University:
Michigan State University
Course:
ERG 160 | Success in Science, Technology, Engineering, and Mathematics
Academic year:

2024
Views:

294

Pages:

2
Author:

slypery5hy

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 103. Free-Body Diagram: The location of the bucket of sand will be ( xS , c, zS ). ( ) The weight W is W = mg = (170 kg ) 9.81 m/s 2 = 1667.7 N ΣFy = 0: N = ΣM O = 0: or 3N − W − WS = 0 1 (W + WS ) 3 (1) rOA × N + rOB × N + rOC × N + rOG × W − rOS × WS = 0 ( 0.3 m ) i + (1.2 m ) k  × Nj + ( 2.4 m ) i + ( 0.9 m ) k  × Nj + ( 0.6 m ) i × Nj + (1.2 m ) i + ( 0.6 m ) k  × ( −W ) j + ( xS i + zS k ) × ( −WS ) j = 0 or ( 0.3 m ) N k − (1.2 m ) N i + ( 2.4 m ) N k − ( 0.9 m ) N i + ( 0.6 m ) N k − (1.2 m )W k + ( 0.6 m )W i − xSWS k + zSWS i = 0 Equating the coefficients of the unit vectors to zero: i: − (1.2 ) N − ( 0.9 ) N + ( 0.6 )W + zSWS = 0 continued COSMOS: Complete Online Solutions Manual Organization System or, using (1) 1  −2.1  (W + WS )  + 0.6W + zSWS = 0 3  WS = k: 0.1W zS − 0.7 m 0.3 N + 2.4 N + 0.6 N − 1.2W − xSWS = 0 or, using (1) 1  3.3  (W + WS )  − 1.2W − xSWS = 0 3   WS = 0.1W 1.1 m − xS (2) For (WS )min , (1) and (2) imply that xS, should be chosen as small as possible and that zS should be chosen as large as possible with the constraint that (1.1 m − xS ) = ( zS − 0.7 m ) xS + zS = 1.8 m. or The smallest xS and the largest zS that satisfy this condition are xS = 0.6 m zS = 1.2 m The corresponding value of WS is: WS = 0.1(1667.7 N ) = 333.54 N 1.1 m − 0.6 m Therefore the smallest mass of the bucket of sand is ( mS )min = 333.54 N = 34.000kg 9.81 m/s 2 or ( mS )min = 34.0 kg