COSMOS Chapter 3 Solution 140

COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 140. First, observe that it is always possible to construct a line perpendicular to a given line so that the constructed line also passes through a given point. Thus, it is possible to align one of the coordinate axes of a rectangular coordinate system with the axis of the wrench while one of the other axes passes through the given point. See Figures a and b. Have R = Rj M = Mj and and are known. The unknown forces A and B can be expressed as A = Axi + Ay j + Azk B = Bxi + By j + Bzk and The distance a is known. It is assumed that force B intersects the xz plane at (x, 0, z). Then for equivalence ∑ Fx : 0 = Ax + Bx (1) ∑ Fy : R = Ay + By (2) ∑ Fz : 0 = Az + Bz (3) ∑ M x : 0 = − zBy (4) ∑ M y : M = −aAz − xBz + zBx (5) ∑ M z : 0 = aAy + xBy (6) Since A and B are made perpendicular, A⋅B = 0 There are eight unknowns: or Ax Bx + Ay By + Az Bz = 0 (7) Ax , Ay , Az , Bx , By , Bz , x, z But only seven independent equations. Therefore, there exists an infinite number of solutions. continued . COSMOS: Complete Online Solutions Manual Organization System 0 = − zBy Next consider Equation (4): If By = 0, Equation (7) becomes Ax Bx + Az Bz = 0 Ax2 + Az2 = 0 Using Equations (1) and (3) this equation becomes Since the components of A must be real, a nontrivial solution is not possible. Thus, it is required that By ≠ 0, so that from Equation (4), z = 0. To obtain one possible solution, arbitrarily let Ax = 0. (Note: Setting Ay , Az , or Bz equal to zero results in unacceptable solutions.) The defining equations then become. 0 = Bx (1)′ R = Ay + By (2) 0 = Az + Bz (3) M = −aAz − xBz (5)′ 0 = aAy + xBy (6) Ay By + Az Bz = 0 (7)′ Then Equation (2) can be written Ay = R − By Equation (3) can be written Bz = − Az x=− Equation (6) can be written aAy By Substituting into Equation (5)′,  R − By M = −aAz −  −a  By  or Az = −   ( − Az )   M By (8) aR Substituting into Equation (7)′, M  M  By  By  = 0 ( R − By ) By +  − aR aR   . COSMOS: Complete Online Solutions Manual Organization System By = or a 2 R3 a2R2 + M 2 Then from Equations (2), (8), and (3) Ay = R − Az = − Bz = a 2 R3 RM 2 = a2R2 + M 2 a2R2 + M 2  M  a 2 R3 aR 2 M = −  2 2 2 2 aR  a R + M  a R2 + M 2 aR 2 M a R2 + M 2 2 In summary A= RM ( Mj − aRk ) a R2 + M 2 B= aR 2 ( aRj + Mk ) a R2 + M 2 2 2 Which shows that it is possible to replace a wrench with two perpendicular forces, one of which is applied at a given point. Lastly, if R > 0 and M > 0, it follows from the equations found for A and B that Ay > 0 and By > 0. From Equation (6), x < 0 (assuming a > 0). Then, as a consequence of letting Ax = 0, force A lies in a plane parallel to the yz plane and to the right of the origin, while force B lies in a plane parallel to the yz plane but to the left of the origin, as shown in the figure below. .