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CHAPTER THREE

CHAPTER THREE. STATICS OF RIGID BODIES. 3.1 INTRODUCTION. INTRODUCTION CONTD. The forces acting on rigid bodies can be internal or external. F 1 , F 2 and F 3 which are applied by an external force on the rigid body are called external forces.

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CHAPTER THREE

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  1. CHAPTER THREE STATICS OF RIGID BODIES

  2. 3.1 INTRODUCTION

  3. INTRODUCTION CONTD. • The forces acting on rigid bodies can be internal or external. F1, F2and F3 which are applied by an external force on the rigid body are called external forces. • P1 and P2which are forces internal to the rigid body are called internal forces.

  4. INTRODUCTION CONCLUDED. • The external forces are completely responsible for the bulk motion of the rigid body. • As far as this bulk motion is concerned, the internal forces are in equilibrium.

  5. 3.2 PRINCIPLE OF TRANSMISSIVITY OF FORCES • External forces generally cause translation i.e. linear motion and/or rotation (motion about a pivot) of the rigid body.

  6. PRINCIPLE OF TRANSMISSIVITY OF FORCES • Principle of transmissivity states that the condition of rest or motion of a rigid body is unaffected if a force, F acting on a point A is moved to act at a new point, B provided that the point B lies on the same line of action of that force. F F A B

  7. 3.3 CROSS OR VECTOR PRODUCT OF TWO VECTORS • The moment of a force will be formulated using Cartesian vectors in the next section. • It is necessary to first expand our knowledge of vector algebra and introduce the cross-product method of vector multiplication.

  8. Cross Products of ForcesP and Q

  9. 3.3.2 Direction • Vector, V has a direction that is perpendicular to the plane containing P and Q such that the direction of V is specified by the right hand rule i.e. curling the fingers of the right hand from vector P (cross) to vector Q, the thumb then points in the direction of V.

  10. Direction of Cross Product

  11. 3.3.3. Laws of Operation • 1. The cummutative law does not apply i.e. P x Q  Q x P • Rather: P x Q = - Q x P • 2. Multiplication by a scalar • a ( P x Q) = (a P) x Q = P x ( a Q) = ( P x Q ) a • 3. The distributive law: • P x (Q + S ) = ( P x Q ) + ( P x S ) • 4. The associative property does not apply to vector products • (P x Q ) x S P ( Q x S )

  12. Cartesian (Rectangular) Vector Formulation

  13. Cartesian Vector FormulationContd. j k i -ve

  14. Cross Product of Two Vectors P and Q

  15. Equation For Product, V in Determinant Form • The equation for V may be written in a more compact determinant form as: • V = P x Q= i j k Px Py Pz Qx Qy Qz

  16. 3.4 MOMENT OF A FORCE ABOUT A POINT • The moment of a force about a point or axis provides a measure of the tendency of the force to cause a body to rotate about the point or axis.

  17. Moment of a Force About a Point • Consider a force, F acting at point A and a point O which lie on the same plane. The position of A is defined by the position vector, r, which joins the reference point O with A

  18. Moment of a Force About a Point • Moment (Mo) of F about O is defined as the vector product of r and F: i.e. Mo = r x F

  19. Direction of Mo • Using the right hand rule, the direction of Mo can be found. Curling the fingers of the right hand to follow the sense of rotation, the thumb points along the moment axis so the direction and sense of the moment vector is upward and perpendicular to the plane containing r and F.

  20. 3.5 VARIGNON’S THEOREM

  21. 3.6 RECTANGULAR COMPONENTS OF THE MOMENT OF A FORCE (CARTESIAN VECTOR FORMULATION)

  22. Convention

  23. Alternative Method • Fx = 100 cos 25o = 90.63 N • Fy = 100 sin 25o = 42.26 N • MB = - (0.204 x 90.63) + (42.26 x 0.0951) • = - 14.47 N m = 14.47 N m Clockwise

  24. Note • Cross or vector product has a distinct advantage over the scalar formulation when solving problems in three dimensions. This is because, it is easier to establish the position vector, r to the force rather than determine the moment-arm distance, d perpendicular to the line of the force. • For two dimensional problems, it is easier to use the Scalar Formulation

  25. 3.7 SCALAR OR DOT PRODUCT OF TWO VECTORS • Scalar product of two vectors, P and Q • P.Q = P Q cos  ..... magnitude Where: 0 <  < 180o The dot product is often referred to as scalar product of vectors, since the result is a scalar, not a vector. Q  P

  26. 3.7.1. Laws of Operation (i) Commutative law: P. Q = Q. P • (ii) Multiplication by a scalar: a (P. Q) = (a P). Q = P. (a Q) = (P. Q) a • (iii) Distributive law: P. ( Q1 + Q2) = P Q1 + P Q2

  27. 3.7.2 Cartesian Vector Formulation

  28. 3.7.3 Applications of Dot Product

  29. Solution Contd.

  30. Solution Using Cosine Law

  31. Moment of a Force About A Specified Axis

  32. Moment of a Force About A Specified Axis Concluded

  33. Solution Concluded

  34. 3.9 MOMENT OF A COUPLE

  35. 3.9.2. Scalar Formulation • Moment of a couple, M has magnitude: • M = F x d = r F sin  • F is the magnitude of one of the forces; • d is the perpendicular distance or moment arm between the forces.

  36. Scalar Formulation Concluded • The direction and sense of the couple moment are determined by the right hand rule, where the thumb indicates the direction when the fingers are curled with the sense of rotation caused by the two forces. • The moment acts perpendicular to the plane containing the two forces.

  37. 3.9.3 Vector Formulation • The moment of a couple can be expressed using the cross product. • M = r x F

  38. 3.9.4 Equivalent Couples • Two couples are equivalent if they produce the same moment. • Since the moment produced by a couple is always perpendicular to the plane containing the couple forces, it is necessary that the forces of equal couples lie either in the same plane or in planes that are parallel to one another.

  39. Resolution of a Force into a Force at B and a couple • A Force, F acting at point A on a rigid body can be resolved to the same force acting on another point B and in the same direction as the original force plus a couple M equal to r x F i.e. moment of F about B • i.e Force in (a) equal to that in (b) equal to that in C M = 25 kN m 5 kN F = 5 kN 5 kN 5 kN B A B 5 kN 5 m (a) (c) (b)

  40. Example • A force and couple act as shown on a square plate of side a = 625 mm. Knowing that P = 300 N, Q = 200 N and  = 50o, replace the given force and couple by a single force applied at a point located (a) on line AB (b) on line AC. In each case, determine the distance for A to the point of application of the force.

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