https://file.aiquickdraw.com/mathbot/file/1728501738183yui30uuc.jpg

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?

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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.

There are two forces on the 2.76 kg box in the overhead view of the figure but only one is shown. For F_1 = 16.6 N, a = 12.0 m/s^2, and θ = 26.6°, find the second force (a) in unit-vector notation and as (b) magnitude and (c) direction. (State the direction as a negative angle measured from the +x direction.)

Learn how to calculate the second force acting on a 2.76 kg box with one known force and acceleration. This problem involves applying Newton's Second Law, vector decomposition, and trigonometry to find the second force in unit-vector form, magnitude, and direction. Suitable for high school physics students.

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