Figure Q4 shows a simple form of roof truss, freely supported at B and C with a vertical load of 120kN at A. Calculate: a. The reactions at B and C b. The forces in AB, BC, and AC

Since there are no external horizontal forces, the horizontal reaction at B must be zero.

$$\sum F_x = 0 \implies R_{Bx} = 0$$

We take moments about B to find $R_{Cy}$. Clockwise moments are positive.

$$\sum M_B = 0$$ $$(120 \, \text{kN})(x_B) - R_{Cy}(BC) = 0$$ $$(120 \, \text{kN})(2 \, \text{m}) - R_{Cy}(8 \, \text{m}) = 0$$ $$240 \, \text{kN} \cdot \text{m} - 8 R_{Cy} = 0$$ $$R_{Cy} = \frac{240}{8} = 30 \, \text{kN}$$

$$\sum F_y = 0$$ $$R_{By} + R_{Cy} - 120 \, \text{kN} = 0$$ $$R_{By} + 30 \, \text{kN} - 120 \, \text{kN} = 0$$ $$R_{By} = 90 \, \text{kN}$$

The reactions are:
At B: $R_{Bx} = \boxed{0 \, \text{kN}}$, $R_{By} = \boxed{90 \, \text{kN} \text{ (upwards)}}$
At C: $R_{Cy} = \boxed{30 \, \text{kN} \text{ (upwards)}}$

Forces at Joint B: $R_{By} = 90 \, \text{kN}$ (up), $F_{AB}$ (tension, up-left at $60^\circ$), $F_{BC}$ (tension, right).

$$\sum F_y = 0$$ $$R_{By} + F_{AB} \sin(60^\circ) = 0$$ $$90 + F_{AB} \frac{\sqrt{3}}{2} = 0$$ $$F_{AB} = -\frac{180}{\sqrt{3}} = -60\sqrt{3} \, \text{kN} \approx -103.92 \, \text{kN}$$

Since $F_{AB}$ is negative, it is in compression.

$$F_{AB} = 60\sqrt{3} \, \text{kN} \text{ (Compression)}$$ $$\sum F_x = 0$$ $$F_{BC} + F_{AB} \cos(60^\circ) = 0$$ $$F_{BC} + (-60\sqrt{3})\left(\frac{1}{2}\right) = 0$$ $$F_{BC} - 30\sqrt{3} = 0$$ $$F_{BC} = 30\sqrt{3} \, \text{kN} \approx 51.96 \, \text{kN}$$

Since $F_{BC}$ is positive, it is in tension.

$$F_{BC} = 30\sqrt{3} \, \text{kN} \text{ (Tension)}$$

Forces at Joint C: $R_{Cy} = 30 \, \text{kN}$ (up), $F_{AC}$ (tension, up-right at $30^\circ$), $F_{BC}$ (tension, left).
We use $F_{BC} = 30\sqrt{3} \, \text{kN}$ (tension), so it pulls away from C, to the left.

$$\sum F_y = 0$$ $$R_{Cy} + F_{AC} \sin(30^\circ) = 0$$ $$30 + F_{AC} \left(\frac{1}{2}\right) = 0$$ $$F_{AC} = -60 \, \text{kN}$$

Since $F_{AC}$ is negative, it is in compression.

$$F_{AC} = 60 \, \text{kN} \text{ (Compression)}$$

Forces at Joint A: $120 \, \text{kN}$ (down), $F_{AB}$ (compression, pushes into A, so its x-component is to the left, y-component is downwards), $F_{AC}$ (compression, pushes into A, so its x-component is to the right, y-component is downwards).

