To solve for the force in member $BC$ (denoted as $F_{BC}$) using the method of sections, we need to follow these steps:

1. Analyze the Truss Structure and Section

The truss is symmetric, and we will use a vertical section that cuts through members $BC$, $GC$, and $CH$. This will allow us to isolate one side of the truss for easier analysis.

2. Free-Body Diagram

We will draw a free-body diagram of either the left or the right portion of the truss (after cutting through the truss). For simplicity, let’s cut the truss through members $BC$, $GC$, and $CH$, and consider the left part of the truss (from $A$ to $G$).

3. Calculate Reaction Forces

We need to calculate the reactions at supports $A$ and $E$.

• $A$ is a pinned support, so it has both horizontal ($A_x$) and vertical ($A_y$) reactions.
• $E$ is a roller support, so it only has a vertical reaction ($E_y$).

Using equilibrium equations for the entire truss:

• Sum of forces in the $x$-direction: $\sum F_x = 0 \implies A_x = 0$ (no horizontal forces are applied).

• Sum of forces in the $y$-direction: $\sum F_y = 0 \implies A_y + E_y = P_1 + P_2$ Substituting $P_1 = 947 \, \text{N}$ and $P_2 = 435 \, \text{N}$, we get: $A_y + E_y = 947 + 435 = 1382 \, \text{N}$

• Moment about point $A$ (taking counterclockwise moments as positive): $\sum M_A = 0 \implies E_y \cdot 3a - P_2 \cdot (2a) - P_1 \cdot a = 0$ Substituting $P_1 = 947 \, \text{N}$, $P_2 = 435 \, \text{N}$, and $a = 3.7 \, \text{m}$: $E_y \cdot (3 \times 3.7) - 435 \cdot (2 \times 3.7) - 947 \cdot 3.7 = 0$ Simplifying: $E_y \cdot 11.1 - 3222.9 - 3503.9 = 0$ $E_y \cdot 11.1 = 6726.8$ $E_y = \frac{6726.8}{11.1} = 606 \, \text{N}$

• From $A_y + E_y = 1382 \, \text{N}$, we find: $A_y = 1382 - 606 = 776 \, \text{N}$

4. Analyze the Section

Now, let's apply the method of sections. We will analyze the left part of the truss and apply the equilibrium equations.

• Sum of forces in the $y$-direction: $\sum F_y = 0 \implies A_y - P_1 - F_{BC} \cdot \sin(\theta) = 0$ where $\theta$ is the angle of member $BC$ with the horizontal.

• Sum of moments about point $G$ (to eliminate forces in members $GC$ and $CH$): $\sum M_G = 0 \implies A_y \cdot a - F_{BC} \cdot a = 0$ Solving for $F_{BC}$, we get: $F_{BC} = A_y = 776 \, \text{N}$

Thus, the magnitude of the force in member $BC$ is $F_{BC} = 776 \, \text{N}$.

Conclusion:

The force in member $BC$ is $776 \, \text{N}$. Since we assumed all members to be in tension, the positive sign confirms that the force is tensile.

Would you like further details on any step, or to ask about another topic?

Here are 5 related questions you might be interested in:

1. How would the analysis change if the load $P_2$ was increased?
2. What if we analyzed the right section of the truss instead?
3. How do we determine if a member is in tension or compression?
4. How is the method of joints different from the method of sections?
5. Can this method be applied to trusses with inclined supports?

Tip: When calculating moments, choose points where multiple unknown forces act to simplify the equations.