Question

Required information Member ABCand brace BDare used to suspend an electric transmission line from a utility pole. Take P= 400 lb. Н 24in. F 24 in В A 30° 30° Р. Determine the internal forces acting on cross sections Eand F, which are located immediately to the left of point A and to the right of point B, respectively. The internal forces acting on cross section Eare as follows: b, and ME b, VE NE= in.-lb The internal forces acting on cross section F are as follows: lb, and MF - NF b, VF in.-lb B

Please help me out with this problem!

Answer 1

given

P = 400 lb

consider the equilibrium force along x - direction

$\Sigma$ F_x = 0

- N_E + 400 cos60 = 0

N_E = 200 lb

consider the equilibrium force along y - direction

$\Sigma$ F_y = 0

V_E - 400 sin60 = 0

V_E = 346.4 lb

at point E the moment equilibrium

$\Sigma$ M_E = 0

M_E = 0

same way as above

consider the equilibrium force along x - direction

$\Sigma$ F_x = 0

- N_F + 400 cos60 = 0

N_F = 200 lb

consider the equilibrium force along y - direction

$\Sigma$ F_y = 0

V_F - 400 sin60 = 0

V_F = 346.4 lb

at point F the moment equilibrium

$\Sigma$ M_F - ( 400 x sin60 ) x 24 = 0

M_F = 8313.6 lb in anti clockwise direction.