
Summation of forces perpendicular to the inclined plane in block B
$N_B = 30 \cos 60^\circ + C \sin 30^\circ$
$N_B = 15 + 0.5C$
Amount of friction in block B at impending motion
$f_B = \mu_{B} N_B = 0.20(15 + 0.5C)$
$f_B = 3 + 0.10C$
$f_B + C \cos 30^\circ = 3 \sin 60^\circ$
$(3 + 0.10C) + C \cos 30^\circ = 30 \sin 60^\circ$
$0.966C = 22.98$
$C = 23.79 \, \text{ kN}$
Summation of vertical forces in block A
$N_A = 40 + C \sin 30^\circ$
$N_A = 40 + 23.79 \sin 30^\circ$
$N_A = 51.895 \, \text{ kN}$
Summation of horizontal forces in block A
$f_A = C \cos 30^\circ$
$f_A = 23.79 \cos 30^\circ$
$f_A = 20.60 \, \text{ kN}$
Coefficient of friction at A
$f_A = \mu_A N_A$
$20.60 = \mu_A(51.895)$
$\mu_A = 0.397$ answer
Comments
Hi sir. Good day. Sir,
Hi sir. Good day. Sir, pacheck naman po, iba po ata ung given na 3kn and 4kn dun sa solution.