A massless string is wrapped around a 10 kg cylinder. the picture will look like the letter b. you pull upward on the string with a force of P = 49N. The bottom of the cylinder s on horizontal table and will roll w/o slipping. using dynamics involving center of mass, find the magnitude and the direction of frictional force (f). (no slippage between string and cylinder) Solution: Torque about cm: RP - Rf = Iα ---> R(P-f) = Iα I = 1/2MR^(2) α = a/R R(P-f) = 1/2MR^(2) (a/R) ---> P-f = 1/2Ma Forces Horz: f = ma Why do we need to consider the forces horizontal?? Isn't it already included in the torque calculation? P-ma = 1/2Ma P = 3/2Ma Ma = 2/3P a = 2/3 (49N) / 10kg ---> 3.2667 N My teacher got 32.667N? but how?