$$\sum F_x = 0$$ $$-F_{AB} \cos(60^\circ) + F_{AC} \cos(30^\circ) = 0$$ $$-(60\sqrt{3})\left(\frac{1}{2}\right) + (60)\left(\frac{\sqrt{3}}{2}\right) = 0$$ $$-30\sqrt{3} + 30\sqrt{3} = 0$$ $$0 = 0 \quad (\text{Consistent})$$ $$\sum F_y = 0$$ $$-120 - F_{AB} \sin(60^\circ) - F_{AC} \sin(30^\circ) = 0$$ $$-120 - (60\sqrt{3})\left(\frac{\sqrt{3}}{2}\right) - (60)\left(\frac{1}{2}\right) = 0$$ $$-120 - (60 \times \frac{3}{2}) - 30 = 0$$ $$-120 - 90 - 30 = 0$$ $$-240 = 0 \quad (\text{Inconsistent})$$

There is an inconsistency in the vertical equilibrium at Joint A. Let's re-examine the directions of forces at Joint A.
For the $120 \, \text{kN}$ downward load to be balanced, the vertical components of $F_{AB}$ and $F_{AC}$ must be upwards. This means $F_{AB}$ and $F_{AC}$ must be in tension (pulling away from A).

Let's re-calculate $F_{AB}$ and $F_{AC}$ using Joint A, assuming they are in tension.
Joint A: $120 \, \text{kN}$ (down), $F_{AB}$ (tension, up-right at $60^\circ$), $F_{AC}$ (tension, up-left at $30^\circ$).

$$\sum F_x = 0$$ $$F_{AB} \cos(60^\circ) - F_{AC} \cos(30^\circ) = 0$$ $$F_{AB} (0.5) = F_{AC} (\sqrt{3}/2) \implies F_{AB} = F_{AC}\sqrt{3} \quad (1)$$ $$\sum F_y = 0$$ $$F_{AB} \sin(60^\circ) + F_{AC} \sin(30^\circ) - 120 = 0$$ $$F_{AB} (\sqrt{3}/2) + F_{AC} (0.5) = 120 \quad (2)$$

Substitute (1) into (2):

$$(F_{AC}\sqrt{3})(\sqrt{3}/2) + F_{AC}(0.5) = 120$$ $$F_{AC}(3/2) + F_{AC}(1/2) = 120$$ $$2F_{AC} = 120 \implies F_{AC} = 60 \, \text{kN}$$

Substitute $F_{AC}$ back into (1):

$$F_{AB} = (60)\sqrt{3} = 60\sqrt{3} \, \text{kN}$$

Since both are positive, $F_{AB} = 60\sqrt{3} \, \text{kN}$ (Tension) and $F_{AC} = 60 \, \text{kN}$ (Tension).

Now, let's re-check Joint B and C with these new values for $F_{AB}$ and $F_{AC}$, and $F_{BC} = 30\sqrt{3} \, \text{kN}$ (Tension).

Joint B: $R_{By} = 90 \, \text{kN}$ (up), $F_{AB} = 60\sqrt{3} \, \text{kN}$ (tension, pulls away from B, so its x-component is to the left, y-component is upwards), $F_{BC} = 30\sqrt{3} \, \text{kN}$ (tension, pulls away from B, so to the right).

$$\sum F_y = 0$$ $$R_{By} + F_{AB} \sin(60^\circ) = 0$$ $$90 + (60\sqrt{3})(\sqrt{3}/2) = 0$$ $$90 + 90 = 0 \implies 180 = 0 \quad (\text{Inconsistent})$$

There is a fundamental contradiction in the problem setup or my interpretation of the forces.
Let's re-examine the angles. The angles $60^\circ$ and $30^\circ$ are given at the base.
If the load is at A, and the supports are at B and C, then the members AB and AC are typically in compression for a downward load at the apex of a simple truss.

Let's assume the initial calculation for $F_{AB}$ and $F_{AC}$ being in compression was correct, and the error was in the FBD for Joint A.
If $F_{AB}$ is compression, it pushes into A. Its x-component is to the left, y-component is downwards.
If $F_{AC}$ is compression, it pushes into A. Its x-component is to the right, y-component is downwards.
Then $\sum F_y = -120 - F_{AB}\sin(60^\circ) - F_{AC}\sin(30^\circ) = 0$. This implies $-120 = \text{positive value}$, which is impossible.

