Explanation Given information: The mass ( m ) of the rod is 20 kg. Show the free body diagram as in Figure (1). Express the force vector F A of A as follows: F A = − A x i + A y j + A z k (1) Here, 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 position vector for point BC . r B C = r C − r B = ( x C i + y C j + z C k ) − ( x B i + y B j + z B k ) = ( x C − x B ) i + ( y C − y B ) j + ( z C − z B ) k (2) Here, the position vector at point A is r A , the position vector at point B is r 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 , the distance of point C from the origin along the z direction is z C , 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 , and the distance of point B from the origin along the z direction is z B . Determine the modulus of the position vector at point BC . | r B C | = ( x C − x B ) 2 + ( y C − y B ) 2 + ( z C − z B ) 2 (3) Express the tension in cable BC as follows: T B C = T B C ( r B C | r B C | ) (4) Express the normal reaction at point B as follows: N B = N B i (5) Determine the position vector at point AG . r A G = r G − r A = ( x G i + y G j + z G k ) − ( x A i + y A j + z A k ) = ( x G − x A ) i + ( y G − y A ) j + ( z G − z A ) k (6) Here, the position vector at point G is r G , the position vector at point A is r A , the distance of point G from the origin along the x direction is x G , the distance of point G from the origin along the y direction is y G , the distance of point G from the origin along the z direction is z G , the distance of point A from the origin along the x direction is x A , the distance of point A from the origin along the y direction is y A , and the distance of point A from the origin along the z direction is z A . Determine the position vector at point AB . r A B = r B − r A = ( x B i + y B j + z B k ) − ( x A i + y A j + z A k ) = ( x B − x A ) i + ( y B − y A ) j + ( z B − z A ) k (7) Use Figure (1). Express the force equation of equilibrium as follows: ∑ F = 0 F A + T B C + N B + ( W ) k = 0 (8) Here, the acceleration due to gravity is g . Calculation: Substitute 0 m for x C , 0 m for x B , 0 m for y C , 1.0 m for y B , 2.5 m for z C , and 2.0 m for z B in Equation (2). r B C = ( 0 − 0 ) i + ( 0 − 1.0 ) j + ( 2.5 − 2.0 ) k = ( − 1.0 j + 0.5 k ) m Substitute 0 m for x C , 0 m for x B , 0 m for y C , 1.0 m for y B , 2.5 m for z C , and 2.0 m for z B in Equation (3). | r B C | = ( 0 − 0 ) 2 + ( 0 − 1 ) 2 + ( 2.5 − 2 ) 2 = 1 2 + 0.5 2 = 1.25 m Substitute ( − 1.0 j + 0.5 k ) m for r B C and 1.25 m for | r B C | in Equation (4). T B C = T B C ( − 1.0 j + 0.5 k 1.25 ) = − 1.0 1.25 T B C j+ 0.5 1.25 T B C k Substitute 0.75 m for x G , 1.5 m for x A , 0.5 m for y G , 0 m for y A , 1.0 m for z G , and 0 m for z A in Equation (6). r A G = ( 0.75 − 1.5 ) i + ( 0.5 − 0 ) j + ( 1.0 − 0 ) k = ( − 0.75 i + 0.5 j + 1.0 k ) m Substitute 0 m for x B , 1.5 m for x A , 1.0 m for y B , 0 m for y A , 2.0 m for z B , and 0 m for z A in Equation (7). r A B = ( 0 − 1.5 ) i + ( 1.0 − 0 ) j + ( 2.0 − 0 ) k = ( − 1.5 i + 1.0 j + 2.0 k ) m Substitute − A x i + A y j + A z k for F A , 20 kg for m , and 9.81 m / s 2 for g , − 1.0 1.25 T B C j+ 0.5 1.25 T B C k for T B C , N B i for N B , in Equation (8). ( − A x i + A y j + A z k ) + ( − 1.0 1.25 T B C j+ 0.5 1.25 T B C k ) + ( N B i ) + 20 ( 9.81 ) = 0 ( − A x i + N B ) i + ( A y − 1 1.25 T B C ) j + ( A z + 0.5 1.25 T B C − 196.2 ) k = 0 Equate i, j, and k components in the equation as follows: − A x + N B = 0 (9) A y − 1 1

Statics and Mechanics of Materials (5th Edition)

5th Edition

ISBN: 9780134382593

Author: Russell C. Hibbeler

Publisher: PEARSON

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Free Body Diagram

The system of forces acting on a body tends to either rotate it or make the body undergo motion. In general, when an external agent exerts a force on a body, the body undergoes translational motion and rotational motion. The body has six degrees of freed…

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Chapter 4.6, Problem 30P

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Find the components of reaction at point A, the tension in the cable, and the reaction at point B.

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Chapter 4.6, Problem 29P

Chapter 4.6, Problem 31P

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Chapter 4 Solutions

Statics and Mechanics of Materials (5th Edition)

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Find the components of reaction at point A , the tension in the cable, and the reaction at point B .

Solution Summary: The author shows the free body diagram as in Figure (1). The force vector F_A of A is i, j, and k.

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Find the components of reaction at point A, the tension in the cable, and the reaction at point B.

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Chapter 4 Solutions