Presentation on theme: "2 - 1 STATICS OF PARTICLES Forces are vector quantities; they add according to the parallelogram law. The magnitude and direction of the resultant R of."— Presentation transcript:
2 - 1 STATICS OF PARTICLES Forces are vector quantities; they add according to the parallelogram law. The magnitude and direction of the resultant R of two forces P and Q can be determined either graphically or by trigonometry. P R Q A
2 - 2 A Q P F Any given force acting on a particle can be resolved into two or more components, i.e.., it can be replaced by two or more forces which have the same effect on the particle. A force F can be resolved into two components P and Q by drawing a parallelogram which has F for its diagonal; the components P and Q are then represented by the two adjacent sides of the parallelogram and can be determined either graphically or by trigonometry.
2 - 3 x y F x = F x i F y = F y j F i j A force F is said to have been resolved into two rectangular components if its components are directed along the coordinate axes. Introducing the unit vectors i and j along the x and y axes, F = F x i + F y j F x = F cos F y = F sin tan = FyFy FxFx F = F x + F y 22
2 - 4 When three or more coplanar forces act on a particle, the rectangular components of their resultant R can be obtained by adding algebraically the corresponding components of the given forces. R x = R x R y = R y The magnitude and direction of R can be determined from tan = RyRy RxRx R = R x + R y 22
2 - 5 x y z A B C D E O FxFx FyFy FzFz F x y z A B C D E O FxFx FyFy FzFz F xx x y z A B C D E O FxFx FyFy FzFz F yy zz A force F in three-dimensional space can be resolved into components F x = F cos x F y = F cos y F z = F cos z
2 - 6 x y z Fy jFy j Fx iFx i Fz kFz k (Magnitude = 1) cos x i cos z k cos y j F = F The cosines of x, y, and z are known as the direction cosines of the force F. Using the unit vectors i, j, and k, we write F = F x i + F y j + F z k or F = F (cos x i + cos y j + cos z k )
2 - 7 x y z Fy jFy j Fx iFx i Fz kFz k (Magnitude = 1) cos x i cos z k cos y j F = F = cos x i + cos y j + cos z k Since the magnitude of is unity, we have cos 2 x + cos 2 y + cos 2 z = 1 F = F x + F y + F z 2 2 2 cos x = FxFFxF cos y = FyFFyF cos z = FzFFzF In addition,
2 - 8 x y z M (x 1, y 1, z 1 ) N (x 2, y 2, z 2 ) d x = x 2 - x 1 dz = z 2 - z 1 < 0 d y = y 2 - y 1 A force vector F in three-dimensions is defined by its magnitude F and two points M and N along its line of action. The vector MN joining points and N is MN = d x i + d y j + d z k = = ( d x i + d y j + d z k ) MN 1d1d F The unit vector along the line of action of the force is
2 - 9 x y z M (x 1, y 1, z 1 ) N (x 2, y 2, z 2 ) d x = x 2 - x 1 d z = z 2 - z 1 < 0 d y = y 2 - y 1 A force F is defined as the product of F and. Therefore, F = F = ( d x i + d y j + d z k ) FdFd d = d x + d y + d z 2 2 2 From this it follows that F x = Fd x d F y = Fd y d F z = Fd z d
2 - 10 When two or more forces act on a particle in three- dimensions, the rectangular components of their resultant R is obtained by adding the corresponding components of the given forces. The particle is in equilibrium when the resultant of all forces acting on it is zero. R x = F x R y = F y R z = F z
2 - 11 F x = 0 F y = 0 F z = 0 In two-dimensions, only two of these equations are needed To solve a problem involving a particle in equilibrium, draw a free-body diagram showing all the forces acting on the particle. The conditions which must be satisfied for particle equilibrium are F x = 0 F y = 0
2 - 12 q V = P x Q P Q The vector product of two vectors is defined as V = P x Q The vector product of P and Q forms a vector which is perpendicular to both P and Q, of magnitude V = PQ sin This vector is directed in such a way that a person located at the tip of V observes as counterclockwise the rotation through which brings vector P in line with vector Q. The three vectors P, Q, and V - taken in that order - form a right-hand triad. It follows that Q x P = - (P x Q)
2 - 13 ik j 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 i x j = k, j x k = i, k x i = j, i x k = - j, j x i = - k, k x j = - i 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 The determinant containing each component of P and Q is expanded to define the vector V, as well as its scalar components
2 - 14 P = P x i + P y j + P z k Q = Q x i + Q y j + Q z k V = P x Q = iPxQxiPxQx jPyQyjPyQy kPzQzkPzQz = V x i + V y j + V z k 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
2 - 15 O d A F MoMo r 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 . 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.
