Question

800mm The 20kg homogeneous cylinder of 800-mm diameter rests against the surfaces shown. B The coefficient of static friction is 0.5 for AB and 0.6 for AC. The coefficient of kinetic friction is 0.4 for AB and 0.5 for AC. 300 (i) 5 points: Draw the free body diagram of the cylinder (ii) 15 points: Calculate the magnitude of the clockwise couple M required to initiate the rotation. (iii) 15 points: Suppose that the contact at the surface AB is smooth and the clockwise couple M = 200 N.m. Determine the magnitude and the direction of the friction at the surface AC.

Answer 1

800mm Page No. Date Pages B m 6o A ☺ M Σ. 7 to Go C CNB 30° 30° N mg (1) To initiate rotation both to a fe should attain their limiting value, ie, tas NB and Mesne, respectively. {Mos: Coefficient of static friction for AB les coefficient of Static friction for Ac} For the condition when the moment has initiated we write moment equation just initiated about o EM=0 D M - MB, NBR-Mcs Nc R=0

Page No. Date Page 2 NB sin 30 + MBA -) ME + (*) We also have E Fn = 0 tus Ne cos Go =) √3 Ne - NB + 0.5x3 NB + 0.681 Ne 2 = => 1.166 NC 0.067 NB=0 NA Cos3. 2 2 رن - Also, Efy =) Nc cos 6 + NB Cos 30 t u NB sin 30 - Mc Nesin Go = a - mg - Nc + √3 NB + 0.581 No - 0.6X3 Nc - 20X9.81 = 0 2 2 2 2 =) -0.0196 Nct 1.116 NB - 196.2=0 using (;) & (ii) 0.0196 x 0.067 N3 +1.116 NB-16.) = 0 - Na = 1+5.984 N

Page No. Date Page 3 Ne = 10.112 N > M = (MBS NB + Mc Nc) R 0.6X 10.112) 400 0.5 X 175.984 +0.6x10.112.) x 400 1ooo M = 37.624 N-m In this case, O Coefficient of friction for AB = 0 Coefficient of friction for AC = Mca Coefficient of For M= 200 N-m the cylinder notates → we use Kinetic friction for Ac' Kinetic friction

4 the force force equations 2 =) Nc + 3 N.3 -0.5*3 Nc - 196.2=0 Again writing I Fr=o => Ncsin to NB sin 30 + McK Nc cos 6o=0 =) Nc - NB + 0.5% / Nc=0 = 1.116 Nc - 0.5 NB=0 Ety=0 =) => Nccos 60° + NB cos 30 - Maa Nesingo = 0.067 Nc +0.866 NB - 196.2=0 -mg = 0 2 2

Page No. Date Page 5 We have 1.116 Nc - 0.5 No = 0 0.067 Ne to.866 N3 196.2 = 0 =) 0.067X0.5 NB + 0.866 NB 146.2 T.116 Ng = 218.969 N Nc = 98.104 N 7 Magnitude of friction at the surface = Mc Nc Nc = 49.052 N (towards A) АСС