Question: Determine the horizontal and vertical component of force at A, B, and C.
a = 20 + last 2 digits of your student number, in kN
b = 1.75 + (second to the last digit of your student number times 0.9), in meters
a = 27 kN
b = 1.75 m
Graph shape: a curved arch beam spans from A at the left support to C at the right support, with a peak at B near the center; there are two downward point loads labeled a, one on the left rising side and one on the right descending side, and a horizontal tie/member between A and the right side near C. position_hint = top-left
Determine reactions at A and C and the cable force
3 global Eqs
1 hinge condition
15 kN
2 m
A
B
C
Aₓ
Aᵧ
Cᵧ
0.5 m
2 m
2 m
1 m
15 kN
10 kN
2 m
A
B
C
0.5 m
2 m
2 m
1 m
15 kN
2 m
A
B
Aₓ
Aᵧ
Bₓ
Bᵧ
T
0.5 m
2 m
top-left: A curved cable/arch with supports at A and C and a crown point B; a 15 kN downward load is applied near the left side and a 10 kN downward load near the right side, with reaction arrows Aₓ left, Aᵧ up, and Cᵧ up; dimensions shown are 0.5 m, 2 m, 2 m, and 1 m; position_hint = top-left
top-right: The same arch/cable diagram with A on the left support, B at the top center, and C at the right support; downward loads are 15 kN near the left and 10 kN near the right; the span is marked 0.5 m, 2 m, 2 m, and 1 m; position_hint = top-right
bottom-left: A reduced left-half diagram of the arch/cable showing support A, crown B, a 15 kN downward load, and reaction components Aₓ, Aᵧ at A plus Bₓ, Bᵧ at B and horizontal cable force T; dimensions include 0.5 m and 2 m; position_hint = bottom-left
User: the first picture is the problem that need to solve while the 2nd picture is a free body diagram and a guide
1. **Problem Statement:**
Determine the horizontal and vertical components of the forces at points A, B, and C for the curved arch beam with given loads and dimensions.
2. **Given Data:**
- Load at left and right sides: $15\ \text{kN}$ and $10\ \text{kN}$ downward respectively.
- Distances: $0.5\ \text{m}$, $2\ \text{m}$, $2\ \text{m}$, and $1\ \text{m}$ as shown.
- Unknown reaction forces: $A_x$, $A_y$, $C_y$, and cable force $T$.
3. **Equilibrium Equations:**
For static equilibrium, sum of forces and moments must be zero.
- Sum of horizontal forces: $$\sum F_x = 0$$
- Sum of vertical forces: $$\sum F_y = 0$$
- Sum of moments about a point (e.g., A): $$\sum M_A = 0$$
4. **Step 1: Sum of horizontal forces**
$$A_x - T = 0 \implies A_x = T$$
5. **Step 2: Sum of vertical forces**
$$A_y + C_y - 15 - 10 = 0 \implies A_y + C_y = 25$$
6. **Step 3: Sum of moments about A**
Taking moments counterclockwise as positive:
$$-15 \times 0.5 - 10 \times (0.5 + 2 + 2) + C_y \times (0.5 + 2 + 2 + 1) = 0$$
Calculate distances:
$$-15 \times 0.5 - 10 \times 4.5 + C_y \times 5.5 = 0$$
$$-7.5 - 45 + 5.5 C_y = 0$$
$$5.5 C_y = 52.5$$
$$C_y = \frac{52.5}{5.5} = 9.545\ \text{kN}$$
7. **Step 4: Calculate $A_y$**
$$A_y = 25 - C_y = 25 - 9.545 = 15.455\ \text{kN}$$
8. **Step 5: Hinge condition at B**
At hinge B, horizontal and vertical components must satisfy:
$$B_x + A_x = 0$$
$$B_y + A_y - 15 = 0$$
9. **Step 6: Calculate cable force $T$**
From step 4, $A_x = T$ and from hinge condition $B_x = -A_x = -T$.
10. **Step 7: Calculate $B_y$**
$$B_y = 15 - A_y = 15 - 15.455 = -0.455\ \text{kN}$$ (downward)
11. **Summary of forces:**
- $A_x = T$ (unknown horizontal cable force)
- $A_y = 15.455\ \text{kN}$ (upward)
- $C_y = 9.545\ \text{kN}$ (upward)
- $B_x = -T$
- $B_y = -0.455\ \text{kN}$ (downward)
12. **Interpretation:**
The horizontal cable force $T$ equals $A_x$ and balances $B_x$. Vertical reactions at supports balance the loads.
**Final answers:**
$$A_x = T$$
$$A_y = 15.455\ \text{kN}$$
$$C_y = 9.545\ \text{kN}$$
$$B_x = -T$$
$$B_y = -0.455\ \text{kN}$$