(a) Explanation The free body diagram for arm C D is given below Write the expression for moment at C ∑ M C = F × d ⊥ (I) Here, F is the force, d ⊥ is the perpendicular distance between C and point where force is acting, and M C is the moment at C At equilibrium the moment at C will be zero ∑ M C = F × d ⊥ = 0 (II) Write the complete expression for the total moment at C using figure ∑ M C = B ( d 1 ) − T ( d 2 ) cos θ = 0 (III) Here, d 1 is vertical the distance from B to C , T is the force at D , d 2 is the vertical distance from C to D ,and θ is the angle made by C and the normal passing through C . Consider the case of crank A B as shown in figure 3 The moment at A is ∑ M A = M − B ( d 1 ) sin ( θ + 30 ° ) (IV) Conclusion: The required values can be calculated by considering the geometry of the free body diagram. Consider the triangle B A E in figure 3, the side B E is B E = 100 cos 60 ° = 86.6 mm Thus side C E is C E = 400 + 100 sin 60 ° = 450 mm The tangent of angle θ using figure 3 tan θ = 86.6 450 θ = tan − 1 86.6 450 = 10.89 ° Form figure 3 the length of line B C is B C = ( 450 ) 2 + ( 86 (b) To determine The couple M applied to the crank A B to mechanism in equilibrium when α = 90 ° .

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Chapter 6.4, Problem 6.134P

(a)

To determine

The couple M applied to the crank AB to mechanism in equilibrium when α=60°.

(b)

To determine

The couple M applied to the crank AB to mechanism in equilibrium when α=90°.

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Chapter 6.4, Problem 6.133P

Chapter 6.4, Problem 6.135P

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The couple M applied to the crank A B to mechanism in equilibrium when α = 60 ° .

Solution Summary: The free body diagram for arm CD is given below. Write the expression for moment at C using figure 3.