Question

1. The block shown in the figure has a mass of 612 Kg. The static coefficient of friction between the block an the supporting surface is 0.33 while the kinetic coefficient of friction is 0.26. Ignore the size of the block. For P is parallel to the incline while the 1000N force is vertical. a) If P = 2000 N, the shown block is (circle the correct answer) (5 points) a) in equilibrium b) moving up the incline c) is about to move up the incline (impending motion) d) moving down the incline e) is about to move down the incline (impending motion) b) If P = 4000 N, the shown block is (circle the correct answer) (5 points) a) in equilibrium b) moving up the incline c) is about to move up the incline (impending motion) d) moving down the incline e) is about to move down the incline (impending motion) 1000 N 60 degrees

Answer 1

1.) m= 612 kg Ms=0.33 MK = 0.26 1000 N Qo) P= 2000 N. N = low coso & ng coso V = loow Cos60° + 6124908 x cos 60° = 3499 N P Now p 2000N. B Fad = Foore down the inclineo P F. BoD f - 1000 sin 60° + 6060N 612 x 98 x Sin 60° = 1ooocosm & low sino ng sino mg cose こ zooo t 0.33 X 3499 = 3155N S. block has tendency to onome down the incline Fau Force up the incline. 2000 + MSN so block will down the inclineo moving down the Incline. move b.) Still Fau < Fed As Fad = 6060N Facu 4000 + 1155 = 5155 N so @ block will be moving down the incline.