Six forces lying in a plane and forming angles of `60^(@)` relative to one another are applied to the center of a homogenous sphere with a mass m = 6 kg. These forces are radially outward and consecutively IN, 2N, 3N, 4N, 5N and 6N. The acceleration of the sphere is

Options

A. Zero
B. 1/2 `m//s^(2)`
C. `1 m//s^(2)`
D. `2 m//s^(2)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the Forces
We have six forces acting on the sphere:
- F1 = 1 N
- F2 = 2 N
- F3 = 3 N
- F4 = 4 N
- F5 = 5 N
- F6 = 6 N

These forces are applied at angles of 60 degrees relative to each other.

### Step 2: Calculate the Resultant of Opposite Forces
Since the forces are radially outward and arranged in a circular pattern, we can group them into pairs that are opposite to each other.

1. **Pair F1 and F4**:
- Resultant (R1) = F4 - F1 = 4 N - 1 N = 3 N

2. **Pair F2 and F5**:
- Resultant (R2) = F5 - F2 = 5 N - 2 N = 3 N

3. **Pair F3 and F6**:
- Resultant (R3) = F6 - F3 = 6 N - 3 N = 3 N

### Step 3: Combine the Resultant Forces
Now we have three resultant forces (R1, R2, R3), each equal to 3 N. Since these resultant forces are also at angles of 60 degrees to each other, we can find the net resultant force (R_net) using vector addition.

### Step 4: Calculate the Net Resultant Force
Since the angle between each resultant force is 60 degrees, we can use the formula for the resultant of two forces at an angle θ:

\[ R = \sqrt{R_1^2 + R_2^2 + 2R_1R_2 \cos(θ)} \]

However, since we have three equal forces (R1, R2, R3) of 3 N each, we can simplify the calculation. The resultant of three equal forces at 120 degrees can be computed as:

\[ R_{net} = 3 + 3 + 3 = 9 \text{ N (in the direction of the resultant)} \]

### Step 5: Calculate the Acceleration of the Sphere
Using Newton's second law, we can find the acceleration (a) of the sphere:

\[ F = ma \implies a = \frac{F}{m} \]

Where:
- F = net resultant force = 9 N
- m = mass of the sphere = 6 kg

Substituting the values:

\[ a = \frac{9 \text{ N}}{6 \text{ kg}} = 1.5 \text{ m/s}^2 \]

### Final Answer
The acceleration of the sphere is **1.5 m/s²**.

---