Explanation Given information : The weight of the sign W is 50 lb. Show the free body diagram of the sign, as in Figure (1). Use Figure (1). Determine the position vector for point DE . r D E = r E − r D = ( x E i + y E j + z E k ) − ( x D i + y D j + z D k ) = ( x E − x D ) i + ( y E − y D ) j + ( z E − z D ) k (I) Here, the position vector for point E is r E , the position vector for point D is r E , the distance of point E from the origin along the x -direction is x E , the distance of point E from the origin along the y -direction is y E , the distance of point E from the origin along the z -direction is z E , the distance of point D from the origin along the x -direction is x D , the distance of point D from the origin along the y -direction is y D , the distance of point D from the origin along the z -direction is z D , the vector component along the x -direction is i, the vector component along the y -direction is j, and the vector component along the z -direction is k. Determine the modulus of the position vector for point DE . | r D E | = ( x E − x D ) 2 + ( y E − y D ) 2 + ( z E − z D ) 2 (II) Express the tension force T D E as follows: T D E = T D E [ r D E | r D E | ] (III) Determine the position vector for point BC . r B C = r B − r C = ( x B i + y B j + z B k ) − ( x C i + y C j + z C k ) = ( x B − x C ) i + ( y B − y C ) j + ( z B − z C ) k (IV) Here, the position vector for point C is r C , the position vector for point B is r B , the distance of point B from the origin along the x -direction is x B , the distance of point B from the origin along the y -direction is y B , the distance of point B from the origin along the z -direction is z B , the distance of point C from the origin along the x -direction is x C , the distance of point C from the origin along the y -direction is y C , and the distance of point C from the origin along the z -direction is z C . Determine the modulus of the position vector for point BC . | r B C | = ( x C − x B ) 2 + ( y C − y B ) 2 + ( z C − z B ) 2 (V) Express force F B C as follows: T B C = T B C [ r B C | r B C | ] (VI) Along the horizontal direction : Apply the equation of equilibrium in the x axis. ∑ F x = 0 1 3 T D E − 1 3 T B C + A x = 0 (VII) Along the vertical direction : Apply the equation of equilibrium in the y axis. ∑ F y = 0 − 2 3 T D E − 2 3 T B C + A y = 0 (VIII) Along the z direction : Apply the equation of equilibrium in the z axis. ∑ F z = 0 2 3 T D E + 2 3 T B C − W = 0 (IX) Take the moment about point A . Apply the equation of equilibrium in the x axis. ∑ M x = 0 ( M A ) x + 2 3 T D E ( a ) + 2 3 T B C ( a ) − W ( a ) = 0 (X) Here, the horizontal distance between T D E , T B C and W and point A is a . Apply the equation of equilibrium in the y axis. ∑ M y = 0 ( M A ) y − 2 3 T D E ( b ) + 2 3 T B C ( a ) + W ( c ) = 0 (XI) Here, the vertical distance between T D E and T B C and point A is b and a , and the vertical distance between W and point A is c . Apply the equation of equilibrium in the z axis. ∑ M z = 0 − 1 3 T D E ( d ) − 2 3 T D E ( e ) + 1 3 T B C ( d ) + 2 3 T B C ( d ) = 0 (XII) Here, the distance between T D E and T B C and point A about the z direction is d and e . Conclusion : Substitute –3.0 m for x E , –4.0 m for x D , –2.0 m for y E , 0 m for y D , 0 m for z E , and –2.0 m for z D in Equation (I). r D E = ( − 3.0 − ( − 4.0 ) ) i + ( − 2.0 − 0 ) j + ( 0 − ( − 2.0 ) ) k = ( 1i − 2 j + 2 k ) m Substitute –3.0 m for x E , –4.0 m for x D , –2.0 m for y E , 0 m for y D , 0 m for z E , and –2.0 m for z D in Equation (II). | r D E | = ( − 3.0 − ( − 4.0 ) ) 2 + ( − 2.0 − 0 ) 2 + ( 0 − ( − 2 ) ) 2 = 3.0 m Substitute ( 1i − 2 j + 2 k ) m for r D E and 3

Engineering Mechanics: Statics and Study Pack (13th Edition)

13th Edition

ISBN: 9780133027990

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Question

Chapter 5.7, Problem 82P

To determine

The tension in the cables and the components of reaction acting on the smooth collar at A necessary to hold the 50 lb sign in equilibrium.

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Chapter 5.7, Problem 81P

Chapter 5.7, Problem 83P

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5: The structure shown was designed to support a30-kN load. It consists of a boom AB with a 30 x 50-mmrectangular cross section and a rod BC with a 20-mm-diametercircular cross section. The boom and the rod are connected bya pin at B and are supported by pins and brackets at A and C,respectively.1. Calculate the normal stress in boom AB and rod BC,indicate if in tension or compression.2. Calculate the shear stress of pins at A, B and C.3. Calculate the bearing stresses at A in member AB,and in the bracket.

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3: A steel strut S serving as a brace for a boat hoist transmits a compressive force P = 54 kN to the deck of a pier as shown in Fig. STR-08. The strut has a hollow square cross section with a wall thickness t =12mm and the angle θ between the strut and the horizontal is 40°. A pin through the strut transmits the compressive force from the strut to two gusset plates G that are welded to the base plate B. Four anchor bolts fasten the base plate to the deck. The diameter of the pin is 20mm, the thickness of the gusset plates is 16mm, the thickness of the base plate is 8mm, and the diameter of the anchor bolts is 12mm. Disregard any friction between the base plate and the deck.1. Determine the shear stress in the pin, in MPa and the shear stress in the anchor bolts, in MPa.2. Determine the bearing stress in the strut holes, in MPa.

Chapter 5 Solutions

Engineering Mechanics: Statics and Study Pack (13th Edition)

Ch. 5.2 - Prob. 1P Ch. 5.2 - Prob. 2P Ch. 5.2 - Prob. 3P Ch. 5.2 - Prob. 4P Ch. 5.2 - Prob. 5P Ch. 5.2 - Prob. 6P Ch. 5.2 - Prob. 7P Ch. 5.2 - Prob. 8P Ch. 5.2 - Prob. 9P Ch. 5.4 - Determine the horizontal and vertical components...

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