Question

Please help with this 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 Pis 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 1

Answer 1

Given: Coefficient of static friction, l;= 0.33 Coefficient of kinetic friction, U, = 0.26 Р 1600 N Ni Weight of Block, W = 61249.81 Wlcoso W sing = 6003.72 N W=612x 9.81 = 6003.72N 0=60° on Let Ny be the normal force act acting the block ar to plane. Equilibrium of forces perpendicular to plane. N, - a 1000 Corot Wcoso N, 11oo x cox 60 + 6003.72 x cos 60' * (W+1000). sino N , I 3501.86 N (W + 100o).Coso Forces along the plane are - 1. (Wx1000) sino (6003.72 +1000) sin 60° 6065.4 N.

Part - (a)

If P=2000 N

since,

Now in Question - a P = 2000 N a < 6065.4 N (down-slope force) So, block will move downward & hence a friction force will aut upward. P= 2000N so. total upward force along slope P+ fx = 2000 + 0.26X 3501.86 2910,48 N Total downward force force along slope 6065 fix = M, N, 6065,4 N

Total downward force > Total upward force along the slope,

Correct option = (d) Shown block is moving down the incline

Part (b)

If P= 4000 N

so again,

Question P: 4000 N. Again P= 4000 N c downward force (6065.4 N) So, block will try to move downward. But a friction force acting Sep along the slope in upward direction, try to stop Let, say static friction build up here, then Max. static friction force which may along the slope upward Is, max = Uso Ni - 0.33 X 3501.86 P= 4000 N fs= Ms. act 6065.4 N 1155. 62 - 4ooo + 1155.62 = 5155.62 N i. Tofal upward force P+ fs = 4ooo + 1155.62 - Total downward force = 6065.4 N Downward force > upward force Hence block will be moving downward.

Correct option = (d) Block is moving down the incline