(a) Explanation The arrangement is shown in Fig. P6.87. The free-body diagram of the entire frame is given in Figure 1. Write the expressions for equilibrium for the given system using the conditions on moments M and force F . The counter clockwise moments at B sum up to zero. Σ M B = 0 (I) The sum of y component of all forces is equal to zero. Σ F y = 0 (II) The sum of x component of all forces is equal to zero. Σ F x = 0 (III) Conclusion: Apply the condition in equation (I) to the free-body diagram in Figure 1 and solve for A y . − A y ( 5 in . ) − ( 60 lb ) ( 4 in . ) = 0 A y = − 48 lb Apply the condition in equation (II) to the free-body diagram in Figure 1 and solve for B y . B y − 60 lb − 48 lb = 0 B y = + 108 lb Apply the condition in equation (III) to the free-body diagram in Figure 1. A x + B x = 0 (IV) The load is applied at D . Since B C E is a two-force member (with forces applied at B and C only), the reaction at B is directed along C B . Therefore, the x component of B which is represented as B x is eqial to zero. B x = 0 Substitute 0 for B x in equation (IV) to find A x . A x + 0 = 0 A x = 0 Write the expression for the magnitude of the reaction A and reaction B (b) To determine The components of reaction at A and B if the 60 lb load is applied at E .

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Chapter 6.3, Problem 6.87P

(a)

To determine

The components of reaction at A and B if the 60 lb load is applied vertically downward at D.

(b)

To determine

The components of reaction at A and B if the 60 lb load is applied at E.

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Chapter 6.3, Problem 6.86P

Chapter 6.3, Problem 6.88P

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