... from the publisher. Engineering Mechanics - Statics Chapter 8M0Rrμp0p0Rr−⎛⎜⎝⎞⎟⎠r2⌠⎮⎮⌡d02πθ1⌠⎮⌡d=π 6 μp0R 3 =π 6 μ3PπR2⎛⎜⎝⎞⎟⎠R 3 =μPR2=MμPR2=Thus,Problem ... Engineering Mechanics - Statics Chapter 8Given:a 2in=b 3in=P 500 lb=M 3lbft=Solution:MabμkPb2⎛⎜⎝⎞⎟⎠ 3 a2⎛⎜⎝⎞⎟⎠ 3 −b2⎛⎜⎝⎞⎟⎠2a2⎛⎜⎝⎞⎟⎠2−⎡⎢⎢⎢⎢⎣⎤⎥⎥⎥⎥⎦=aμkP2b⎛⎜⎝⎞⎟⎠b 3 a 3 −b2a2−⎛⎜⎜⎝⎞⎟⎟⎠=μk2Mb ... AxMr tanφsθ+()= Ax 32 9 N=+→ΣFx = 0;Ax2Tcosβ()− 0= T12Axcosβ()⎛⎜⎝⎞⎟⎠= T 232 . 36 N=CyTsinβ()= ByCy=ByCy⎛⎜⎝⎞⎟⎠ 164 .3 164 .3 ⎛⎜⎝⎞⎟⎠N=842© 2007...