Explanation Given information: The tractor boom supports the uniform mass of 500 kg in the bucket which has a center of mass at G . The load is supported equally on each side of the tractor by a similar mechanism. Assumption: Consider the state of member as tension where the force is pulling the member and as compression where the force is pushing the member. Consider the force indicating the right side as positive and the left side as negative in horizontal components of forces. Consider the force indicating upward is taken as positive and downward as negative in vertical components of forces. Consider the clockwise moment as negative and the anti-clockwise moment as positive wherever applicable. Apply the equation of equilibrium wherever applicable. Acceleration due to gravity is g = 9.81 m/sec 2 . Conversion: Convert the mass into weight using the relation: W = m g Substitute 500 kg for m and 9.81 m/sec 2 for g . W = ( 500 × 9.81 ) N Bucket: Show the free-body diagram of the bucket as in Figure 1. Using Figure 1, Moment about point E : Determine the force in the member AB by taking the moment about point E . ∑ M E = 0 500 × 9.81 2 × ( 0.1 ) − F A B ( 0.25 ) = 0 (I) Along the vertical direction: Determine the vertical reaction at point A by resolving the vertical component of forces. ∑ F y = 0 E y − 500 × 9.81 2 = 0 (II) Along the horizontal direction: Determine the horizontal reaction at point A by resolving the horizontal component of forces. ∑ F x = 0 F A B − E x = 0 (III) Resultant force at point E : Determine the resultant force at point E using the equation: F E = ( E x ) 2 + ( E y ) 2 (IV) Conclusion: Solve Equation (I). 245.25 − 0.25 F A B = 0 F A B = 981 N Solve Equation (II). E y = 2 , 452.5 N Substitute 981 N for F A B in Equation (III). 981 − E x = 0 E x = 981 N Substitute 981 N for E x and 2,452.5 N for E y in Equation (IV). F E = 981 2 + 2 , 452.5 2 = 2 , 641.42 N = 2 , 641.42 N × 1 kN 1,000 N = 2.64 kN Thus, the force in the hydraulic cylinder AB is F A B = 981 N _ . Thus, the resultant force at pin E is F E = 2.64 kN _ . Entire segment: Show the free-body diagram of the entire segment as in Figure 2. Using Figure 2, Determine the value of θ : tan θ = Opposite side Adjacent side (V) Moment about point F : Determine the force in the member CD by taking the moment about point F . ∑ M F = 0 500 × 9.81 2 × 2.80 − F C D cos θ ( 0.7 ) + F C D sin θ ( 1

13th Edition

ISBN: 9780132915540

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 6.6, Problem 102P

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The force in each hydraulic cylinder AB and CD and the resultant force at pins E and F.

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Chapter 6.6, Problem 101P

Chapter 6.6, Problem 103P

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Chapter 6 Solutions

Engineering Mechanics: Statics

Ch. 6.3 - State if the members are in tension or...Ch. 6.3 - State if the members are in tension or...Ch. 6.3 - Prob. 3FP Ch. 6.3 - Determine the greatest load P that can be applied...Ch. 6.3 - Identify the zero-force members in the truss....Ch. 6.3 - State if the members are in tension or...Ch. 6.3 - Prob. 1P Ch. 6.3 - Prob. 2P Ch. 6.3 - Determine the force in each member of the truss,...Ch. 6.3 - Determine the force in each member of the truss,...

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The force in each hydraulic cylinder AB and CD and the resultant force at pins E and F .

The force in each hydraulic cylinder AB and CD and the resultant force at pins E and F.

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Chapter 6 Solutions