Explanation The direction of force towards up and right is considered as positive and the direction of torque counter clockwise is considered as positive. The free-body diagram corresponding to the arrangement in Fig. P6.76 is given in Figure 1. The force in the member B D is represented as F B D and the components of reaction at C are represented by C x and C y . It can be observe that the force that B D exerts on A B C is directed along the line B D since B D is a two-force member. Write the expressions for equilibrium for the given system using the conditions on moments M and force F . The counter clockwise moments at C sum up to zero. Σ M C = 0 (I) The sum of y component of all forces is equal to zero. Σ F y = 0 (II) The counter clockwise moments at B sum up to zero. Σ M B = 0 (III) Conclusion: Apply the condition in equation (I) to the free-body diagram in Figure 1 and solve for F B D . ( 90 lb ) ( 12 in . ) − ( 4 5 F B D ) ( 4.5 in . ) = 0 F B D = + 300 lb | F B D | = 300 lb Tension Apply the condition in equation (II) to the free-body diagram in Figure 1

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ISBN: 9781259977268

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek

Publisher: McGraw-Hill Education

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Chapter 6.3, Problem 6.76P

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The force in the member BD and the components of reaction at C.

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