The frame is supported by a pin at A and and rope DE. Member BC weighs 360 lbs, and member AD weighs 540 lbs. Determine the support reactions for the frame.

The frame is supported by a pin at A and and rope DE. Member BC weighs 360 lbs, and member AD weighs 540 lbs. Determine the support reactions for the frame.

- March 16, 2022

Draw the FBD of the member BC.

.

From the problem figure, calculate the $\alpha$ .

$\alpha=tan^{-1}\frac{6}{12}$

$\alpha=26.565\degree$

Consider the equilibrium conditions.

$(1)\textup{ }\Sigma F_y=0$
$\therefore B_y-360-700+Tsin\alpha=0$
$\therefore B_y+0.447T=1060.........(1)$

$(2)\textup{ }\Sigma F_x=0$
$\therefore B_x-Tcos\alpha=0$
$\therefore B_x=0.894T..........(2)$

$(3)\textup{ }\Sigma M_B=0$
$(360\times 6)+(700\times 12)-(Tsin26.565\degree \times 12)=0$
$T=1967.743 \textup{ lb}$

Substitute T in equation (1).

$B_y=1060-0.447(1967.743)$
$B_y=180.419\textup{ lb}$

Substitute T in equation (2).

$B_x=0.894(1967.743)$
$B_x=1759.162\textup{ lb}$

Draw the FBD of the member AD.

From the problem figure, calculate the $\beta$ .

$\beta=tan^{-1}\frac{12}{18}$

$\beta=33.69\degree$

Consider the equilibrium conditions.

$(1)\textup{ }\Sigma F_y=0$

$A_y+B_y-540-Tsin\alpha-Tcos\beta=0$
$A_y+180.419-540-1967.743sin26.565\degree -1967.743cos33.69\degree$
$A_y=2876.843\textup{ lb}$

$(2)\textup{ }\Sigma F_x=0$
$B_x+A_x+Tcos\alpha -Tsin\beta =0$
$A_x+1759.162+1967.743cos26.565\degree -1967.743sin33.69\degree =0$

$A_x=-2427.660\textup{ lb}$

As shown in the problem figure,

$E_y=Tcos\beta=1967.743cos33.69\degree$
$E_y=1637.262\textup{ lb}$

and

$E_x=Tcos\beta=1967.743sin33.69\degree$
$E_x=1091.505\textup{ lb}$

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