Question

Determine F_CG and F_GH in (lb), the magnitudes of the forces in members CG and GH, respectively, using the method of sections. Assume for your calculations that each member is in tension, and include in your response the sign of each force that you obtain by applying this assumption.
Express your answers numerically in pounds to three significant figures separated by a comma.

Answer 1

For the entire truss, by equilibrium of horizontal forces;

$\\\sum F_x =0 \\\\\Rightarrow A_x - P_2 = 0 \\\\\Rightarrow A_x-200 = 0 \\\\\Rightarrow A_x = 200\;\;lb$

For the entire truss, take moment about E;

$\\\sum M_E = 0 \\\\\Rightarrow P_2(a)+P_1(2a) - A_y(3a) = 0 \\\\\Rightarrow 200(1)+494(2) - A_y(3) = 0 \\\\\Rightarrow A_y = 396\;\;lb$

Take section as shown below;

y P, (494 Lb O=tanta lo G 10.50 70-63-4350 FGH No FEG A |200 lb B Foco 39626 a/2 a/2

Take equilibrium of vertical forces;

$\\\sum F_y = 0 \\\\\Rightarrow 396 -494 - F_{CG}sin63.435^o = 0 \\\\\Rightarrow F_{CG} = -109.567\;\;lb$

...(Answer)

Take moment about C;

$\\\sum M_C = 0 \\\\\Rightarrow 494(0.5a)-396(1.5a) - F_{GH}(a) = 0 \\\\\Rightarrow F_{GH} = -347\;\;lb$

...(Answer)