This means that for a downward load at A, at least one of the members AB or AC must have an upward vertical component to counteract the load. This implies at least one of them must be in tension.

Let's re-evaluate the problem statement and diagram.
The diagram shows a triangular truss ABC. A vertical load of $120 \, \text{kN}$ is at A.
B and C are supports.
If B is a pin and C is a roller, the reactions $R_{By}=90 \, \text{kN}$ and $R_{Cy}=30 \, \text{kN}$ are correct.

Let's use the Method of Sections to find $F_{AB}$ and $F_{AC}$.
To find $F_{AB}$, cut the truss through AC, BC. Consider the left section. Take moments about C.
The forces $F_{AC}$ and $R_{Cy}$ pass through C.
The forces are $R_{By}$, $R_{Bx}$, $F_{AB}$, $F_{BC}$, and the $120 \, \text{kN}$ load.
Consider the entire truss.
To find $F_{AB}$, cut through AC and BC. Consider the left part.
Take moments about C.
$R_{By} \times 8 - 120 \times 6 - F_{AB,y} \times x_C = 0$
This is getting complicated. Let's stick to Method of Joints and find the error.

The error must be in the direction assumption for the forces at the joints.
Let's assume $F_{AB}$ and $F_{AC}$ are in compression, and $F_{BC}$ is in tension.
This is the typical scenario for a simple truss with a downward load at the apex.

Re-check Joint B:
$R_{By} = 90 \, \text{kN}$ (up)
$F_{AB}$ (compression, pushes into B, so its x-component is to the right, y-component is downwards).
$F_{BC}$ (tension, pulls away from B, so to the right).
$\sum F_y = 0$: $R_{By} - F_{AB} \sin(60^\circ) = 0 \implies 90 - F_{AB}(\sqrt{3}/2) = 0 \implies F_{AB} = 180/\sqrt{3} = 60\sqrt{3} \, \text{kN}$ (Compression). This is consistent.
$\sum F_x = 0$: $F_{AB} \cos(60^\circ) + F_{BC} = 0 \implies (60\sqrt{3})(1/2) + F_{BC} = 0 \implies 30\sqrt{3} + F_{BC} = 0 \implies F_{BC} = -30\sqrt{3} \, \text{kN}$.
This means $F_{BC}$ is in compression, not tension.

Let's correct $F_{BC}$ to be compression.
$F_{BC} = 30\sqrt{3} \, \text{kN}$ (Compression).

Re-check Joint C:
$R_{Cy} = 30 \, \text{kN}$ (up)
$F_{AC}$ (compression, pushes into C, so its x-component is to the left, y-component is downwards).
$F_{BC}$ (compression, pushes into C, so to the right).
$\sum F_y = 0$: $R_{Cy} - F_{AC} \sin(30^\circ) = 0 \implies 30 - F_{AC}(1/2) = 0 \implies F_{AC} = 60 \, \text{kN}$ (Compression). This is consistent.
$\sum F_x = 0$: $-F_{AC} \cos(30^\circ) + F_{BC} = 0 \implies -(60)(\sqrt{3}/2) + (30\sqrt{3}) = 0 \implies -30\sqrt{3} + 30\sqrt{3} = 0$. This is consistent.

Now, re-check Joint A with these forces:
$F_{AB} = 60\sqrt{3} \, \text{kN}$ (Compression) - pushes into A, so its x-component is to the left, y-component is downwards.
$F_{AC} = 60 \, \text{kN}$ (Compression) - pushes into A, so its x-component is to the right, y-component is downwards.
$120 \, \text{kN}$ (down).

$\sum F_x = 0$:
$-F_{AB} \cos(60^\circ) + F_{AC} \cos(30^\circ) = 0$
$-(60\sqrt{3})(1/2) + (60)(\sqrt{3}/2) = 0$
$-30\sqrt{3} + 30\sqrt{3} = 0$. Consistent.

$\sum F_y = 0$:
$-120 - F_{AB} \sin(60^\circ