Ch 4. Moments/Equivalent Systems Multimedia Engineering Statics Moment2-D Scalar Moment3-D Scalar Moment3-D Vector Couples and Equiv. System Distributed Loads, Intro
 Chapter 1. Basics 2. Vectors 3. Forces 4. Moments 5. Rigid Bodies 6. Structures 7. Centroids/Inertia 8. Internal Loads 9. Friction 10. Work & Energy Appendix Basic Math Units Sections Search eBooks Dynamics Statics Mechanics Fluids Thermodynamics Math Author(s): Kurt Gramoll ©Kurt Gramoll

 STATICS - CASE STUDY SOLUTION Direction of Forces Although scalars can be used to solve this problem, vector notation will be used. The standard directions will be used for the coordinate system. The two rocket forces that turn the firework are      F1 = 0i - Tj + 0k      F2 = -Tsinαi + Tcosαj + 0k where T is the trust, 0.5 lb and α is the misaligned rocket angle of 10o. The position vectors from point O to each force's point of application are      r1 = -3i + 0j + 0k      r2 = 3i + 0j + 0k Solution of a) Equivalent Force and Moment   Calculating Equivalent Force The system of two forces can be replaced by an equivalent system consisting of a force FR and couple MR applied at point O. The force FR is found by summing the individual forces.      FR = ΣF = F1 + F2           = -Tsinαi + T(cosα - 1)j + 0k Substituting the known values, α = 10o and T = 0.5 lb, gives      FR = -0.08682i - 0.007596j + 0k lb The moment MR is the sum of the individual moments.      MRo = ΣMo = r1× F1 + r2× F2 The cross products can be written as determinants, giving,               = 0i - 0j + 3T(1 + cosα)k Substituting the known values gives      MRo = 0i - 0j + 2.977k in-lb Solution of b) Single Force Resultant Since all the forces act in the same plane, there is a single force resultant for this problem. The resultant force is the same FR as before but it is located at a different point Q, which must satisfy the equation      MRo = rOQ × FR Writing this equation in determinant form and substituting known vectors gives,            2.977k = [0 - rOQz (-0.007596)]i                 - [0 - rOQz (-0.08682)]j +                   + [(rOQx (-0.007596) - rOQy(-0.08682)]k From the i and j terms, rOQz must be zero. However, the k term gives      2.977 = -0.007596 rOQx + 0.08682 rOQy From inspecting the diagram, it is noted that point Q was chosen to be directly vertical from O. Thus, the rOQx must be zero. This gives,      2.977 = 0.08682 rOQy Therefore      rOQ = 0i + 34.29j + 0k  in Applying the single force FR at the point Q has the same effect as the two original forces F1 and F2.

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