Question

Question 2 (35 points) 800mm The 20kg homogeneous cylinder of 800-mm diameter rests against the surfaces shown. 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 30- (1) 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

(i)

M С B RC Rb 60 30

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(II)

By applying three equation which we use for the finding the reaction force in equlibrium condtion

$\small \\\sum F_X=0 \\\\R_Csin(60)+\mu_{bs}R_B cos(30)+\mu_{cs}R_C cos(60)=R_Bsin(30) \\\\0.866 R_C+0.5*0.866R_B+0.6*0.5R_C=0.5R_B \\\\1.166R_C-0.067R_B=0-----(1)$

Fy = 0 Rccos(60) + Rocos(30) + posRosin(30) = -McsRcsin(60) + mg 0.5Rc +0.866R8 +0.5 * 0.5Rp = 0.6 * 0.866Rc + 20 * 9.81 -0.0196Rc + 1.116R) = 196.2 - -(1)

by solving equation 1 and 2

$\small \\\\R_B=175.9845 N \\\\R_C=10.112 \ N$

Now calculating the moment

$\small \\\\M=R(\mu_{BS}R_b+\mu_{CS}R_c) \\\\M=0.4(0.5*175.984+0.6*10.112) \\\\M=37.62 \ Nm$

(III)as given that surafe AB is smooth

$\small \\\mu_{AB}=0 \\\\\mu_{AC}=0.5$

$\small \\\sum F_X=0 \\\\R_Csin(60)+\mu_{bk}R_B cos(30)+\mu_{ck}R_C cos(60)=R_Bsin(30) \\\\0.866 R_C+0*0.866R_B+0.5*0.5R_C=0.5R_B \\\\1.166R_C-0.5R_B=0-----(3)$

$\small \\\sum F_y=0 \\\\R_Ccos(60)+R_Bcos(30)+\mu_{bk}R_B sin(30)=-\mu_{ck}R_C sin(60)+mg \\\\0.5 R_C+0.866R_B+0*0.5R_B=0.5*0.866R_C+20*9.81 \\\\-0.0067R_C+0.866R_b=196.2-----(4)$

by solving equation 1 and 2

$\small \\\\R_B=218.969 N \\\\R_C=98.104 \ N$

Now calculating the moment

$\small \\\\F_A_C=(\mu_{Bk}R_b+\mu_{Ck}R_c) \\\\M=(0*218.9694+0.5*98.104) \\\\F_{AC}=49.05 \ N$

Since M is posetive so it act towards A