Problem 532 | Friction

$\mu = \tan \phi$

$\mu = \tan 15^\circ$
 

 

Summation of forces on block A normal to the 30° incline
$N_A = 2 \cos 30^\circ + C \cos 60^\circ$

$N_A = 2 \cos 30^\circ + 0.5C$
 

Amount of friction under block A
$f_A = \mu N_A = \tan 15^\circ (2 \cos 30^\circ + 0.5C)$

$f_A = 2 \cos 30^\circ \tan 15^\circ + 0.5C \tan 15^\circ$
 

Summation of forces on block A parallel to the 30° incline
$f_A + 2 \sin 30^\circ = C \sin 60^\circ$

$(2 \cos 30^\circ \tan 15^\circ + 0.5C \tan 15^\circ) + 1 = C \sin 60^\circ$

$2 \cos 30^\circ \tan 15^\circ + 1 = C \sin 60^\circ - 0.5C \tan 15^\circ$

$(\sin 60^\circ - 0.5 \tan 15^\circ)C = 2 \cos 30^\circ \tan 15^\circ + 1$

$C = \dfrac{2 \cos 30^\circ \tan 15^\circ + 1}{\sin 60^\circ - 0.5 \tan 15^\circ}$

$C = 2 \, \text{ kN}$
 

Summation of forces on block B normal to the 45° incline
$N_B = 2 \cos 45^\circ + C \cos 45^\circ$

$N_B = 2 \cos 45^\circ + 2 \cos 45^\circ$

$N_B = 2.8284 \, \text{ kN}$
 

Amount of friction under block B
$f_B = \mu N_B = \tan 15^\circ (2.8284)$

$f_B = 0.7578 \, \text{ kN}$
 

Summation of forces on block B parallel to the 45° incline
$P + 2 \sin 45^\circ = f_B + C \sin 45^\circ$

$P + 2 \sin 45^\circ = 0.7578 + 2 \sin 45^\circ$

$P = 0.7578 \, \text{ kN}$           answer

Sum of all vertical forces in black A
$R_A \cos 45^\circ = 2$

$R_A = 2\sqrt{2} \, \text{ kN}$
 

Sum of all horizontal forces in block A
$C = R_A \sin 45^\circ$

$C = 2\sqrt{2} \sin 45^\circ$

$C = 2 \, \text{ kN}$
 

 

Sum of all forces normal to the incline in block B
$R_B \cos 15^\circ = C \cos 45^\circ + 2 \cos 45^\circ$

$R_B \cos 15^\circ = 2 \cos 45^\circ + 2 \cos 45^\circ$

$R_B = 2.9282 \, \text{ kN}$
 

Sum of all forces parallel to the incline in block B
$P + 2 \sin 45^\circ = C \sin 45^\circ + R_B \sin 15^\circ$

$P + 2 \sin 45^\circ = 2 \sin 45^\circ + 2.9282 \sin 15^\circ$

$P = 0.7578 \, \text{ kN}$           answer