Problem 525 | Friction

Coefficient of friction
$\mu = \tan \phi = \tan 20^\circ$

$\mu = 0.364$
 

Friction forces at each end of the ladder
$f_A = \mu N_A = 0.364N_A$

$f_B = \mu N_B = 0.364N_B$
 

$\Sigma F_H = 0$

$N_B = f_A$

$N_B = 0.364N_A$
 

$\Sigma F_V = 0$

$N_A + f_B = W$

$N_A + 0.364N_B = W$

$N_A + 0.364(0.364N_A) = W$

$1.1325N_A = W$

$N_A = 0.883W$
 

Thus,
$f_A = 0.364(0.883W)$

$f_A = 0.3214W$
 

$\Sigma M_B = 0$

$W(2.4 \cos \theta) + f_A(4.8 \sin \theta) = N_A(4.8 \cos \theta)$

$W \cos \theta + 2f_A \sin \theta = 2N_A \cos \theta$

$W + 2f_A \tan \theta = 2N_A$

$W + 2(0.3214W) \tan \theta = 2(0.883W)$

$1 + 0.6428 \tan \theta = 1.766$

$0.6428 \tan \theta = 0.766$

$\tan \theta = 1.191~661~481$

$\theta = 50^\circ$           answer

$\sin 20^\circ = \dfrac{R_B}{W}$

$R_B = W \sin 20^\circ$
 

$R_{BH} = R_B \cos 20^\circ = W \sin 20^\circ \cos 20^\circ$

$R_{BV} = R_B \sin 20^\circ = W \sin 20^\circ \sin 20^\circ = W \sin^2 20^\circ$
 

 

$\Sigma M_A = 0$

$R_{BH} (4.8 \sin \theta) + R_{BV} (4.8 \cos \theta) = W(2.4 \cos \theta)$

$4.8R_{BH} \sin \theta = 2.4W \cos \theta - 4.8R_{BV} \cos \theta$

$2R_{BH} \sin \theta = W \cos \theta - 2R_{BV} \cos \theta$

$2R_{BH} \sin \theta = (W - 2R_{BV}) \cos \theta$

$\dfrac{\sin \theta}{\cos \theta} = \dfrac{W - 2R_{BV}}{2R_{BH}}$

$\tan \theta = \dfrac{W - 2R_{BV}}{2R_{BH}}$

$\tan \theta = \dfrac{W - 2W \sin^2 20^\circ}{2W \sin 20^\circ \cos 20^\circ}$

$\tan \theta = \dfrac{1 - 2 \sin^2 20^\circ}{2 \sin 20^\circ \cos 20^\circ}$

$\theta = \arctan \left( \dfrac{1 - 2 \sin^2 20^\circ}{2 \sin 20^\circ \cos 20^\circ} \right)$

$\theta = \arctan 1.191753593$

$\theta = 50^\circ$           answer