Let's analyze the problem step by step.

We are given:

• A 2.76 kg box.
• Force $\mathbf{F}_1 = 16.6 \, \text{N}$ directed along the $+x$-axis.
• The box has an acceleration $a = 12.0 \, \text{m/s}^2$ at an angle $\theta = 26.6^\circ$ relative to the $+x$-axis.

We are tasked with finding the second force $\mathbf{F}_2$, which in this case is likely causing the resultant acceleration at an angle.

Step 1: Calculate Net Force on the Box

Using Newton's second law, we know the total force acting on the box is:

$\mathbf{F}_{\text{net}} = m \cdot a$

Where:

• $m = 2.76 \, \text{kg}$ is the mass of the box.
• $a = 12.0 \, \text{m/s}^2$ is the magnitude of the acceleration.

Thus, the net force $\mathbf{F}_{\text{net}}$ is:

$\mathbf{F}_{\text{net}} = 2.76 \times 12.0 = 33.12 \, \text{N}$

Step 2: Break Down the Net Force into Components

Since the acceleration is at an angle $\theta = 26.6^\circ$ from the $+x$-axis, we can break the net force into components:

• In the $x$-direction: $F_{\text{net}, x} = 33.12 \cos(26.6^\circ)$

• In the $y$-direction: $F_{\text{net}, y} = 33.12 \sin(26.6^\circ)$

Let's compute these:

$F_{\text{net}, x} = 33.12 \times \cos(26.6^\circ) \approx 33.12 \times 0.894 = 29.62 \, \text{N}$

$F_{\text{net}, y} = 33.12 \times \sin(26.6^\circ) \approx 33.12 \times 0.448 = 14.84 \, \text{N}$

Step 3: Find $\mathbf{F}_2$

The net force is the vector sum of the forces $\mathbf{F}_1$ and $\mathbf{F}_2$. In vector form:

$\mathbf{F}_{\text{net}} = \mathbf{F}_1 + \mathbf{F}_2$

Since $\mathbf{F}_1$ is along the $x$-axis, we can write:

$\mathbf{F}_1 = (16.6 \, \hat{i}, 0 \, \hat{j})$

So:

$\mathbf{F}_2 = \mathbf{F}_{\text{net}} - \mathbf{F}_1$

In component form:

• $F_{2, x} = F_{\text{net}, x} - F_{1, x} = 29.62 - 16.6 = 13.02 \, \text{N}$
• $F_{2, y} = F_{\text{net}, y} = 14.84 \, \text{N}$

Thus, in unit-vector notation:

$\mathbf{F}_2 = (13.02 \, \hat{i} + 14.84 \, \hat{j}) \, \text{N}$

Step 4: Magnitude of $\mathbf{F}_2$

The magnitude of $\mathbf{F}_2$ is:

$|\mathbf{F}_2| = \sqrt{(13.02)^2 + (14.84)^2}$ $|\mathbf{F}_2| = \sqrt{169.52 + 220.32} = \sqrt{389.84} \approx 19.74 \, \text{N}$

Step 5: Direction of $\mathbf{F}_2$

The direction of $\mathbf{F}_2$ relative to the $+x$-axis is:

$\theta_2 = \tan^{-1} \left( \frac{14.84}{13.02} \right)$ $\theta_2 \approx \tan^{-1}(1.14) \approx 48.5^\circ$

Since the force is above the $+x$-axis, the direction is positive.

Final Answer:

• (a) In unit-vector notation: $\mathbf{F}_2 = 13.02 \, \hat{i} + 14.84 \, \hat{j} \, \text{N}$
• (b) Magnitude: $19.74 \, \text{N}$
• (c) Direction: $48.5^\circ$ above the $+x$-axis.

Would you like more details on any part of this solution, or clarification?

Here are 5 related questions:

1. How would the solution change if $\mathbf{F}_1$ were not along the $x$-axis?
2. What if the angle $\theta$ were negative, indicating motion below the $x$-axis?
3. How do forces combine vectorially in 3D instead of 2D?
4. What happens if friction is introduced to the surface?
5. How can we verify the net force using work-energy principles?

Tip: Breaking forces into components using trigonometry is key for solving 2D force problems like this one.