Four forces of magnitude P, 2P, 3P and 4P act along the four sides of a square ABCD in cyclic order. Use the vector method to find the magnitude of resultant force.

Explanation

To solve the problem of finding the resultant force acting on a square ABCD due to the forces of magnitudes P, 2P, 3P, and 4P, we will use vector addition. The forces act along the sides of the square in a cyclic order.

### Step-by-Step Solution:

1. **Identify the Forces and Directions**:
- Let’s denote the forces acting on the square:
- Force \( F_1 = P \) acting to the right along side AB.
- Force \( F_2 = 2P \) acting upwards along side BC.
- Force \( F_3 = 3P \) acting to the left along side CD.
- Force \( F_4 = 4P \) acting downwards along side DA.

2. **Assign Coordinate Directions**:
- We will use a Cartesian coordinate system:
- Right (along AB) is the positive x-direction.
- Up (along BC) is the positive y-direction.
- Left (along CD) is the negative x-direction.
- Down (along DA) is the negative y-direction.

3. **Express Forces as Vectors**:
- Convert each force into vector form:
- \( \vec{F_1} = P \hat{i} \)
- \( \vec{F_2} = 2P \hat{j} \)
- \( \vec{F_3} = -3P \hat{i} \)
- \( \vec{F_4} = -4P \hat{j} \)

4. **Sum the Forces**:
- The resultant force \( \vec{R} \) can be found by summing all the forces:
\[ \vec{R} = \vec{F_1} + \vec{F_2} + \vec{F_3} + \vec{F_4} \]
\[ \vec{R} = (P - 3P) \hat{i} + (2P - 4P) \hat{j} \]
\[ \vec{R} = -2P \hat{i} - 2P \hat{j} \]

5. **Calculate the Magnitude of the Resultant Force**:
- The magnitude of the resultant vector \( \vec{R} \) is given by:
\[ |\vec{R}| = \sqrt{(-2P)^2 + (-2P)^2} \]
\[ |\vec{R}| = \sqrt{4P^2 + 4P^2} = \sqrt{8P^2} = 2\sqrt{2}P \]

### Final Result:
The magnitude of the resultant force is \( 2\sqrt{2}P \).

---