1. **State the problem:** We have forces of magnitudes 2N, 4N, 6N, 1N, and 5N acting along the sides AB, BC, CD, DE, and EA respectively of a rectangular pentagon with side length 1.5m. AB is taken as the reference X-axis. We need to find the magnitude and direction of the resultant force. 2. **Identify directions of forces:** Since the pentagon is rectangular, the sides are oriented at right angles. Taking AB along the positive X-axis: - Force on AB (2N) acts along +X. - Force on BC (4N) acts along +Y. - Force on CD (6N) acts along -X. - Force on DE (1N) acts along -Y. - Force on EA (5N) acts along +X (since EA closes the pentagon and is parallel to AB). 3. **Resolve forces into components:** - Along X-axis: $2 + (-6) + 5 = 2 - 6 + 5 = 1$ N - Along Y-axis: $4 + (-1) = 3$ N 4. **Calculate magnitude of resultant force:** $$ R = \sqrt{(1)^2 + (3)^2} = \sqrt{1 + 9} = \sqrt{10} \approx 3.16 \text{ N} $$ 5. **Calculate direction of resultant force:** Angle $\theta$ with respect to AB (X-axis) is given by: $$ \theta = \tan^{-1}\left(\frac{3}{1}\right) = \tan^{-1}(3) \approx 71.57^\circ $$ **Final answer:** The magnitude of the resultant force is approximately 3.16 N and it acts at an angle of about 71.57 degrees above the AB side (X-axis).