Problem: Forces of magnitude 2P, 4P, 3P and P act on a particle in directions parallel to the sides AB, BC, CD and DE of a regular hexagon. Find the magnitude and direction of the resultant.

Solution:

To solve this problem, we need to use vector addition. We can represent each force as a vector and add them using the parallelogram law of vector addition.

Step 1: Draw a diagram of the hexagon and the forces acting on it.

Step 2: Represent each force as a vector with its tail at the starting point of the force and its head at the ending point of the force. Label each vector with its magnitude and direction.

Step 3: Using the parallelogram law of vector addition, draw the diagonal of the parallelogram formed by the vectors 2P and 4P. This diagonal represents the resultant of these two vectors.

Step 4: Draw the diagonal of the parallelogram formed by the vectors 3P and P. This diagonal represents the resultant of these two vectors.

Step 5: Add the two resultant vectors using vector addition. The magnitude and direction of the resulting vector is the magnitude and direction of the resultant force.

Step 6: Use trigonometry to determine the magnitude and direction of the resultant force.

The magnitude of the resultant force is given by:

|F| = sqrt[(2P)^2 + (4P)^2 + 2(2P)(4P)cos(120) + (3P)^2 + P^2 + 2(3P)(P)cos(120)]

Simplifying this expression gives:

|F| = sqrt(38P^2)

|F| = 2sqrt(19)P

The direction of the resultant force is given by:

theta = arctan[(3P + Psin(120))/(2P + 4Pcos(120))]

Simplifying this expression gives:

theta = arctan[(2sqrt(3) + sqrt(3)/2)/(3/2)]

theta = arctan(4sqrt(3)/3)

theta = 71.57 degrees (rounded to two decimal places)

Therefore, the magnitude of the resultant force is 2sqrt(19)P and its direction is 71.57 degrees.