COSMOS Chapter 3 Solution 141

COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 141. First, choose a rectangular coordinate system where one axis coincides with the axis of the wrench and another axis intersects the prescribed line of action ( AA′ ) . Note that it has been assumed that the line of action of force B intersects the xz plane at point P ( x, 0, z ) . Denoting the known direction of line AA′ by λ A = λxi + λ y j + λzk it follows that force A can be expressed as ( A = Aλ A = A λxi + λ y j + λzk ) Force B can be expressed as B = Bxi + By j + Bzk Next, observe that since the axis of the wrench and the prescribed line of action AA′ are known, it follows that the distance a can be determined. In the following solution, it is assumed that a is known. Then, for equivalence ΣFx : 0 = Aλx + Bx (1) ΣFy : R = Aλ y + By (2) ΣFz : 0 = Aλz + Bz (3) ΣM x : 0 = − zBy Since there are six unknowns obtain a solution. . (4) ΣM y : M = −aAλz + zBx − xBz (5) ΣM z : 0 = aAλ y + xBy (6) ( A, Bx , By , Bz , x, z ) and six independent equations, it will be possible to COSMOS: Complete Online Solutions Manual Organization System Case 1: Let z = 0 to satisfy Equation (4) Aλ y = R − By Now Equation (2) Bz = − Aλz Equation (3) x=− Equation (6)  a = −  By  aAλ y By   R − By   ( ) Substitution into Equation (5)   a M = −aAλz −  −    By ∴ A=−    R − By ( − Aλz )     ( ) 1 M  B λz  aR  y Substitution into Equation (2) R=− 1 M  By λ y + By λz  aR  ∴ By = Then λz aR 2 λz aR − λ y M R aR λy − λz M λx MR Bx = − Aλx = λz aR − λ y M A=− MR λz aR − λ y M Bz = − Aλz = = λz MR λz aR − λ y M In summary A= B= and P λA aR λy − λz M R ( λ Mi + λz aRj + λz Mk ) λz aR − λ y M x    λz aR − λ y M R  x = a 1 −  = a 1 − R    By  λz aR 2         or x = Note that for this case, the lines of action of both A and B intersect the x axis. continued . λy M λz R COSMOS: Complete Online Solutions Manual Organization System Case 2: Let By = 0 to satisfy Equation (4) A= Now Equation (2) R λy Equation (1) λ  Bx = − R  x   λy    Equation (3) λ Bz = − R  z  λy  aAλ y = 0 Equation (6)     which requires a = 0 Substitution into Equation (5)  λ M = z −R  x   λ y   λ    − x −R  z     λ y    or This last expression is the equation for the line of action of force B. In summary  R A=  λy   λ A    R B=  λy    ( −λxi − λzk )   Assuming that λx , λ y , λz > 0, the equivalent force system is as shown below. Note that the component of A in the xz plane is parallel to B. . M  λz x − λx z =   λ y  R