Problem 514 | Friction

$a = 10 \cos 60^\circ = 5 \, \text{ cm}$

$b = 10 - a = 5 \, \text{ cm}$
 

$\Sigma M_A = 0$

$Pb = W_{cyl}a$

$P(5) = 10(5)$

$P = 10 \, \text{ kN}$           answer
 

$\Sigma M_O = 0$

$10f = 10P$

$f = P$

$f = 10 \, \text{ kN}$
 

$\Sigma F_y = 0$

$N = 10 \cos 30^\circ + P \cos 30^\circ$

$N = 10 \cos 30^\circ + 10 \cos 30^\circ$

$N = 17.32 \, \text{ kN}$
 

$f = \mu N$

$10 = \mu (17.32)$

$\mu = 0.577$           answer