Question

Problem 17.76 The 16.5-kg roll of paper has a radius of gyration kA = 120 mm about an axis passing through point A. It is pin supported at both ends by two brackets AB. The roll rests on the floor, for which the coefficient of kinetic friction is = 0.2. A horizontal force F = 60 N is applied to the end of the paper. (Figure 1) Figure 1 of 1 300 mm B 400 mm

Answer 1

mg, T, N passes through A.

F = 60N = 120 mm KA A to 8= 3oomm m=16.5kg Mis = 0.2 B x = initial angular acceleration sind = cosa= 4 5 315 T ac 13oomm Et f = TSend 400mm mg ng + Toosk =N f f MIRON f = 0.2 (mg + T cosa) Fto-ring o.ring + 0.2 T cosa Tsina Go + 32-34 + 0-2x TX (0.6) = = T(0.8) T= 92.34 92.34 135.794N 0.8 -0.12 0.68 Fr - ofr = Inolit f = 0.2(16.5*9.8+ 135.794 x(0.6) 0-6)) f = 48.629 N x = 110371* 0-3 I roll Inou - MA KA = 0.2376 kg. m² x = 140 857 rad/s2

So, torque due to mg, T, N is zero