Explanation The free-body diagram is shown in figure 1. The sum of the moments of the forces along AE must be zero. Σ M → A E = 0 λ ^ A E ⋅ ( r → D / A × T → D F ) + λ ^ A E ⋅ ( r → B / A × P → ) = 0 (I) Here, Σ M A E is the sum of the moments of forces along AE, λ ^ A E is the unit vector along AE, T → D F is the tension, r → D / A and r → B / A are position vectors and P → is the weight. Write the expression for λ ^ A E . λ ^ A E = A E → A E (II) Here, A E is the magnitude of the vector A E → . Write the expression of A E → . A E → = ( 480 mm ) i ^ + ( 160 mm ) j ^ − ( 240 mm ) k ^ Find the magnitude of A E → . A E = ( 480 mm ) 2 + ( 160 mm ) 2 + ( − 240 mm ) 2 = 560 mm Substitute [ ( 480 mm ) i ^ + ( 160 mm ) j ^ − ( 240 mm ) k ^ ] for A E → and 560 mm for A E in equation (II) to find λ ^ A E . λ ^ A E = ( 480 mm ) i ^ + ( 160 mm ) j ^ − ( 240 mm ) k ^ 560 mm = 6 i ^ + 2 j ^ − 3 k ^ 7 (III) Write the expressions for the position vectors

Vector Mechanics for Engineers: Statics and Dynamics

12th Edition

ISBN: 9781259638091

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial …

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Chapter 4.3, Problem 4.138P

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The tension in the wire when 640 N load is applied at B.

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Chapter 4.3, Problem 4.137P

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Vector Mechanics for Engineers: Statics and Dynamics

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The tension in the wire when 640 N load is applied at B.

Solution Summary: The author explains the tension in the wire when 640N load is applied at B. The sum of the moments of forces along AE must be zero.

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