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Chapter 06, Problem 019 A 14 N horizontal force F pushes a block weighing 3.3 N against a vertical wall (see the figure). The

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let us Take a It Weight of block - 33N ocR = lo mfpeé) force, oy block = 14N = 3.30 = 3.3N Me = 0.74 . Up = 0.26 firstly, how@ frictional force fie Balauced by gracistettoval MeN +MAN) zing acceration = 14–604 0.96 2 a = 4.48 ufpec? [ag So, block wil_ forces u y-disection:- y = f + fx W = UN 4 UxN-W = 10:36+$.04 – 3.3 3 12:1 fy = 12.1 forces aloug x-diretion & Eha = PN Efn

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