Explanation Given information : The given force vector (F) is { 6 i + 8 j + 10 k } N . The given moment vector ( M O ) is { − 1 4i + 8j + 2 k } N ⋅ m . The x coordinate value is 1 m. Determine the position vector. r = x i + y j + z k (I) Determine the y and z coordinates using the relation. M O = r × F (II) Determine the perpendicular distance d . M O = F d (III) Conclusion : Substitute 1 m for x in Equation (I). r = x i + y j + z k = ( 1i + y j + z k ) m Substitute { − 1 4i + 8j + 2 k } N ⋅ m for M O , ( 1i + y j + z k ) m for r , and { 6 i + 8 j + 10 k } N for F in Equation (II). { − 1 4i + 8j + 2 k } = | i j k 1 y z 6 8 10 | = { i ( 10 y − 8 z ) − j ( 10 − 6 z ) + k ( 8 − 6 y ) } = { i ( 10 y − 8 z ) + j ( − 10 + 6 z ) + k ( 8 − 6 y ) } Equate the j component on both sides

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Chapter 4.4, Problem 46P

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The y and z coordinates and the perpendicular distance d from point O to the line of action of F.

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Chapter 4.4, Problem 45P

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(I) [40 Points] Using centered finite difference approximations as done in class, solve the equation for O: d20 dx² + 0.010+ Q=0 subject to the boundary conditions shown in the stencil below. Do this for two values of Q: (a) Q = 0.3, and (b) Q= √(0.5 + 2x)e-sinx (cos(5x)+x-0.5√1.006-x| + e −43*|1+.001+x* | * sin (1.5 − x) + (cosx+0.001 + ex-1250+ sin (1-0.9x)|) * x - 4.68x4. For Case (a) (that is, Q = 0.3), use the stencil in Fig. 1. For Case (b), calculate with both the stencils in Fig. 1 and Fig 2. For all the three cases, show a table as well as a plot of O versus x. Discuss your results. Use MATLAB and hand in the MATLAB codes. 1 0=0 x=0 2 3 4 0=1 x=1 Fig 1 1 2 3 4 5 6 7 8 9 10 11 0=0 x=0 0=1 x=1 Fig 2

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The y and z coordinates and the perpendicular distance d from point O to the line of action of F .

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