywhere in space without changing its meaning, provided that its magnitude and direction are kept intact (couple has same moment about every point in space).

12. 12 100mm 150mm 100mm 135 N 90 N 100mm 135 N 3.14 Equivalent Couples Figure shows three couples which act successively on the same rectangular box. As seen in the preceding section, the only motion a couple can impart to a rigid body is a rotation. Since each of the three couples shown has the same moment M ( same direction and same magnitude M = 135 N. m ), we can expect the three couples to have the same effect on the box.

13. 13 Consider two intersecting planes P 1 and P 2 with each containing a couple Resultants of the vectors also form a couple By Varigon’s theorem Sum of two couples is also a couple that is equal to the vector sum of the two couples 3.14 ADDITION OF COUPLES

14. 14 A couple can be represented by a vector with magnitude and direction equal to the moment of the couple. Couple vectors obey the law of addition of vectors. Couple vectors are free vectors, i.e., the point of application is not significant. Couple vectors may be resolved into component vectors. 3.15 COUPLES CAN BE REPRESENTED BY VECTORS

15. 15 Force vector F can not be simply moved to O without modifying its action on the body. Attaching equal and opposite force vectors at O produces no net effect on the body. The three forces may be replaced by an equivalent force vector and couple vector, i.e, a force-couple system. 3.16 Resolution of a given force into a force at O and A Couple

16. 16 Moving F from A to a different point O’ requires the addition of a different couple vector M O’ The moments of F about O and O’ are related, Moving the force-couple system from O to O’ requires the addition of the moment of the force at O about O’.

17. 17 Sample problem 3.6 Determine the components of the single couple equivalent to the two couples shown. 135 N 90 N x 300mm 225mm 175mm 225mm z

18. 18 x z y Mx = -(60.75 N.m)i My = +(27.0 N.m)j Mz = +(20.25 N.m)k Our computations will be simplified if we attach two equal and opposite 90N forces at A. This enables us to replace the original 90N force couple by two new 90N force couples, one of which lies in the zx plane and the other in a plane parallel to the xy plane. The three couples shown in the adjoining sketch can be represented by three couple vectors Mx, My and Mz directed along the coordinate axes. The corresponding moments are Mx = - (135N) (0.45m) = - 60.75N.m My = +( 90 N) (0.3 m) = + 27N.m Mz = +( 90 N) (0.225m) = + 20.25N.m 90 N 135 N 225mm 300mm 175mm y z Solution These thre moments represent the components of the single couple M equivalent to the given. We write M = -(60.75 N.m)i + (27.0 N.m)j + (20.25 N.m)k

19. 19 Replace the couple and force shown by an equivalent single force applied to the lever. Determine the distance from the shaft to the point of application of this force. Sample problem 3.7 400 N 260 N

20. 20 (i) Step 1 The given force and couple are replaced by an equivalent force-couple system at O. The force F = - (400N)j is moved to O, and a moment couple is added to the system, which is the moment about point O of the force in its original position. M O = x F = (0.150mi + 0.260mj) x (-400N)j = -60k M O = - (60 Nm)k Solution 400 N 200 N F = - (400N)j - (24 N.m)k - (60 N.m)k - (400 N.)j

21. 21 Example 2 (i) Step 2 This couple added to couple moment of –(24Nm)k produced due to two 200N forces, thus the couple moment is now –(84Nm)k. We then apply F at point C. - (84 Nm)k = x F = [(OC) cos 60  i + (OC) sin 60  j] x (-400)j = - (OC) cos 60  (400 N)k OC cos 60  = 0.210 m = 210 mm OC = 420 mm - (400 N) j - (84 N.m) k - (400 N) j

22. 22 The effect of a couple does not depend on its location, thus the couple moment –(24 Nm)k can be moved to B. -(24 Nm)k = x F = - BC cos 60  (400N) k We conclude that (BC) cos 60˚ = 0.060m = 60mm BC =120mm OC = OB + BC = 300 mm + 120 mm OC = 420 mm

23. 23 A system of forces may be replaced by a collection of force-couple systems acting a given point O The force and couple vectors may be combined into a resultant force vector and a resultant couple vector, The force-couple system at O may be moved to O’ with the addition of the moment of R about O’, Two systems of forces are equivalent if they can be reduced to the same force-couple system. 3.17 Reduction of A System of Forces To One Force and One Couple

24. 24 Equivalent System MOVING A FORCE ITS LINE OF ACTION ON Sliding VECTOR

25. 25 MOVING A FORCE OF ITS LINE OF ACTION OFF Free VECTOR

26. 26 Resultant of a Force and Couple System

27. 27 3.20 Further Reduction of a Force and Couple System If R and M RP are perpendicular to each other, the force-couple system at P can be further reduced to a single resultant force. Will be the case for system consisting either : (a) concurrent force (b) coplanar force (c) parallel forces (a) concurrent force

28. 28 (b) Coplanar force systems (c) Parallel force systems

29. 29 The moment of the resultant force about the grip the moment of all the forces about the grip F R d = F 1 d 1 + F 2 d 2 + F 3 d 3

30. 30 Sample problems 3.8 A 4.80m long beam is subjected to the forces shown. Reduce the given system of forces to a)an equivalent force couple system at A b) single force or resultant 150N600N 100N 250N

31. 31 use F R d = F 1 d 1 + F 2 d 2 + F 3 d 3 ; solution

32. 32 = - 600(1.6m) + 100(2.8m) – 250(4.8) = - 1880 Nm + F R = 150 - 600 + 100 - 250 = - 600 N = 600 N  M A = F 1 d 1 + F 2 d 2 + F 3 d 3 F1F1 F2F2 F3F3 solution 150N 600N 100N 250N

33. 33 A B d F R = 600 [N] F R d  =  M A (600) d = 1880 Nm d = 1880/600 = 3.13 m Thus, the single force equivalent to the given system; F R = 600 [N], d = 3.13 [m] solution

34. 34 Example Three stage lights are mounted on a pipe as shown. The lights at A and B each weigh 16.4 N, while the one at C weighs 14 N. (a)If d = 0.625 m, determine the distance from D to the line of action of the resultant of the weights of the three lights. (b)Determine the value of d so that the resultant of the weights passes through the midpoint of the pipe.

35. 35 Solution 16.4 N 14 N D ABC = D LR 2.1 m 0.85 md 0.25 m For equivalence, F : - 16.4 – 16.4 – 14 = - R R = 46.8 N : - (0.25)(16.4) – (1.10)(16.4) – (1.10 + d)(14) = - L (46.8) 37.54 + 14d = 46.8 L »Eq 1

36. 36 Example 3 (cont) (a)For d = 0.625 m, substitute the value of d in Eq 1, 37.54 + 14 (0.625) = 46.8 L L = 0.9891 m The resultant passes through a point 0.9891 m to the right of D (b) For L = 1.05 m, substitute the value of L in Eq. 1, 37.54 + 14d = 46.8 (1.05) d = 0.8275 m

37. 37 THE END