PLEASE CHANGE TO \"140-LB\" AND \'P=33LB) Determine the acceleration of the 150-

Question

PLEASE CHANGE TO "140-LB" AND 'P=33LB)

Determine the acceleration of the 150-1b cabinet and the normal reaction under the legs A and B if P = 35 1b. The coefficients of static and kinetic friction between the cabinel and the plane are mu s = 0.2 and mu s = 0.15, respectively. The cabinets center of gravity is located at G.

Explanation / Answer

Total normal reaction = 140 lb P= 33 lb therefore mu * R = 0.2 * 140 = 28 P is greater than MUs*R P > MUs*R Let R1 and R2 = vertical rxn force and f1 and f2 = friction force acting on the block Kinetic Friction case put moment about A = 0 we get 33*4 + 140*1 = R2 * 2 R2 = 136 lb R1 = 140 - 136 = 4lb f1 = MUk * R1 = 0.15 * 4 =0.6 f2 = MUk *R2 = 0.15* 136 = 20.4 Net Force = 33-(20.4+0.6) = 12lb Hence acceleration = 12*g/140 = 0.84 m/s^2......ans

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