Explanation Given information: The mass of each bar ( m ) is 3 kg . The length of each bar ( l ) is 450 mm . Calculation: Consider the acceleration due to gravity ( g ) is 9.81 m / s 2 . Refer the problem 16.21. The acceleration of all points of vertical rod AB ( a ) is 6.3057 m/s 2 at an angle of 40 ° . Calculate the acceleration rod AB at x direction ( a x ) : a x = 6.3057 × cos 40 ° = 4.8304 m / s 2 → Calculate the acceleration rod AB at y direction ( a y ) : a y = 6.3057 × sin 40 ° = 4.0532 m / s 2 → Calculate the mass per unit length as m l : Substitute 3 kg for m and 450 mm for l . m l = 3 450 mm × 10 − 3 m 1 mm = 6.6667 kg / m Calculate the effective force per length at x direction: m l a x Substitute 6.6667 kg / m for m l and 4.8304 m / s 2 for a x . m l a x = 6.6667 × 4.8304 = 32.203 N / m → Calculate the effective force per length at y direction: m l a y Substitute 6

Vector Mechanics for Engineers: Statics and Dynamics

11th Edition

ISBN: 9780073398242

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

Publisher: McGraw-Hill Education

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Chapter 16.1, Problem 16.21P

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Draw the shear and bending-moment diagrams for the vertical rod AB.

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Chapter 16.1, Problem 16.20P

Chapter 16.1, Problem 16.22P

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