COSMOS Chapter 8 Solution 42

COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 42. FBD pulley: Note that φSA = tan −1 µ SA = tan −1 ( 0.5 ) = 26.565° < 30°, Cable is needed to keep A from sliding downward. ΣFy = 0: 2T − WB = 0, T = WB , 2 WB = 2T (1) FBD block A: (a) For minimum WB , there will be impending slip of block A downward, and FA = µ SA N A as shown. ΣFy′ = 0: N A − WA cos 30° = 0, N A = WA cos 30° = 23.544 N cos 30° = 20.390 N ( ) WA = ( 2.4 kg ) 9.81 m/s 2 = 23.544 N FA = ( 0.50 )( 20.390 N ) = 10.195 N FBD block C: ΣFx′ = 0: T − WA sin 30° + FA = 0 T = ( 23.544 N ) sin 30° − 10.195 N = 1.577 N From (1) WB = 2T = 3.154 N, mB = 3.154 N ( 9.81 m/s 2 ) = 0.322 kg, mB min = 322 g
ΣFy = 0: NC − WC = 0, NC = 58.86 N FC max = µ SC NC = 0.30 ( 58.86 N ) = 17.658 N Since T = 1.577 N < FC max , block B doesn’t slip and above answer for mB min is correct. Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System (b) For mB max assume impending slip of block C to left, FC = Fmax ΣFx = 0: − T + FC = 0, From (1) WB = 2T = 35.316 N, T = FC = FC max = 17.658 N mB = WB 35.316 N = = 3.6 kg g 9.81 m/s 2 From FBD block A, ΣFx = 0: T − WA sin 30° + FA = 0, FA = WA sin 30° − T FA = ( 23.544 N ) sin 30° − 17.658 N = − 5.886, Since FA < FA max , A does not slip Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. FAmax = 10.195 N M B max = 3.6 kg