Problem 530 | Friction

The plank impends to the right
$\dfrac{R_B}{\sin 25^\circ} = \dfrac{P}{\sin 105^\circ}$

$R_B = 0.4375P$
 

 

$\Sigma M_A = 0$

$Px = (R_B \cos 50^\circ)(10)$

$Px = (0.4375P \cos 50^\circ)(10)$

$x = 2.81 \, \text{ ft}$           answer
 

The plank impends to the left
$\dfrac{R_A}{\sin 10^\circ} = \dfrac{P}{\sin 105^\circ}$

$R_A = 0.1798P$
 

 

$\Sigma M_B= 0$

$Py = (R_A \cos 65^\circ)(10)$

$Py = (0.1798P \cos 65^\circ)(10)$

$y = 0.76 \, \text{ ft}$           answer