the single force equivalent to the two forces acting on the mule has a magnitude of approximately 194 N and is directed at an angle of approximately 64.4° counterclockwise from the positive x-axis.
To find the single force equivalent to the two forces pulling on the mule, we can use the principle of vector addition. Here's how:
1. Resolve the forces into horizontal and vertical components:
Force 1 (F1):
* Horizontal component (F1x) = F1 * cos(60°) = 140 N * cos(60°) ≈ 70 N
* Vertical component (F1y) = F1 * sin(60°) = 140 N * sin(60°) ≈ 122 N
Force 2 (F2):
* Horizontal component (F2x) = F2 * cos(75°) ≈ 16.9 N
* Vertical component (F2y) = F2 * sin(75°) ≈ 56.8 N
2. Add the horizontal and vertical components separately:
* Total horizontal component (Fx) = F1x + F2x ≈ 70 N + 16.9 N ≈ 86.9 N
* Total vertical component (Fy) = F1y + F2y ≈ 122 N + 56.8 N ≈ 178.8 N
3. Calculate the magnitude of the resultant force (F):
F = √(Fx² + Fy²) ≈ √(86.9² + 178.8²) ≈ 194 N
4. Determine the direction of the resultant force (θ):
θ = tan⁻¹(Fy / Fx) ≈ tan⁻¹(178.8 / 86.9) ≈ 64.4°
Therefore, the single force equivalent to the two forces acting on the mule has a magnitude of approximately 194 N and is directed at an angle of approximately 64.4° counterclockwise from the positive x-axis.
Note: This solution assumes that the two forces are acting in the same plane. If the forces are not in the same plane, a more complex analysis would be required.