Chapter 2 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. R P A Q Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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

F = Fx i + Fy j Fx = F cos q Fy = F sin q Fy tan q = Fx F = Fx + Fy 2 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 = Fx i + Fy j y Fx = F cos q Fy = F sin q tan q = Fy Fx Fy = Fy j F j F = Fx + Fy 2 q x i Fx = Fx i

Rx = S Rx Ry = S Ry Ry tan q = R = Rx + Ry 2 Rx 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. Rx = S Rx Ry = S Ry The magnitude and direction of R can be determined from tan q = Ry Rx R = Rx + Ry 2

Fx = F cos qx Fy = F cos qy Fz = F cos qz Fy Fy qy F F qx Fx Fx Fz Fz B B Fy Fy qy A F A F D D O O qx Fx Fx x x Fz Fz E E C C z z y A force F in three-dimensional space can be resolved into components B Fy F Fx = F cos qx Fy = F cos qy A D O Fz = F cos qz Fx x qz E Fz C z

F = F (cosqx i + cosqy j + cosqz k ) l (Magnitude = 1) The cosines of qx , qy , and qz are known as the direction cosines of the force F. Using the unit vectors i , j, and k, we write Fy j F = F l cos qy j Fx i cos qz k x Fz k F = Fx i + Fy j + Fz k cos qx i z or F = F (cosqx i + cosqy j + cosqz k )

l = cosqx i + cosqy j cos2qx + cos2qy F = Fx + Fy + Fz 2 cosqx = Fx F l (Magnitude = 1) cos qy j l = cosqx i + cosqy j + cosqz k Fy j cos qz k F = F l Since the magnitude of l is unity, we have Fx i x cos2qx + cos2qy + cos2qz = 1 Fz k z cos qx i In addition, F = Fx + Fy + Fz 2 cosqx = Fx F cosqy = Fy F cosqz = Fz F

MN = dx i + dy j + dz k MN 1 l = = ( dx i + dy j + dz k ) d y 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 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 = dx i + dy j + dz k The unit vector l along the line of action of the force is l = = ( dx i + dy j + dz k ) MN 1 d

F = F l = ( dx i + dy j + dz k ) F d Fdx d Fdy d Fdz d Fx = Fy = Fz = d = dx + dy + dz 2 2 2 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 F = F l = ( dx i + dy j + dz k ) F d From this it follows that Fdx d Fdy d Fdz d Fx = Fy = Fz =

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. Rx = S Fx Ry = S Fy Rz = S Fz The particle is in equilibrium when the resultant of all forces acting on it is zero.

S Fx = 0 S Fy = 0 S Fz = 0 S Fx = 0 S Fy = 0 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 S Fx = 0 S Fy = 0 S Fz = 0 In two-dimensions , only two of these equations are needed S Fx = 0 S Fy = 0