Question

1. The block shown in the figure has a mass of 612 Kg. The static coefficient of friction between the block and the supporting surface is 0.33 while the kinetic coefficient of friction is 0.26. Ignore the size of the block. Force 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

Note: In both cases block is moving down the incline plane so friction on the block will be kinetic friction.

Here m= 612 kg Ms = 0.33, lk = 0.26 ce un N P for From figure, normal force the block by incline, mg (os 60° + 1000C0360° - 612*9.8 x + 500 3498.8 N N = 1000 EOS 6094 1000N friction 1000 sing mgsingo on the Maximum friction mg cos 60° mg block 60°C fs - Ms N = 0.33× 3498.8 1154.604 N 1154.6N kinetic friction the block fr MK N = 0.26x3498.8 fk = 909.688N = 909.7N Dowan ward force on the block = 100D sin 60° + sin 60° ET = LOUD * 13 + 612x9.8 x 3 mg 2 2 Hers 7 Fi > P F = 6060 IN. (a) If Home, P = 2000N Now, So friction on the block will be in upward direction as shown in figure. Now, (P + fs) o CF Hence block will be moving down the incline plane Correct option 16.) Here P = 4000N Again, F. >P So friction on the block will be in upward direction shown in the figure (p + fs ) Hence the shown block is moving down the incline plane Correct option is @ as Also, <

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