COSMOS Chapter 3 Solution 127

COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 127. For the smallest weight on the trailer so that the resultant force of the four weights acts over the axle at the intersection with the center line of the trailer, the added 0.6 × 0.6 × 1.2-m box should be placed adjacent to one of the edges of the trailer with the 0.6 × 0.6-m side on the bottom. The edges to be considered are based on the location of the resultant for the three given weights. Have ΣF : − ( 200 N ) j − ( 400 N ) j − (180 N ) j = R ∴ R = − ( 780 N ) j Have ΣM z : ( 200 N )( 0.3 m ) + ( 400 N )(1.7 m ) + (180 N )(1.7 m ) = ( 780 N )( x ) ∴ x = 1.34103 m Have ΣM x : ( 200 N )( 0.3 m ) + ( 400 N )( 0.6 m ) + (180 N )( 2.4 m ) = ( 780 N )( z ) ∴ z = 0.93846 m From the statement of the problem, it is known that the resultant of R from the original loading and the lightest load W passes through G, the point of intersection of the two center lines. Thus, ΣM G = 0. Further, since the lightest load W is to be as small as possible, the fourth box should be placed as far from G as possible without the box overhanging the trailer. These two requirements imply ( 0.3 m ≤ x ≤ 1 m ) (1.8 m ≤ z ≤ 3.7 m ) continued . COSMOS: Complete Online Solutions Manual Organization System Let x = 0.3 m, ΣM Gz : ( 200 N )( 0.7 m ) − ( 400 N )( 0.7 m ) − (180 N )( 0.7 m ) + W ( 0.7 m ) = 0 ∴ W = 380 N ΣM Gx : − ( 200 N )(1.5 m ) − ( 400 N )(1.2 m ) + (180 N )( 0.6 m ) + ( 380 N )( z − 1.8 m ) = 0 ∴ z = 3.5684 m < 3.7 m Let z = 3.7 m, ∴ acceptable ΣM Gx : − ( 200 N )(1.5 m ) − ( 400 N )(1.2 m ) + (180 N )( 0.6 m ) + W (1.7 m ) = 0 ∴ W = 395.29 N > 380 N Since the weight W found for x = 0.3 m is less than W found for z = 3.7 m, x = 0.3 m results in the smallest weight W. or W = 380 N at . ( 0.3 m, 0, 3.57 m )