Explanation Concept used: Draw the diagram for the system as shown below: Write the expression for the position vector for point B with respect to C. d → C B = ( 160 i ^ + 144 j ^ − 192 k ^ ) mm Here, d → C B is the position vector for point B with respect to C. Write the expression for the position vector for point G with respect to F. d → G F = ( − 160 i ^ ) mm Here, d → G F is the position vector for point G with respect to F. Write the expression for the position vector for point D with respect to E. d → E D = ( 192 k ^ ) mm Here, d → E D is the position vector for point D with respect to E. Write the expression for the position vector for point F with respect to B. d → B F = ( − 160 i ^ − 72 j ^ + 96 k ^ ) mm Here, d → B F is the position vector for point F with respect to B. Write the expression for the magnitude of position vector for point F with respect to B. | d B F | = ( − 160 ) 2 + ( − 72 ) 2 + ( 96 ) 2 = 40000 = 200 mm Here, | d B F | is the magnitude of position vector for point F with respect to B. Write the expression for unit vector along line BF. n ^ B F = d → B F | d B F | ……. (1) Here, n ^ B F is the unit vector along line BF. Write the expression for representation of force in vector. F → 2 = | F 2 | ⋅ n ^ B F …… (2) Here, F → 2 is the representation of force in vector and | F 2 | is the magnitude of applied force. Write the expression for equivalent couple. M = d → E D × F → 1 + d → G F × F → 2 + d → C B × F → 3 …… (3) Here, M is the equivalent couple. Write the expression for magnitude of equivalent couple. | M | = M x 2 + M y 2 + M z 2 …… (4) Here, | M | is the magnitude of equivalent couple, M x is the x component of equivalent couple, M y is the y component of equivalent couple and M z is the z component of equivalent couple. Write the expression for direction of equivalent couple from x axis. θ x = cos − 1 ( M x | M | ) …… (5) Here, θ x is the direction of equivalent couple from x axis. Write the expression for direction of equivalent couple from y axis. θ y = cos − 1 ( M y | M | ) …… (6) Here, θ y is the direction of equivalent couple from y axis. Write the expression for direction of equivalent couple from z axis. θ z = cos − 1 ( M z | M | ) …… (7) Here, θ z is the direction of equivalent couple from z axis. Calculation: Substitute ( − 160 i ^ − 72 j ^ + 96 k ^ ) mm for d → B F and 200 mm for | d B F | in equation (1) n ^ B F = ( − 160 i ^ − 72 j ^ + 96 k ^ ) 200 = ( − 0.8 i ^ − 0.36 j ^ + 0.48 k ^ ) Substitute ( − 0.8 i ^ − 0.36 j ^ + 0.48 k ^ ) for n ^ B F and 50 N for | F 2 | in equation (2) F → 2 = 50 ( − 0

2nd Edition

ISBN: 9780073398167

Author: Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David Mazurek

Publisher: McGraw-Hill Education

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Chapter 3.3, Problem 58P

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By replacing the couple with equivalent couple and specify its magnitude and direction.

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Chapter 3.3, Problem 57P

Chapter 3.3, Problem 59P

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Chapter 3 Solutions

Statics and Mechanics of Materials

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