1. Slide #: 1 Chapter 3 Rigid Bodies : Equivalent Systems of Forces

2. Slide #: 2 Introduction In previous lessons, the bodies were assumed to be particles. Actually the bodies are a combination of a large number of particles. particle body

3. Slide #: 3 Introduction translation rotation A force may cause a body to move in a straight path that is called Another force may cause the body to move differently, for example, The external forces may cause a motion of translation or rotation or both. The point of application of force determines the effect.

4. Slide #: 4 Principle of Transmissibility : Equivalent Forces F F’ According to the principal of transmissibility, the effect of an external force on a rigid body remains unchanged if that force is moved along its line of action. Two forces acting on the rigid body at two different points have the same effect on that body if they have the same magnitude, same direction, and same line of action. Two such forces are said to be equivalent.

5. Slide #: 5 Moving the point of application of the force F to the rear bumper does not affect the motion or the other forces acting on the truck. Principle of Transmissibility : Equivalent Forces

6. Slide #: 6 Vector Product of Two Vectors Concept of the moment of a force about a point is more easily understood through applications of the vector product or cross product. Vector product of two vectors P and Q is defined as the vector V which satisfies the following conditions: 1.Line of action of V is perpendicular to plane containing P and Q. 2. Magnitude of V is 3. Direction of V is obtained from the right-hand rule.

7. Slide #: 7 Vector Product of Two Vectors Vector products: -are not commutative, -are distributive, -are not associative,

8. Slide #: 8 Vector Product of Unit Vectors It follows from the definition of the vector product of two vectors that the vector products of unit vectors i, j, and k are: i x i = j x j = k x k = 0

9. Slide #: 9 Vector Products in Terms of Rectangular Components The rectangular components of the vector product V of two vectors P and Q are determined as follows: Given P = P x i + P y j + P z k Q = Q x i + Q y j + Q z k V = (P y Q z - P z Q y ) i + (P z Q x - P x Q z ) j + (P x Q y - P y Q x ) k V = P x Q V = V x i + V y j + V z k

10. Slide #: 10 where V x = P y Q z - P z Q y V y = P z Q x - P x Q z V z = P x Q y - P y Q x V = P x Q = V x i + V y j + V z k V = P x Q = iPxQxiPxQx jPyQyjPyQy kPzQzkPzQz Expansion of Determinant Vector Products in Terms of Rectangular Components

11. Slide #: 11 V = P x Q = ijk P = P x i + P y j + P z k Q = Q x i + Q y j + Q z kPxPx PyPy PzPz QxQx QyQy QzQz Expansion of Determinant

12. Slide #: 12 V = P x Q = ijk P = 4 i + 3 j + 2 k Q = 6 i + 1 j + 5 k432615 Expansion of Determinant - Example

13. Slide #: 13 V = P x Q = i46i46 j31j31 k25k25 Calculation of Determinant : Method -1 i46i46 j31j31

14. Slide #: 14 V = P x Q = i46i46 j31j31 k25k25 Calculation of Determinant : Method -1 i46i46 j31j31 123 456 V = P x Q = + + - - - 123456

15. Slide #: 15 V = P x Q = i46i46 j31j31 k25k25 Calculation of Determinant : Method -1 i46i46 j31j31 15 i12 j4 k 18 k2 i20 j V = P x Q = 13 i - 8 j -14 k

16. Slide #: 16 Calculation of Determinant : Method -2 D = a 11 a 21 a 31 a 12 a 22 a 32 a 13 a 23 a 33 a 11 a 12 a 13 a 21 a 31 a 22 a 32 a 23 a 33 = D 11 D 11 : Minor of a 11 in D a 22 a 32 a 23 a 33 D 11 = = a 22 * a 33 – a 32 * a 23

17. Slide #: 17 Calculation of Determinant : Method -2 D = a 11 a 21 a 31 a 12 a 22 a 32 a 13 a 23 a 33 a 11 a 12 a 13 a 21 a 31 a 23 a 33 = D 12 D 12 : Minor of a 12 in D a 22 a 32 a 21 a 31 a 23 a 33 D 12 = = a 21 * a 33 – a 31 * a 23

18. Slide #: 18 Calculation of Determinant : Method -2 D = a 11 a 21 a 31 a 12 a 22 a 32 a 13 a 23 a 33 a 11 a 12 a 13 = D 13 D 13 : Minor of a 13 in D a 22 a 32 a 21 a 31 a 23 a 33 D 13 = = a 21 * a 32 – a 31 * a 22 a 22 a 32 a 21 a 31

19. Slide #: 19 Calculation of Determinant : Method -2 D = a 11 a 21 a 31 a 12 a 22 a 32 a 13 a 23 a 33 a 11 a 12 a 13 D = a 11 D 11 - a 12 D 12 + a 13 D 13 a 22 a 32 a 21 a 31 a 23 a 33 D= a 11 (a 22 * a 33 – a 32 * a 23 ) - a 12 (a 21 * a 33 – a 31 * a 23 ) + a 13 (a 21 * a 32 – a 31 * a 22 )

20. Slide #: 20 V = P x Q = i46i46 j31j31 k25k25 Calculation of Determinant : Method -2 V = P x Q = 3131 2525 i 4646 3131 k - 4646 2525 j + V = P x Q = i (15-2) – j (20-12) + k (4-18) V = P x Q = 13 i – 8 j – 14 k

21. Slide #: 21 Moment of a Force About a Point The moment of force F about point O is defined as the vector product M O = r x F where r is the position vector drawn from point O to the point of application of the force F. The angle between the lines of action of r and F is . A force vector is defined by its magnitude and direction. Its effect on the rigid body also depends on it point of application.

