Three boys are pulling a heavy trolled by means of three ropes. The boy in the middle is exerting a pull of `100N`. The other two boys ,whose ropes both make an angle `30^(@)` with the central rope, are pulling with forces of `50sqrt(3)N` and `100sqrt(3)N`. Find the magnitudes of the resultant pull on the trolltey.

Explanation

To solve the problem of finding the resultant pull on the trolley by the three boys, we will break down the forces into their components and then combine them to find the resultant.

### Step 1: Identify the Forces
- The middle boy exerts a force of \( F_m = 100 \, N \) in the positive x-direction.
- The left boy exerts a force of \( F_l = 50\sqrt{3} \, N \) at an angle of \( 30^\circ \) to the left of the middle boy.
- The right boy exerts a force of \( F_r = 100\sqrt{3} \, N \) at an angle of \( 30^\circ \) to the right of the middle boy.

### Step 2: Break the Forces into Components
We will break down the forces exerted by the left and right boys into their x and y components.

**Left Boy's Force:**
- \( F_{lx} = F_l \cos(30^\circ) = 50\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 75 \, N \) (to the left, which we consider negative)
- \( F_{ly} = F_l \sin(30^\circ) = 50\sqrt{3} \cdot \frac{1}{2} = 25\sqrt{3} \, N \) (upwards)

**Right Boy's Force:**
- \( F_{rx} = F_r \cos(30^\circ) = 100\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 150 \, N \) (to the right, which we consider positive)
- \( F_{ry} = F_r \sin(30^\circ) = 100\sqrt{3} \cdot \frac{1}{2} = 50\sqrt{3} \, N \) (upwards)

### Step 3: Calculate the Total Components
Now, we will sum up the x and y components of all the forces.

**Total x-component:**
\[ F_{Rx} = F_m + F_{lx} + F_{rx} = 100 + (-75) + 150 = 175 \, N \]

**Total y-component:**
\[ F_{Ry} = F_{ly} + F_{ry} = 25\sqrt{3} + 50\sqrt{3} = 75\sqrt{3} \, N \]

### Step 4: Calculate the Magnitude of the Resultant Force
The magnitude of the resultant force \( R \) can be calculated using the Pythagorean theorem:
\[ R = \sqrt{(F_{Rx})^2 + (F_{Ry})^2} = \sqrt{(175)^2 + (75\sqrt{3})^2} \]
Calculating \( (75\sqrt{3})^2 \):
\[ (75\sqrt{3})^2 = 75^2 \cdot 3 = 5625 \cdot 3 = 16875 \]
Now substituting back:
\[ R = \sqrt{30625 + 16875} = \sqrt{47500} \]
Calculating the square root:
\[ R \approx 218.2 \, N \]

### Final Result
The magnitude of the resultant pull on the trolley is approximately \( 218.2 \, N \).

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