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Mg Now, that we have drawn the free body diagram, consider the information they've given us.­ The problem has indicated the system is in equilibrium, so this should be a hint to use Newton's second Law. Knowing that, we can write the following two equations: ~:F=O t=O Lers first take a look at the sum of all forces in the horizontal component for ~F. FN -FTsin30o= 0 FN = FT/2 2FN = FT Now that we have the tension in terms of the normal force (and vice versa), we can now consider friction in our problem. The equation for torque would be written as the following -fR +(1/2)(Rcos300)FT = 0 . ·Note: The signs depend on whether this force Is maklOl the system turn clockwise (-) or counterclockwise (+). -j.LFNR+(1/2)("3/2)RFT = 0 j.L{FT/2)R = (1/2)("3/2)RFr Cancel R, Fr, and ~ from both sides. j.L = "3/2 -r;u I"r B~ , .J Y'I 0.) ~k f' I!. 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