The frame is supported by a pin at A and rope DE Member BC w

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

Solution

To find the reaction forces at E and D we must consider equilibrium forces at E and A,

sigma Fv = 0

Re + Ra - Fde sin 56.29 - Fdc sin 26.57 - 540 = 0 ------- eqn 1

Fde + Fdc = 360 x 9.81 + 700 x 9.81 ------------------------- eqn 2

Sigma Fh = 0

He + Ha - Fde cos 56.29 + Fdc cos 26.57 = 0 -------------- eqn 3

sigma Me = 0

Ra x 12 - 540 x 12 - Fdc sin 26.57 x 24 - Fdc cos 26.57 x 12 = 0 -------- eqn 4

Member ABD:

Mb = 0

Ha x 12 + Fde cos 56.29 x 6 = 0-------------- eqn 5

Member BC:

Mb = Fde cos 56.29 x 6 + Fdc sin 26.57 x 12

Mb = 0

Fde = Fdc sin 26.57 x 12 / cos 56.29 x 6 =  5.36 / 3.32 = 1.61 Fdc

Using eqn 2 find Fdc,

Fdc = (360 x 9.81 + 700 x 9.81 ) / 2.61 = 3984.13 lbf

Fde = 6414 lbf

Using eqn 4 find Ra,

Ra x 12 - 540 x 12 - Fdc sin 26.57 x 24 - Fdc cos 26.57 x 12 = 0

Ra = (540x12 + 3984 sin 26.57 x 24 + 3984 cos 26.57 x 12 ) / 12

Ra = 3774.2 lbf

Using eqn 1 find Re,

Re = - 3774 + 6414 sin 56.29 + 3984 sin 26.57 + 540

Re = 3883.53 lbf

Ha = -6414 cos 56.29 x 0.5 = (-) 1779.85 lbf

He = - 1779.85 + 6414 cos 56.29 -3984 cos 26.57 = -1783.38 lbf