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500 N 4 = 0.30 -0.25 Determine whether the block shown is in equilibrium and find the magnitude and direction of the friction
500 N U=0,30 41 = 0.25 20°/
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Answer #1

500 N Ms 0.30 р K = 0.25 20° P = 400 N N psino f. B.D. of Blockin P Pos20° 50081220 20 500 GS 20° 20 500 fs N = P sn 20% + 5friction force = lk N Hence, 0-25 X 686.65 M 151-6625 f 151.66 N in downward direction of slope.

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