Explanation Take all vectors along the x axis and y axis as positive. The Diagram of component of force F along A B is shown in figure1. From the figure 1, relation between force and distance given. ( F A B ) x 0.65 in . = ( F A B ) y 0.55 in (I) The free body diagram of the blade A C E . From figure 2, write the expression for the moment at C . ∑ M C = ∑ F × D ⊥ (II) Here, ∑ M C is the net moment at C , F is the force, and D ⊥ is the perpendicular distance between the C and the point where force is experienced. Above equation implies that net moment at any point is the sum of product of each force acting on the system and perpendicular distance of the force and the point. Thus, the expression of net clockwise moment M C at point C . ∑ M C = 0 (III) Here, ∑ M C is the sum of all moment of force at C . From the figure2, write the equilibrium expression for the total anticlockwise moment acting at C . ( 300 lb ) ( 1.6 in . ) − ( F A B ) x ( 0.5 in . ) − ( F A B ) y ( 1.4 in . ) = 0 (IV) Here, ( F A B ) x is the component of force along A B and ( F A B ) y is the component of force along A B . The free body diagram of the blade lower handle is shown in figure 3. From figure 3, write the expression for the moment at D . ∑ M D = ∑ F × D ⊥ (III) Here, ∑ M D is the net moment at d , F is the force, and D ⊥ is the perpendicular distance between the D and the point where force is experienced. Above equation implies that net moment at any point is the sum of product of each force acting on the system and perpendicular distance of the force and the point. Thus, the expression of net clockwise moment M D at point D . ∑ M D = 0 Here, ∑ M D is the sum of all moment of force at D . From the figure3, write the equilibrium expression for the total anticlockwise moment acting at D . ( F A B ) y ( 0

Loose Leaf for Vector Mechanics for Engineers: Statics and Dynamics

12th Edition

ISBN: 9781259977206

Author: BEER, Ferdinand P., Johnston Jr., E. Russell, Mazurek, David, Cornwell, Phillip J., SELF, Brian

Publisher: McGraw-Hill Education

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Chapter 6, Problem 6.175RP

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The tension in the fusible link DE and the force exerted on member BCE at B.

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Chapter 6, Problem 6.174RP

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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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Loose Leaf for Vector Mechanics for Engineers: Statics and Dynamics

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The tension in the fusible link D E and the force exerted on member B C E at B .

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