Six forces lying in a plane and forming angles of `60^(@)` relative to one another are applied to the centre of a homogeneous sphere with a mass `m =6kg`. These forces are radially outward and consecutively `1N,2N,3N,4N,5N` and `6N` The acceleration of the sphere is .

Options

A. `0`
B. `1//2 m//s^(2)`
C. `1m//s^(2)`
D. `2m//s^(2)`

Explanation

To solve the problem, we need to find the net acceleration of a homogeneous sphere subjected to six forces of varying magnitudes that are applied radially outward at angles of 60 degrees to each other. The forces are 1N, 2N, 3N, 4N, 5N, and 6N, and the mass of the sphere is 6 kg.

### Step-by-Step Solution:

1. **Identify the Forces and Their Directions:**
The forces are applied at angles of 60 degrees to each other. We can denote the forces as:
- F1 = 1N
- F2 = 2N
- F3 = 3N
- F4 = 4N
- F5 = 5N
- F6 = 6N

2. **Calculate the Net Force:**
Since the forces are symmetrically distributed at 60 degrees, we can analyze the forces in pairs. We can group them based on their opposite pairs:
- F1 and F4 (1N and 4N)
- F2 and F5 (2N and 5N)
- F3 and F6 (3N and 6N)

Each pair will cancel out some of the forces:
- F1 (1N) and F4 (4N) will yield a resultant force of 3N in the direction of F4.
- F2 (2N) and F5 (5N) will yield a resultant force of 3N in the direction of F5.
- F3 (3N) and F6 (6N) will yield a resultant force of 3N in the direction of F6.

3. **Find the Resultant Forces:**
After cancellation, we have three resultant forces of 3N each, acting at angles of 60 degrees to each other. We can represent these forces in a coordinate system:
- F1' = 3N (along x-axis)
- F2' = 3N (at 60 degrees to F1')
- F3' = 3N (at 120 degrees to F1')

4. **Calculate the Components of the Resultant Forces:**
We can find the x and y components of these forces:
- For F1':
- \( F_{x1} = 3N \)
- \( F_{y1} = 0 \)

- For F2':
- \( F_{x2} = 3 \cos(60^\circ) = 3 \times \frac{1}{2} = 1.5N \)
- \( F_{y2} = 3 \sin(60^\circ) = 3 \times \frac{\sqrt{3}}{2} = 2.598N \)

- For F3':
- \( F_{x3} = 3 \cos(120^\circ) = 3 \times \left(-\frac{1}{2}\right) = -1.5N \)
- \( F_{y3} = 3 \sin(120^\circ) = 3 \times \frac{\sqrt{3}}{2} = 2.598N \)

5. **Sum the Components:**
Now, we can sum the x and y components:
- Total \( F_x = F_{x1} + F_{x2} + F_{x3} = 3 + 1.5 - 1.5 = 3N \)
- Total \( F_y = F_{y1} + F_{y2} + F_{y3} = 0 + 2.598 + 2.598 = 5.196N \)

6. **Calculate the Magnitude of the Net Force:**
The magnitude of the net force \( F_{net} \) can be calculated using the Pythagorean theorem:
\[ F_{net} = \sqrt{(F_x)^2 + (F_y)^2} = \sqrt{(3)^2 + (5.196)^2} = \sqrt{9 + 27} = \sqrt{36} = 6N \]

7. **Calculate the Acceleration:**
Finally, we can find the acceleration \( a \) using Newton's second law:
\[ a = \frac{F_{net}}{m} = \frac{6N}{6kg} = 1 \, m/s^2 \]

### Final Answer:
The acceleration of the sphere is \( 1 \, m/s^2 \).

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