COSMOS Chapter 3 Solution 137

COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 137. First, reduce the given force system to a force-couple at the origin. ΣF : FA + FG = R Have  ( 4 in.) i + ( 6 in.) j − (12 in.) k  ∴ R = (10 lb ) k + 14 lb   = ( 4 lb ) i + ( 6 lb ) j − ( 2 lb ) k 14 in.   R= and Have 56 lb ΣM O : ∑ ( rO × F ) + ∑ M C = MOR { } M OR = (12 in.) j × (10 lb ) k  + (16 in.) i × ( 4 lb ) i + ( 6 lb ) j − (12 lb ) k   (16 in.) i − (12 in.) j   ( 4 in.) i − (12 in.) j + ( 6 in.) k  + ( 84 lb ⋅ in.)   + ( 120 lb ⋅ in.)   20 in. 14 in.     ∴ M 0R = ( 221.49 lb ⋅ in.) i + ( 38.743 lb ⋅ in.) j + (147.429 lb ⋅ in.) k = (18.4572 lb ⋅ ft ) i + ( 3.2286 lb ⋅ ft ) j + (12.2858 lb ⋅ ft ) k . COSMOS: Complete Online Solutions Manual Organization System The force-couple at O can be replaced by a single force if the direction of R is perpendicular to M OR . To be perpendicular R ⋅ M OR = 0 Have R ⋅ M OR = ( 4i + 6 j − 2k ) ⋅ (18.4572i + 3.2286 j + 12.2858k ) = 0? = 73.829 + 19.3716 − 24.572 ≠0 ∴ System cannot be reduced to a single equivalent force. To reduce to an equivalent wrench, the moment component along the line of action of P is found. M1 = λ R ⋅ M OR λR = R R  ( 4i + 6 j − 2k )  =  ⋅ (18.4572i + 3.2286 j + 12.2858k ) 56   = 9.1709 lb ⋅ ft M1 = M1λ R = ( 9.1709 lb ⋅ ft )( 0.53452i + 0.80178 j − 0.26726k ) and And pitch p= M1 9.1709 lb ⋅ ft = = 1.22551 ft R 56 lb or p = 1.226 ft Have M 2 = M OR − M1 = (18.4572i + 3.2286 j + 12.2858k ) − ( 9.1709 )( 0.53452i + 0.80178 j − 0.26726k ) = (13.5552 lb ⋅ ft ) i − ( 4.1244 lb ⋅ ft ) j + (14.7368 lb ⋅ ft ) k Require M 2 = rQ/O × R (13.5552i − 4.1244 j + 14.7368k ) = ( yj + zk ) × ( 4i + 6 j − 2k ) = − ( 2 y + 6z ) i + ( 4z ) j − ( 4 y ) k From j: −4.1244 = 4 z From k: 14.7368 = −4 y or or z = −1.0311 ft y = −3.6842 ft ∴ line of action of the wrench intersects the yz plane at y = −3.68 ft, z = 1.031 ft .