Problem 520 | Friction

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

$\tan \phi_B = \mu_B$

$\tan \phi_B = 0.20$

$\phi_B = 11.31^\circ$
 

 

$60^\circ - \phi_B = 48.69^\circ$

$60^\circ + \phi_B = 71.31^\circ$
 

Sine law to the force polygon B
$\dfrac{C}{\sin (60^\circ - \phi_B)} = \dfrac{30}{\sin (60^\circ + \phi_B)}$

$\dfrac{C}{\sin 48.69^\circ} = \dfrac{30}{\sin 71.31^\circ}$

$C = 23.79 \, \text{ kN}$
 

Sine law to force polygon A
$\dfrac{40}{\sin (60^\circ - \phi_A)} = \dfrac{C}{\sin \phi_A}$

$\dfrac{40}{\sin (60^\circ - \phi_A)} = \dfrac{23.79}{\sin \phi_A}$

$40 \sin \phi_A = 23.79 \sin (60^\circ - \phi_A)$

$40 \sin \phi_A = 23.79 (\sin 60^\circ \cos \phi_A - \cos 60^\circ \sin \phi_A)$

$40 \sin \phi_A = 20.60 \cos \phi_A - 11.895 \sin \phi_A$

$51.895 \sin \phi_A = 20.60 \cos \phi_A$

$\tan \phi_A = 0.3970$

$\mu_A = 0.397$           answer