Question

MENT RESOURCES ment 10 Question 5 calculate the couple applied to the lower of the two 27-kg cylinders which will alow them to roll slowly down the incline. The coefficients af static and kinetic friction for all contacting surfaces are - 0.57 and H =0.51. 360 mm dia. Score Results by Study 28 Answer: M- N- 28 the telecase Type here to search

Please help me solve this but explain in simple terms. I am not sure how to do it.

Answer 1

Draw the free body diagram. mg mg 180 mm | FD ND FB 180mm N. SofA For 1st Cylinder Taking moments about C. EMg = 0 -M+(F2x0.180)–(F0.180) = 0 M=(F:- F) 10.180 ........ (1) Apply equations of equilibrium. EF = 0 F2-N, - mg sin 0 = 0 Fi-No-27(9.81)sin18° = 0) F:-N, = 81.84 ........(2) EF, = 0 NA-Fy - mg cos 0 = 0 NA-F, = 27(9.81)cos18° N2-F) = 251.9 -.-...-.-(3) For 2nd cylinder Taking moments about C, EM = 0 (Fg 0.180) – (F, 0.180)= 0 Fg = FD .......(4)

EF = 0 Fg +N, -mg sin 0 = 0 Fo+N, -27(9.81)sin 18° = 0) Fo+N, = 81.84 MiN, +N, -81.84 = 0) N, (0.51+1)=81.84 No = 54.20 N Fo=0.51(54.20) = 27.644 N F3 = 27.644 N EF, = 0 Ng +Fo-mg cos 0 = 0 Ng+27.644 – 27(9.81)cos 18° = 0 Na =224.26 N Coefficient of friction at B, NB 27.644 224.26 = 0.123 <0.57 Hence the cylinder 2 does not slips at B From equation (2) F2-N, = 81.814 F - 54.20 = 81.814 F = 136.014 N From (3) NA - Fo = 251.9 NA - 27.644 = 251.9 N = 279.54N

Coefficient of friction at A, 136.014 279.54 = 0).48 <0.57 Hence the cylinder 1 does not slips at A Our assumption in correct From equation (1), M=(F4-FD)0.180 = (136.014 – 27.644)*0.180 M=19.5 N-m