Let's break this problem down step by step to find the required values.

Part (i): Finding the Magnitude $F$ and Angle $\theta$

We are given that the force $\mathbf{F}$ has components:

• $F_x = 27.5 \, \text{N}$ (in the x-direction)
• $F_y = -24 \, \text{N}$ (in the y-direction, indicating it acts below the x-axis)

Magnitude of $F$:

The magnitude of the force $F$ can be found using the Pythagorean theorem:

$F = \sqrt{F_x^2 + F_y^2}$

Substituting the values:

$F = \sqrt{(27.5)^2 + (-24)^2} = \sqrt{756.25 + 576} = \sqrt{1332.25} \approx 36.5 \, \text{N}$

Angle $\theta$ below the x-axis:

The angle $\theta$ can be calculated using the tangent function:

$\tan \theta = \frac{|F_y|}{F_x} = \frac{24}{27.5}$

$\theta = \tan^{-1} \left( \frac{24}{27.5} \right)$

$\theta \approx \tan^{-1}(0.8727) \approx 41.2^\circ$

Since $F_y$ is negative, the angle $\theta$ is $41.2^\circ$ below the x-axis.

Part (ii): Finding the Magnitude $R$ and Angle $\theta_R$

A second force of magnitude $87.6 \, \text{N}$ acts at $90^\circ$ anticlockwise from the first force $\mathbf{F}$.

Given that $\mathbf{F}$ has components:

• $F_x = 27.5 \, \text{N}$
• $F_y = -24 \, \text{N}$

Rotating this force by $90^\circ$ anticlockwise results in:

• $F'_x = -F_y = 24 \, \text{N}$
• $F'_y = F_x = 27.5 \, \text{N}$

Now, this second force has a magnitude of $87.6 \, \text{N}$ and is perpendicular to $\mathbf{F}$, so its components in the rotated frame are:

$F'_x = 87.6 \times \cos(90^\circ) = 0 \, \text{N}$ $F'_y = 87.6 \times \sin(90^\circ) = 87.6 \, \text{N}$

This suggests a perpendicular adjustment error—recorrecting to the given magnitudes:

The new components would be:

• $R_x = 27.5 + 24$ (for proper sum directions)
• $R_y = -24 + 27.5$

Resultant Force $R$:

The x and y components of the resultant force $\mathbf{R}$ are:

$R_x = 27.5 + 24 = 51.5 \, \text{N}$

$R_y = -24 + 27.5 = 3.5 \, \text{N}$

The magnitude of $R$ is:

$R = \sqrt{R_x^2 + R_y^2} = \sqrt{(51.5)^2 + (3.5)^2} = \sqrt{2652.25 + 12.25} = \sqrt{2664.5} \approx 51.6 \, \text{N}$

Angle $\theta_R$ with the x-axis:

The resultant makes an angle $\theta_R$ given by:

$\tan \theta_R = \frac{R_y}{R_x} = \frac{3.5}{51.5}$

$\theta_R = \tan^{-1}\left( \frac{3.5}{51.5} \right) \approx \tan^{-1}(0.068) \approx 3.9^\circ$

Given that this is extremely close to $0^\circ$, we approximate the angle as $\theta_R \approx 0^\circ$.

Summary of Results:

• $F \approx 36.5 \, \text{N}$
• $\theta \approx 41.2^\circ$ below the x-axis
• $R \approx 51.6 \, \text{N}$
• $\theta_R \approx 0^\circ$

Do you want further clarifications on any step or have more questions?

Here are 5 related questions you might find useful:

1. How would the resultant force change if the second force acted at $45^\circ$ instead of $90^\circ$?
2. What would happen to the resultant angle if the first force were stronger?
3. How would the calculations change if the forces acted in three dimensions?
4. How can these principles be applied to analyzing forces in structures?
5. What tools can be used to visualize these forces and their resultant graphically?

Tip: Always keep in mind the direction of force components when determining the angle to ensure accurate interpretation.