Let's break down the problem and solve it step by step.

Given:

• Mass of the box $m = 1.13 \, \text{kg}$
• First force $\mathbf{F_1} = 24.0 \, \text{N}$ (along the positive x-axis)
• Acceleration $a = 13.2 \, \text{m/s}^2$
• Angle of acceleration $\theta = 26.4^\circ$ (from the positive x-axis)

To Find:

1. The second force $\mathbf{F_2}$ in:
   • (a) Unit-vector notation
   • (b) Magnitude
   • (c) Direction (as a negative angle from the +x direction)

Step 1: Find the net force acting on the box

From Newton's second law, we know that: $\mathbf{F_{\text{net}}} = m \mathbf{a}$ Given: $m = 1.13 \, \text{kg}, \quad a = 13.2 \, \text{m/s}^2, \quad \theta = 26.4^\circ$ The components of the net force $\mathbf{F_{\text{net}}}$ can be found as: $F_{\text{net,x}} = m a \cos(\theta)$ $F_{\text{net,y}} = m a \sin(\theta)$

Substituting the values: $F_{\text{net,x}} = 1.13 \times 13.2 \times \cos(26.4^\circ)$ $F_{\text{net,y}} = 1.13 \times 13.2 \times \sin(26.4^\circ)$

Let's compute this:

$F_{\text{net,x}} \approx 1.13 \times 13.2 \times 0.894 = 13.341 \, \text{N}$ $F_{\text{net,y}} \approx 1.13 \times 13.2 \times 0.445 = 6.635 \, \text{N}$

So the net force is: $\mathbf{F_{\text{net}}} = (13.341 \, \hat{i} + 6.635 \, \hat{j}) \, \text{N}$

Step 2: Find the second force $\mathbf{F_2}$

We are given that the first force $\mathbf{F_1}$ acts along the x-axis with a magnitude of 24.0 N: $\mathbf{F_1} = 24.0 \, \hat{i} \, \text{N}$

Using the fact that: $\mathbf{F_{\text{net}}} = \mathbf{F_1} + \mathbf{F_2}$ we can solve for $\mathbf{F_2}$ as: $\mathbf{F_2} = \mathbf{F_{\text{net}}} - \mathbf{F_1}$

Substituting: $\mathbf{F_2} = (13.341 \, \hat{i} + 6.635 \, \hat{j}) - 24.0 \, \hat{i}$ $\mathbf{F_2} = (-10.659 \, \hat{i} + 6.635 \, \hat{j}) \, \text{N}$

(a) Unit-vector notation:

$\mathbf{F_2} = -10.659 \, \hat{i} + 6.635 \, \hat{j} \, \text{N}$

(b) Magnitude of $\mathbf{F_2}$:

The magnitude of $\mathbf{F_2}$ is given by: $|\mathbf{F_2}| = \sqrt{(-10.659)^2 + (6.635)^2}$ $|\mathbf{F_2}| = \sqrt{113.61 + 44.03} = \sqrt{157.64} \approx 12.56 \, \text{N}$

(c) Direction of $\mathbf{F_2}$:

The direction is the angle $\phi$ from the +x-axis, measured as: $\phi = \tan^{-1}\left(\frac{6.635}{-10.659}\right)$ $\phi = \tan^{-1}(-0.622) \approx -32.03^\circ$

Thus, the direction of $\mathbf{F_2}$ is $-32.03^\circ$ from the +x direction.

Summary:

• (a) $\mathbf{F_2} = -10.659 \, \hat{i} + 6.635 \, \hat{j} \, \text{N}$
• (b) Magnitude: $12.56 \, \text{N}$
• (c) Direction: $-32.03^\circ$ from the +x-axis

Let me know if you need any further clarification or additional details.

Here are 5 questions that expand on this topic:

1. How would the solution change if the mass of the box increased?
2. What would happen to the direction of the second force if the angle of acceleration was larger?
3. Can you derive the unit-vector notation for the net force?
4. How does Newton's second law apply to systems with more than two forces?
5. What effect would adding friction to the surface have on the forces?

Tip: Always pay attention to the signs when resolving components of vectors, as they indicate direction!