2 - 16 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 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
2 - 17 x y Fx iFx i Fz kFz k Fy jFy j O r In the more general case of the moment about an arbitrary point B of a force F applied at A, we have M B = r A/B x F = ixA/BFxixA/BFx jyA/BFyjyA/BFy kzA/BFzkzA/BFz where z B (x B, y B, z B ) A (x A, y A, z A ) r A/B = x A/B i + y A/B j + z A/B k and x A/B = x A - x B y A/B = y A - y B z A/B = z A - z B
2 - 18 x y Fx iFx i F Fy jFy j O M B = (x A - x B )F y + (y A - y B ) F x z B A (x A - x B ) i rA/BrA/B (y A - y B ) j In the case of problems involving only two dimensions, the force F can be assumed to lie in the xy plane. Its moment about point B is perpendicular to that plane. It can be completely defined by the scalar M B = M B k The right-hand rule is useful for defining the direction of the moment as either into or out of the plane (positive or negative k direction).
2 - 19 P Q The scalar product of two vectors P and Q is denoted as P Q,and is defined as The scalar product of P and Q is expressed in terms of the rectangular components of the two vectors as P Q = PQ cos P Q = P x Q x + P y Q y + P z Q z where is the angle between the two vectors
2 - 20 x y z O L A xx P zz yy The projection of a vector P on an axis OL can be obtained by forming the scalar product of P and the unit vector along OL. P OL = P Using rectangular components, P OL = P x cos x + P y cos y + P z cos z
2 - 21 The mixed triple product of three vectors S, P, and Q is S (P x Q ) = SxPxQxSxPxQx SyPyQySyPyQy SzPzQzSzPzQz The elements of the determinant are the rectangular components of the three vectors.
2 - 22 x F O The moment of a force F about an axis OL is the projection OC on OL of the moment M O of the force F. This can be written as a mixed triple product. z A (x, y, z) y MOMO L C r xxFxxxFx yyFyyyFy zzFzzzFz M OL = M O = (r x F) = x, y, z = direction cosines of axis OL x, y, z = components of r F x, F y, F z = components of F
2 - 23 d F - F M Two forces F and - F having the same magnitude, parallel lines of action, and opposite sense are said to form a couple. The moment of a couple is independent of the point about which it is computed; it is a vector M perpendicular to the plane of the couple and equal in magnitude to the product Fd.
2 - 24 Two couples having the same moment M are equivalent (they have the same effect on a given rigid body). x y z d F - F x y z O O M (M = Fd) x y z O MxMx MyMy MzMz M
2 - 25 Any force F acting at a point A of a rigid body can be replaced by a force-couple system at an arbitrary point O, consisting of the force F applied at O and a couple of moment M O equal to the moment about point O of the force F in its original position. The force vector F and the couple vector M O are always perpendicular to each other. O A r F O A F MOMO
2 - 26 O A1A1 r1r1 F1F1 O M1M1 r2r2 A2A2 F2F2 r3r3 A3A3 F3F3 F1F1 M2M2 M3M3 F2F2 F3F3 R M RORO O Any system of forces can be reduced to a force-couple system at a given point O. First, each of the forces of the system is replaced by an equivalent force-couple system at O. Then all of the forces are added to obtain a resultant force R, and all of couples are added to obtain a resultant couple vector M O. In general, the resultant force R and the couple vector M O will not be perpendicular to each other. R R
2 - 27 O A1A1 r1r1 F1F1 r2r2 A2A2 F2F2 r3r3 A3A3 F3F3 R M RORO O As far as rigid bodies are concerned, two systems of forces, F 1, F 2, F 3..., and F’ 1, F’ 2, F’ 3..., are equivalent if, and only if, F = F’ and M o = M o ’
2 - 28 O A1A1 r1r1 F1F1 r2r2 A2A2 F2F2 A3A3 F3F3 R M RORO O This is the case for systems consisting of either (a) concurrent forces, (b) coplanar forces, or (c) parallel forces. If the resultant force and couple are directed along the same line, the force-couple system is termed a wrench. R If the resultant force R and the resultant couple vector M O are perpendicular to each other, the force-couple system at O can be further reduced to a single resultant force.