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 resolvedinto two components Pand Q by drawing aparallelogram which hasF for its diagonal; thecomponents P and Qare then represented bythe two adjacent sidesof the parallelogramand can be determinedeither graphically or bytrigonometry.QFAP
3 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 rectangularcomponents if its components are directed along the coordinateaxes. Introducing the unit vectors i and j along the x and y axes,F = Fx i + Fy jyFx = F cos q Fy = F sin qtan q =FyFxFy = Fy jFjF = Fx + Fy2qxiFx = Fx i
4 Rx = S Rx Ry = S Ry Ry tan q = R = Rx + Ry 2 Rx When three or more coplanar forces act on a particle, therectangular components of their resultant R can be obtainedby adding algebraically the corresponding components of thegiven forces.Rx = S Rx Ry = S RyThe magnitude and direction of R can be determined fromtan q =RyRxR = Rx + Ry2
5 Fx = F cos qx Fy = F cos qy Fz = F cos qz Fy Fy qy F F qx Fx Fx Fz Fz BBFyFyqyAFAFDDOOqxFxFxxxFzFzEECCzzyA force F in three-dimensional space can be resolved into componentsBFyFFx = F cos qx Fy = F cos qyADOFz = F cos qzFxxqzEFzCz
6 F = F (cosqx i + cosqy j + cosqz k ) l (Magnitude = 1)The cosines ofqx , qy , and qzare known as thedirection cosines of the force F. Using the unit vectors i , j, and k, we writeFy jF = F lcos qy jFx icos qz kxFz kF = Fx i + Fy j + Fz kcos qx izorF = F (cosqx i + cosqy j + cosqz k )
7 l = cosqx i + cosqy j cos2qx + cos2qy F = Fx + Fy + Fz 2 cosqx = Fx F l (Magnitude = 1)cos qy jl = cosqx i + cosqy j+ cosqz kFy jcos qz kF = F lSince the magnitudeof l is unity, we haveFx ixcos2qx + cos2qy+ cos2qz = 1Fz kzcos qx iIn addition,F = Fx + Fy + Fz2cosqx =FxFcosqy =FyFcosqz =FzF
8 MN = dx i + dy j + dz k MN 1 l = = ( dx i + dy j + dz k ) d y A force vector Fin three-dimensionsis defined by itsmagnitude F andtwo points M andN along its line ofaction. The vectorMN joining pointsM and N isN (x2, y2, z2)Fdy = y2 - y1ldz = z2 - z1< 0dx = x2 - x1M (x1, y1, z1)xzMN = dx i + dy j + dz kThe unit vector l along the line of action of the force isl = = ( dx i + dy j + dz k )MN1d
9 F = F l = ( dx i + dy j + dz k ) F d Fdx d Fdy d Fdz d Fx = Fy = Fz = d = dx + dy + dz222N (x2, y2, z2)dy = y2 - y1A force F isdefined as theproduct of F andl. Therefore,dz = z2 - z1< 0dx = x2 - x1M (x1, y1, z1)xzF = F l = ( dx i + dy j + dz k )FdFrom this it follows thatFdxdFdydFdzdFx =Fy =Fz =
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.Rx = S FxRy = S FyRz = S FzThe particle is in equilibrium when the resultant of all forces acting on it is zero.
11 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 areS Fx = S Fy = S Fz = 0In two-dimensions , only two of these equations are neededS Fx = S Fy = 0
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