Problem 338 | Equilibrium of Parallel Force System

From FBD of beam CD
$\Sigma F_V = 0$

$R_C + R_3 = P$

$R_C + 0.5R_C = 960$

$R_C = 640 \, \text{ lb}$
 

$R_3 = 0.5(640) = 320 \, \text{ lb}$           answer
 

 

$\Sigma M_C = 0$

$12R_3 = 960x$

$12(320) = 960x$

$x = 4 \, \text{ ft}$

Thus, P is 8 ft to the left of D.           answer
 

From the figure above, R_c is at the midspan of AB to produce equal reactions R₁ and R₂. Thus, R₂ and R₃ are 6 ft apart.           answer
 

From FBD of beam AB
$R_1 = 0.5(640) = 320 \, \text{ lb}$           answer

$R_2 = 0.5(640) = 320 \, \text{ lb}$           answer