Question

A force of 5N and a force of 8N act to the same point and are inclined at 45degree to each other. Find the magnitude and direction of the resultant force.

Answer 1

• Magnitude: 12.1 N.
• Direction: 17.0° to the 8 N force.

Explanation

Refer to the diagram attached (created with GeoGebra). Consider the 5 N force in two directions: parallel to the 8 N force and normal to the 8 N force.

• $\displaystyle F_{\text{1, Parallel}} = F_1 \cdot \cos{45^\textdegree} = \dfrac{5\sqrt{2}}{2}\;\text{N}$.
• $\displaystyle F_{\text{1, Normal}} = F_1 \cdot \sin{45^\textdegree} = \dfrac{5\sqrt{2}}{2}\;\text{N}$.

The sum of forces on each direction will be the resultant force on that direction:

• Resultant force parallel to the 8 N force: $(8 + \dfrac{5\sqrt{2}}{2})\;\text{N}$.
• Resultant force normal to the 8 N force: $\dfrac{5\sqrt{2}}{2}\;\text{N}$.

Apply the Pythagorean Theorem to find the magnitude of the resultant force.

$\displaystyle \Sigma F = \sqrt{{(8 + \dfrac{5\sqrt{2}}{2})}^2 + {(\dfrac{5\sqrt{2}}{2})}^2} = 12.1\;\text{N}$ (3 sig. fig.).

The size of the angle between the resultant force and the 8 N force can be found from the tangent value of the angle. Tangent of the angle:

$\displaystyle \dfrac{\Sigma F_\text{Normal}}{\Sigma F_\text{Parallel}} = \dfrac{8 + \dfrac{5\sqrt{2}}{2}}{\dfrac{5\sqrt{2}}{2}} \approx 0.306491$.

Find the size of the angle using inverse tangent:

$\displaystyle \arctan{ \dfrac{\Sigma F_\text{Normal}}{\Sigma F_\text{Parallel}}} = \arctan{0.306491} = 17.0\textdegree$.

In other words, the resultant force is 17.0° relative to the 8 N force.