Explanation Concept used: Redraw the diagram, Consider point C as the origin of the given system. Due to frictionless pulley tension in the cable DAE remain same. Write the expression for force vector at point A. F → A = − 320 j ^ lb …… (1) Here, F → A is the force vector at point A. Write the coordinate of point A as ( 0 , 0 , 48 ) in Write the expression for position vector of point A. r → A = 48 k ^ in Here, r → A is the position vector of point A. Write the coordinate of point B as ( 0 , 0 , 30 ) in Write the expression for position vector of point B. r → B = 30 k ^ in Here, r → B is the position vector of point B. Write the coordinate of point D as ( − 20 , 0 , 0 ) in Write the expression for position vector of point D. r → D = − 20 i ^ in Here, r → D is the position vector of point D. Write the coordinate of point E as ( 0 , 20 , 0 ) in Write the expression for position vector of point E. r → E = 20 j ^ in Here, r → E is the position vector of point E. Write the coordinate of point F as ( 16 , 0 , 0 ) in Write the expression for position vector of point F. r → F = 16 i ^ in Here, r → F is the position vector of point F. Write the expression for position vector of point D with respect to point A. r → D A = ( − 20 ) i ^ + ( 0 − 0 ) j ^ + ( 0 − 48 ) k ^ = ( − 20 i ^ − 48 k ^ ) in Here, r → D A is the position vector of point D with respect to point A. Write the expression for position vector of point E with respect to point A. r → E A = ( 0 − 0 ) i ^ + ( 20 − 0 ) j ^ + ( 0 − 48 ) k ^ = ( 20 j ^ − 48 k ^ ) in Here, r → E A is the position vector of point E with respect to point A. Write the expression for position vector of point F with respect to point B. r → F B = ( 16 − 0 ) i ^ + ( 0 − 0 ) j ^ + ( 0 − 30 ) k ^ = ( 16 i ^ − 30 k ^ ) in Here, r → F B is the position vector of point F with respect to point B. Write the expression for unit vector along point A to point D. r ^ D A = r → D A | r → D A | Here, r ^ D A is the unit vector along point A to point D and | r → D A | is the magnitude of r → D A . Substitute ( − 20 i ^ − 48 k ^ ) in for r → D A in above equation. r ^ D A = ( − 20 i ^ − 48 k ^ ) ( − 20 ) 2 + ( − 48 ) 2 = ( − 20 i ^ − 48 k ^ ) 52 = ( − 0.384 i ^ − 0.923 k ^ ) Write the expression for force in vector form due to tension in AD cable. T → A D = T 1 ⋅ r ^ D A Here, T → A D is the force in vector form due to tension in AD cable and T 1 is the magnitude of tension in cable AD. Substitute ( − 0.384 i ^ − 0.923 k ^ ) for r ^ D A in above equation. T → A D = T 1 ⋅ ( − 0.384 i ^ − 0.923 k ^ ) = ( − 0.384 T 1 i ^ − 0.923 T 1 k ^ ) Write the expression for unit vector along point A to point E. r ^ E A = r → E A | r → E A | Here, r ^ E A is the unit vector along point A to point E and | r → E A | is the magnitude of r → E A . Substitute ( 20 j ^ − 48 k ^ ) in for r → E A in above equation. r ^ E A = ( 20 j ^ − 48 k ^ ) ( 20 ) 2 + ( − 48 ) 2 = ( 20 j ^ − 48 k ^ ) 52 = ( 0.384 j ^ − 0.923 k ^ ) Write the expression for force in vector form due to tension in AE cable. T → A E = T 1 ⋅ r ^ E A Here, T → A E is the force in vector form due to tension in AE cable. Substitute ( 0.384 j ^ − 0.923 k ^ ) in for r ^ D A in above equation. T → A E = T 1 ⋅ ( 0.384 j ^ − 0.923 k ^ ) = ( 0.384 T 1 j ^ − 0.923 T 1 k ^ ) Write the expression for unit vector along point B to point F. r ^ F B = r → F B | r → F B | Here, r ^ F B is the unit vector along point B to point F and | r → F B | is the magnitude of r → F B . Substitute ( 16 i ^ − 30 k ^ ) for r → F B in above equation. r ^ F B = ( 16 i ^ − 30 k ^ ) ( 16 ) 2 + ( − 30 ) 2 = ( 16 i ^ − 30 k ^ ) 34 = ( 0.47 i ^ − 0.882 k ^ ) Write the expression for force in vector form due to tension in BF cable. T → B F = T 2 ⋅ r ^ F B Here, T → B F is the force in vector form due to tension in BF cable and T 2 is the magnitude of tension in cable BF. Substitute ( 0.47 i ^ − 0.882 k ^ ) for r ^ F B in above equation. T → B F = T 2 ⋅ ( 0.47 i ^ − 0.882 k ^ ) = ( 0.47 T 2 i ^ − 0.882 T 2 k ^ ) Write the expression for resultant force. R → = T → A D + T → A E + T → B F + F → C + F → B …… (2) Here, R → is the resultant force and F → C is the reaction force at point C

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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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 62P

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Tension in each cable and reaction at support C.

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Tension in each cable and reaction at support C.

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