I've attached an image depicting this truss.
P1 = 971 N and P2 = 433N
Answer:
BC = 358.67 N
Thus, BC is in tension
Step-by-step explanation:
Assumptions;
Truss members are mass-less
Clockwise is the positive moment
Directions to the right and up are positive directions
Now, let's find the reaction forces at A and E
To find vertical reaction E_y, let's find sum of moments about A and equate to zero.
We will asssume E_y is positive
Thus;
0 = P_1(a) - P_2(a) - E_y(3a)
Making E_y the subject, we have;
E_y = (P_1(a) - P_2(a))/(3a)
a will cancel out, thus;
E_y = (P_1 - P_2)/3
From the attached image, P1 = 971 N and P2 = 433N
Thus;
E_y = (971 - 433)/3
E_y = 179.33 N
Now, to find A_y, let's sum forces in the vertical direction to zero.
We'll assume A_y is positive
Thus;
0 = A_y + E_y - P_1
A_y = P_1 - E_y
A_y = 971 - 179.33
A_y = 791.67 N
To find A_x, let's sum forces in the horizontal direction to zero.
We'll assume A_x is positive.
Thus;
A_x - P_2 = 0
A_x = P_2
A_x = 433 N
A_x = 433 N
Now let's imagine a vertical cut through the three members GH, GC, and BC. We can eliminate the truss members and forces to the right of this cut and replace the cut stubs with equivalent assumed tensors which I will call GH, GC, and BC to represent the forces in each respective member.
Thus;
The angle α from vertical of member GC is; tanθ = (a/2)/a
tanθ = 1/2
θ = tan^(-1)(1/2)
θ = 26.57°
Let's now sum all forces in the vertical direction and equate to zero;
0 = A_y - P_1 - GC•cosθ
GC = (A_y - P_1) / cosθ
GC = (791.67 - 971)/cos26.57
GC = -200.5 N
Thus, GC is in compression
Let's now sum moments about G and equate to zero which will allow us to determine BC.
Thus;
0 = A_y(a) - A_x(a) - BC(a) + GC(0) + GH(0)
BC = A_y - A_x
BC = 791.67 - 433
BC = 358.67 N
Thus, BC is in tension
Let's sum forces in the horizontal direction and equate to zero.
0 = A_x + BC + GCsinθ + GH
-GH = A_x + BC + GCsinθ
-GH = 433 + 358.67 + (-200.5sin26.57)
-GH = 702
GH = -702 N
Thus, GH is in compression