22. Slide #: 22 Moment of a Force About a Point The magnitude of the moment of F about O can be expressed as M O = rF sin  = Fd where d is the perpendicular distance from O to the line of action of F. Magnitude of M O measures the tendency of the force to cause rotation of the body about an axis along M O. The sense of the moment may be determined by the right-hand rule.

23. Slide #: 23 Moment of a Force About a Point Two-dimensional structures have length and breadth but negligible depth and are subjected to forces contained in the plane of the structure. The plane of the structure contains the point O and the force F. M O, the moment of the force about O is perpendicular to the plane. If the force tends to rotate the structure counter clockwise, the sense of the moment vector is out of the plane of the structure and the magnitude of the moment is positive. If the force tends to rotate the structure counterclockwise, the sense of the moment vector is into the plane of the structure and the magnitude of the moment is negative.

24. Slide #: 24 Varignon’s Theorem The moment about a given point O of the resultant of several concurrent forces is equal to the sum of the moments of the various moments about the same point O. Varigon’s Theorem makes it possible to replace the direct determination of the moment of a force F by the moments of two or more component forces of F.

25. Slide #: 25 Rectangular Components of the Moment of a Force x y z Fx iFx i Fz kFz k Fy jFy j x ix i y jy j z kz k O A (x, y, z ) r The moment of F about O,

26. Slide #: 26 The rectangular components of the moment M o of a force F are determined by expanding the determinant of r x F. M o = r x F = ixFxixFx jyFyjyFy kzFzkzFz = M x i + M y j + M z k where M x = y F z - z F y M y = zF x - x F z M z = x F y - y F x Rectangular Components of the Moment of a Force

27. Slide #: 27 Rectangular Components of the Moment of a Force The moment of F about B,

28. Slide #: 28 Moment of a Force About a Point Sample Problem 3.4 The rectangular plate is supported by the brackets at A and B and by a wire CD. Knowing that the tension in the wire is 200 N, determine the moment about A of the force exerted by the wire at C. SOLUTION: The moment M A of the force F exerted by the wire is obtained by evaluating the vector product,

29. Slide #: 29 Moment of a Force About a Point Sample Problem 3.4 SOLUTION:

30. Slide #: 30 Moment of a Force About a Point Sample Problem 3.4 SOLUTION: 0 96 0.08 -128 i - M A = 0.3 -120 j 0.3 -120 0 96 k + 0.08 -128 M A = (0- (96)(0.08)) i – ((0.3)(-128) – (-120)(0.08)) j + (96)(0.3) k M A = - 7.68 i + 28.8 j + 28.8 k

31. Slide #: 31 Moment of a Force About a Point Sample Problem 3.4 SOLUTION:

32. Slide #: 32 Rectangular Components of the Moment of a Force For two-dimensional structures,

33. Slide #: 33 Rectangular Components of the Moment of a Force For two-dimensional structures,

34. Slide #: 34 Moment of a Force About a Point Sample Problem 3.1 A 100-N vertical force is applied to the end of a lever which is attached to a shaft at O. Determine: a)moment about O, b)horizontal force at A which creates the same moment, c)smallest force at A which produces the same moment, d)location for a 240-N vertical force to produce the same moment, e)whether any of the forces from b, c, and d is equivalent to the original force.

35. Slide #: 35 Moment of a Force About a Point Sample Problem 3.1 a) Moment about O is equal to the product of the force and the perpendicular distance between the line of action of the force and O. Since the force tends to rotate the lever clockwise, the moment vector is into the plane of the paper.

36. Slide #: 36 Moment of a Force About a Point Sample Problem 3.1 b)Horizontal force at A that produces the same moment,

37. Slide #: 37 Moment of a Force About a Point Sample Problem 3.1 c) The smallest force A to produce the same moment occurs when the perpendicular distance is a maximum or when F is perpendicular to OA.

38. Slide #: 38 Moment of a Force About a Point Sample Problem 3.1 d) To determine the point of application of vertical 240 N force to produce the same moment,

39. Slide #: 39 Moment of a Force About a Point Sample Problem 3.1 e)Although each of the forces in parts b), c), and d) produces the same moment as the 100 lb force, none are of the same magnitude and sense, or on the same line of action. None of the forces is equivalent to the 100 lb force.

40. Slide #: 40 Scalar Product of Two Vectors 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,

41. Slide #: 41 Scalar Product of Two Vectors : Applications Angle between two vectors: Projection of a vector on a given axis: For an axis defined by a unit vector:

42. Slide #: 42 Mixed Triple Product of Three Vectors Mixed triple product of three vectors, The six mixed triple products formed from S, P, and Q have equal magnitudes but not the same sign,

43. Slide #: 43 Mixed Triple Product of Three Vectors Evaluating the mixed triple product,

44. Slide #: 44 Moment of a Force About a Given Axis Moment M O of a force F applied at the point A about a point O, Scalar moment M OL about an axis OL is the projection of the moment vector M O onto the axis,

45. Slide #: 45 Moment of a Force About a Given Axis Moments of F about the coordinate axes, Moment M OL of the force F about OL measures the tendency of F to make the rigid body rotates about the fixed axis OL

46. Slide #: 46 Moment of a Force About a Given Axis If the axis does not pass through the origin Moment of a force about an arbitrary axis,

47. Slide #: 47 Sample Problem 3.5 a)about A b)about the edge AB and c)about the diagonal AG of the cube. d)Determine the perpendicular distance between AG and FC. A cube is acted on by a force P as shown. Determine the moment of P

48. Slide #: 48 Sample Problem 3.5 Moment of P about A, Moment of P about AB,

49. Slide #: 49 Sample Problem 3.5

50. Slide #: 50 Sample Problem 3.5 Alternative method using determinant