STATICS OF PARTICLES

1. STATICS OF PARTICLES Forces are vectorquantities; 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. R P A Q

2. 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. Q F A P

3. Fy tan q = Fx F = Fx + Fy 2 2 A force F is said to have been resolved into two rectangular componentsif its components are directed along the coordinate axes. Introducing the unit vectors i and j along the x and y axes, F = Fxi + Fyj y Fx = F cos qFy = F sin q Fy = Fyj F j q x i Fx = Fx i

4. Ry tan q = R = Rx + Ry 2 2 Rx Whenthree 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. Rx= SRx Ry= SRy The magnitude and direction of R can be determined from

5. y y B B qy Fy Fy A F A F D D O O Fx x qx Fx x Fz Fz E E C C z z y A force Finthree-dimensional spacecan be resolved into components B Fy F Fx = Fcos qxFy = Fcos qy A D O Fz = Fcos qz Fx x qz E Fz C z

6. y l(Magnitude = 1) The cosines of qx, qy, and qz are known as the direction cosines of the force F. Using the unit vectorsi, j, and k, we write Fyj F = Fl cos qyj Fxi cos qzk x Fzk F = Fxi + Fyj +Fz k cos qxi z or F = F (cosqxi + cosqyj + cosqzk )

7. y F = Fx + Fy+ Fz 2 2 2 Fz F Fy F Fx F cosqz= cosqy = cosqx = l(Magnitude = 1) cos qyj l = cosqxi + cosqyj + cosqzk Fyj cos qzk F = Fl Since the magnitude of l is unity, we have Fxi cos2qx + cos2qy + cos2qz= 1 x Fzk z cos qxi In addition,

8. 1 d MN MN l = = ( dxi + dyj + dzk ) y A force vector F in three-dimensions is defined by its magnitude F and two points Mand N along its line of action. The vector MN joining points M and N is N (x2, y2, z2) F dy = y2 - y1 l dz = z2 - z1 < 0 dx = x2 - x1 M (x1, y1, z1) x z MN = dxi + dyj + dzk The unit vector l along the line of action of the force is

9. F d F = Fl = ( dxi + dyj + dzk ) y 2 2 2 d= dx + dy + dz N (x2, y2, z2) dy = y2 - y1 A force F is defined as the product of F and l. Therefore, dz = z2 - z1 < 0 dx = x2 - x1 M (x1, y1, z1) x z From this it follows that Fdx d Fdz d Fdy d Fx= Fz= Fy=

10. When two or more forces act on a particle inthree-dimensions, the rectangular components of their resultant R is obtained by adding the corresponding components of the given forces. Rx= SFx Ry= SFy Rz= SFz The particle is in equilibrium when the resultant of all forces acting on it is zero.

11. To solve a problem involving a particle in equilibrium, draw a free-body diagramshowing all the forces acting on the particle. The conditions which must be satisfied for particle equilibrium are SFx= 0 SFy= 0 SFz= 0 In two-dimensions, only two of these equations are needed SFx= 0 SFy= 0

12. V = P xQ The vector product of two vectors is defined as Q V = P xQ q P The vector product of P and Q forms a vector which is perpendicular to both P and Q, of magnitude V = PQ sin q This vector is directed in such a way that a person located at the tip of V observes as counterclockwise the rotation through q 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 Qx P = - (PxQ)

13. j It follows from the definition of the vector product of two vectors that the vector products of unit vectors i,j, andkare k i i xi = jxj = kxk = 0 i xj = k,jxk = i , kxi = j , i xk = - j , jxi = - k , k xj = - i The rectangular components of the vector product V of two vectors P and Q are determined as follows: Given P = Pxi + Pyj + Pzk Q = Qxi + Qyj + Qzk The determinant containing each component of P and Q is expanded to define the vector V, as well as its scalar components

14. P = Pxi + Pyj + Pzk Q = Qxi + Qyj + Qzk i Px Qx j Py Qy k Pz Qz V = PxQ = = Vxi + Vyj + Vzk where Vx= PyQz- PzQy Vy = PzQx- PxQz Vz = PxQy- PyQx

15. Mo The moment of force F about point O is defined as the vector product MO = rxF 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 q. r O q d A The magnitude of the moment of F about O can be expressed as MO = rF sin q = Fd where d is the perpendicular distance from O to the line of action of F.

16. y A (x , y, z ) Fyj yj The rectangular components of the moment Mo of a force F are determined by expanding the determinant of r x F. Fxi r Fzk xi O x zk z j y Fy k z Fz i x Fx Mo = rxF = = Mxi + Myj + Mzk Mx = yFz- zFyMy = zFx- xFz Mz = xFy- yFx where

17. y A (x A, yA, z A) Fyj B (x B, yB, z B) In the more general case of the moment about an arbitrary point B of a force F applied at A, we have Fxi r Fzk O x z i xA/B Fx k zA/B Fz j yA/B Fy MB = rA/BxF = rA/B= xA/Bi + yA/Bj + zA/Bk where xA/B= xA- xB yA/B= yA- yB zA/B= zA- zB and

18. F 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 Bis perpendicular to that plane. It can be completely defined by the scalar y Fyj A Fxi rA/B (yA - yB ) j B (xA - xB ) i O x z MB = MBk MB = (xA- xB )Fy + (yA- yB ) Fx The right-hand rule is useful for defining the direction of the moment as either into or out of the plane (positive or negative kdirection).

19. P Q = PQ cos q P Q = PxQx+ PyQy+ PzQz The scalar product of two vectors PandQis denoted as PQ ,and is defined as Q q where q is the angle between the two vectors P The scalar product of P and Q is expressed in terms of the rectangular components of the two vectors as

20. y L The projection of a vector P on an axis OL can be obtained by forming the scalar product of P and the unit vector l along OL. qy A l P qx x O qz z POL= P l Using rectangular components, POL= Pxcos qx + Pycos qy + Pzcos qz

21. The mixed triple product of three vectors S, P, and Q is Sx Px Qx Sy Py Qy Sz Pz Qz S(PxQ ) = The elements of the determinant are the rectangular components of the three vectors.

22. y The moment of a force F about an axis OLis the projection OC onOLof the moment MO of the forceF. This can be written as a mixed triple product. L MO F C l A (x, y, z) r O x z l x x Fx l y y Fy l z z Fz MOL = l MO =l(r xF) = lx, ly , lz = direction cosines of axis OL x, y , z= components of r Fx, Fy , Fz= components of F

23. M - F d F Two forces Fand-Fhaving 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.

24. M y y y - F (M = Fd) M My d F Mx O O O x x x z z Mz z Two couples having the same moment M are equivalent (they have the same effect on a given rigid body).

25. F F MO r A A O O 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 MO equal to the moment about point O of the force F in its original position. The force vector F and the couple vector MO are always perpendicular to each other.

26. R O M F3 F1 F3 F1 R A3 M1 A1 r1 r3 r2 O A2 M2 O M3 O F2 F2 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 MO. In general, the resultant force R and the couple vector MO will not be perpendicular to each other. R R

27. R O M F3 F1 R A3 A1 r1 r3 r2 O A2 O F2 As far as rigid bodies are concerned, two systems of forces, F1, F2, F3 . . . , andF’1, F’2, F’3 . . . , are equivalent if, and only if, S F = S F’ and S Mo = S Mo’

28. F3 F1 If the resultant force R and the resultant couple vector MO are perpendicular to each other, the force-couple system at O can be further reduced to a single resultant force. R A3 R A1 r1 r2 O A2 O R O F2 M 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.