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

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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.

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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

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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

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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 x x y z A B C D E O FxFx FyFy FzFz F y z 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

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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 )

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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 cos x = FxFFxF cos y = FyFFyF cos z = FzFFzF In addition,

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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

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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 From this it follows that F x = Fd x d F y = Fd y d F z = Fd z d

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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

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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

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