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To determine the single resultant force \( R \) that is equivalent to the three forces acting at point \( A \) (i.e., the weight \( W \), tension in cable \( AB \), and tension in cable \( AC \)), we need to follow these steps:

### Step 1: Set up the force vectors
We are given the following dataTo determine the single resultant force \( R \) that is equivalent to the three forces acting at point \( A \) (i.e., the weight \( W \), tension in cable \( AB \), and tension in cable \( AC \)), we need to follow these steps:

### Step 1: Set up the force vectors
We are given the following 
- Weight \( W = 480 \, \text{N} \) acting vertically downward at \( A \).
- Tension in cable \( AB \), \( T_{AB} = 534 \, \text{N} \).
- Tension in cable \( AC \), \( T_{AC} = 712 \, \text{N} \).

We will now determine the components of each force in the \( x \), \( y \), and \( z \) directions.

#### 1.1 Coordinates of points \( A \), \( B \), and \( C \)
- \( A = (3.66, 0, 0) \).
- \( B = (0, 0, 3.05) \).
- \( C = (2.44, 0.915, 0.915) \).

#### 1.2 Position vectors
Let’s define the position vectors:
- \( \vec{AB} = B - A = (0 - 3.66, 0 - 0, 3.05 - 0) = (-3.66, 0, 3.05) \).
- \( \vec{AC} = C - A = (2.44 - 3.66, 0.915 - 0, 0.915 - 0) = (-1.22, 0.915, 0.915) \).

#### 1.3 Normalize the vectors
To get the unit vectors along \( AB \) and \( AC \), we first find their magnitudes:

\[
|\vec{AB}| = \sqrt{(-3.66)^2 + 0^2 + (3.05)^2} = \sqrt{13.3956 + 9.3025} = \sqrt{22.6981} \approx 4.76 \, \text{m}
\]
\[
|\vec{AC}| = \sqrt{(-1.22)^2 + (0.915)^2 + (0.915)^2} = \sqrt{1.4884 + 0.837225 + 0.837225} = \sqrt{3.16285} \approx 1.78 \, \text{m}
\]

Now, the unit vectors:
\[
\hat{u}_{AB} = \frac{\vec{AB}}{|\vec{AB}|} = \frac{(-3.66, 0, 3.05)}{4.76} = (-0.769, 0, 0.641)
\]
\[
\hat{u}_{AC} = \frac{\vec{AC}}{|\vec{AC}|} = \frac{(-1.22, 0.915, 0.915)}{1.78} = (-0.685, 0.514, 0.514)
\]

### Step 2: Resolve forces into components
Now, we resolve the forces along \( AB \), \( AC \), and \( W \).

#### 2.1 Force \( T_{AB} \)
The components of the force \( T_{AB} \) are:
\[
F_{ABx} = T_{AB} \cdot (-0.769) = 534 \cdot (-0.769) = -410.75 \, \text{N}
\]
\[
F_{ABy} = 0 \, \text{N}
\]
\[
F_{ABz} = T_{AB} \cdot 0.641 = 534 \cdot 0.641 = 342.29 \, \text{N}
\]

#### 2.2 Force \( T_{AC} \)
The components of the force \( T_{AC} \) are:
\[
F_{ACx} = T_{AC} \cdot (-0.685) = 712 \cdot (-0.685) = -487.72 \, \text{N}
\]
\[
F_{ACy} = T_{AC} \cdot (0.514) = 712 \cdot 0.514 = 366.67 \, \text{N}
\]
\[
F_{ACz} = T_{AC} \cdot (0.514) = 712 \cdot 0.514 = 366.67 \, \text{N}
\]

#### 2.3 Force due to weight \( W \)
The weight \( W = 480 \, \text{N} \) acts vertically downward along the \( z \)-axis, so its components are:
\[
F_{Wx} = 0 \, \text{N}, \quad F_{Wy} = 0 \, \text{N}, \quad F_{Wz} = -480 \, \text{N}
\]

### Step 3: Calculate the resultant force components
Summing the components of all forces:

\[
R_x = F_{ABx} + F_{ACx} + F_{Wx} = -410.75 + (-487.72) + 0 = -898.47 \, \text{N}
\]
\[
R_y = F_{ABy} + F_{ACy

The horizontal boom carries the weight W = 480 N at A. The boom is supported by the cables AB and AC. The tension in the cables are TAB = 534 N and TAC = 712 N. Determine the single force R that is equivalent to three forces acting on the boom at A.

Learn how to calculate the single resultant force acting on a boom supported by two cables. This problem involves vector resolution, force components, and equilibrium concepts. Suitable for university-level engineering or physics students.

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