Explanation Given information: The symmetric coil tong supports the coil which has a mass of 800 kg and center of mass at G . The coil exerts only vertical reactions at K and L . 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 800 kg for m and 9.81 m/sec 2 for g . W = ( 800 × 9.81 ) N Segment KL : Show the free-body diagram of the segment KL as in Figure 1. Using Figure 1, Moment about point L : Determine the force at point K by taking the moment about point L . ∑ M L = 0 800 × 9.81 × x − F K ( 2 x ) = 0 (I) Conclusion: Solve Equation (I). 7 , 848 x − F K ( 2 x ) = 0 F K = 3 , 924 N Segment BAK : Show the free-body diagram of the segment BAK as in Figure 2. Using Figure 2, Moment about point A : Determine the force in the member BD by taking moment about point A . ∑ M A = 0 F B D sin 45 ° ( 100 ) + F B D cos 45 ° ( 100 ) − F K ( 50 ) = 0 (II) Along the vertical direction: Determine the vertical reaction at point A by resolving the vertical component of forces. ∑ F y = 0 A y − F K − F B D sin 45 ° = 0 (III) Along the horizontal direction: Determine the horizontal reaction at point A by resolving the horizontal component of forces. ∑ F x = 0 A x − F B D cos 45 ° = 0 (IV) Conclusion: Substitute 3,924 N for F K Equation (II). F B D sin 45 ° ( 100 ) + F B D cos 45 ° ( 100 ) − 3 , 924 ( 50 ) = 0 70.71 F B D + 70.71 F B D − 196 , 200 = 0 F B D = 196 , 200 141.42 F B D = 1 , 387.36 N Substitute 3,924 N for F K and 1,387.36 N for F B D in Equation (III). A y − 3 , 924 − 1 , 387.36 sin 45 ° = 0 A y = 4 , 905.01 N Substitute 1,387.36 N for F B D in Equation (IV). A x − 1 , 387.36 cos 45 ° = 0 A x = 981.01 N Segment ACE : Show the free-body diagram of the segment ACE as in Figure 3. Using Figure 3, Moment about point E : Determine the force in the member CF by taking the moment about point E . ∑ M E = 0 A y sin 45 ° ( 700 ) − A x sin 45 ° ( 700 ) − F C F cos 15 ° ( 300 ) = 0 (V) Along the vertical direction: Determine the vertical reaction at point E by resolving the vertical component of forces

13th Edition

ISBN: 9780132915540

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 6.6, Problem 109P

To determine

The horizontal and vertical components of force the linkage exerts on plate DEIJH at points D and E.

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Chapter 6.6, Problem 108P

Chapter 6.6, Problem 110P

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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 horizontal and vertical components of force the linkage exerts on plate DEIJH at points D and E .

The horizontal and vertical components of force the linkage exerts on plate DEIJH at points D and E.

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Chapter 6 Solutions