Explanation Given: The value of P is 20 lb . Concept used: Draw the free body diagram to illustrate the system of forces acting on the system. Write the expression for the position vector of point E with respect to D. d → D E = ( 5 j ^ − 10 k ^ ) in Here, d → D E is the position vector of point E with respect to D. Write the expression for magnitude of position vector E with respect to D. | d D E | = ( 5 ) 2 + ( − 10 ) 2 = 11.18 in Here, | d D E | is the magnitude of position vector E with respect to D. Write the expression for unit vector along line DE. n ^ D E = d → D E | d D E | ....... (1) Here, n ^ D E is the unitvector along line DE. Write the expression to represent the force acting along line DE in vector. F → D E = | F D E | n ^ D E ....... (2) Here, F → D E is the force acting along line DE in vector and | F D E | is the magnitude of force along the line DE. Write the expression for the position vector of point E with respect to O. d → O E = ( 15 i ^ + 5 j ^ ) in Here, d → O E is the position vector of point E with respect to O. Write the expression to represent the force acting at point C in vector. F → C = ( − 16 j ^ − P k ^ ) lb Here, F → C is the force acting at point C in vector. Write the expression for the position vector of point C with respect to O. d → O C = ( 30 i ^ ) in Here, d → O C is the position vector of point C with respect to O. Write the expression for equivalent couple acting on the system. M = d → O E × F → D E + d → O C × F → C ....... (3) Here, M is the equivalent couple acting on the system. 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 ( 5 j ^ − 10 k ^ ) in for d → D E and 11.18 in for | d D E | in equation (1) n ^ D E = ( 5 j ^ − 10 k ^ ) 11.18 = ( 0.4472 j ^ − 0.8944 k ^ ) Substitute ( 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 60P

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A single equivalent couple acting on the system by replacing the applied loads and identify its axis.

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Chapter 3.3, Problem 59P

Chapter 3.3, Problem 61P

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Chapter 3 Solutions

Statics and Mechanics of Materials

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