There are two forces on the 2.00 kg box in the overhead view of but only one is shown.

There are two forces on the 2.00 kg box in the overhead view of but only one is shown. For F1 = 20.0N, a = 12.0 m/s² and ? = 30.0 degrees find the second force (a) in unit-vector notation and as (b) a magnitude and (c) an angle relative to the positive direction of the x axis.

The Correct Answer and Explanation is:

Given:

Mass of the box, m=2.00 kgm = 2.00 \, \text{kg}
Acceleration, a⃗=12.0 m/s2\vec{a} = 12.0 \, \text{m/s}^2
One force, F⃗1=20.0 N\vec{F}_1 = 20.0 \, \text{N} at θ=30.0∘\theta = 30.0^\circ from the x-axis
Need to find the second force F⃗2\vec{F}_2

Step 1: Calculate the Net Force Using Newton’s Second Law

F⃗net=ma⃗=2.00×12.0=24.0 N\vec{F}_{\text{net}} = m \vec{a} = 2.00 \times 12.0 = 24.0 \, \text{N}

This is the total net force acting on the box.

Step 2: Resolve F⃗1\vec{F}_1 Into Components

F⃗1=20.0cos⁡(30∘) i^+20.0sin⁡(30∘) j^=(17.32 i^+10.0 j^) N\vec{F}_1 = 20.0 \cos(30^\circ) \, \hat{i} + 20.0 \sin(30^\circ) \, \hat{j} = (17.32 \, \hat{i} + 10.0 \, \hat{j}) \, \text{N}

Step 3: Let F⃗2=F2x i^+F2y j^\vec{F}_2 = F_{2x} \, \hat{i} + F_{2y} \, \hat{j}

Then: F⃗1+F⃗2=F⃗net⇒F⃗2=F⃗net−F⃗1\vec{F}_1 + \vec{F}_2 = \vec{F}_{\text{net}} \Rightarrow \vec{F}_2 = \vec{F}_{\text{net}} – \vec{F}_1

Let’s assume net force is in some direction (unknown angle, but the magnitude is 24 N). Since only magnitude is given, let’s assume it is entirely along the x-axis for simplicity (unless otherwise told, this is a standard assumption).

So: F⃗net=24.0 i^\vec{F}_{\text{net}} = 24.0 \, \hat{i}

Now subtract: F⃗2=(24.0 i^)−(17.32 i^+10.0 j^)=(6.68 i^−10.0 j^) N\vec{F}_2 = (24.0 \, \hat{i}) – (17.32 \, \hat{i} + 10.0 \, \hat{j}) = (6.68 \, \hat{i} – 10.0 \, \hat{j}) \, \text{N}

(a) Answer in Unit Vector Notation:

F⃗2=6.68 i^−10.0 j^ N\boxed{\vec{F}_2 = 6.68 \, \hat{i} – 10.0 \, \hat{j} \, \text{N}}

(b) Magnitude of F⃗2\vec{F}_2:

∣F⃗2∣=(6.68)2+(−10.0)2=44.6+100=144.6≈12.0 N|\vec{F}_2| = \sqrt{(6.68)^2 + (-10.0)^2} = \sqrt{44.6 + 100} = \sqrt{144.6} \approx \boxed{12.0 \, \text{N}}

(c) Direction of F⃗2\vec{F}_2:

θ=tan⁡−1(−10.06.68)=tan⁡−1(−1.497)≈−56.3∘\theta = \tan^{-1}\left(\frac{-10.0}{6.68}\right) = \tan^{-1}(-1.497) \approx -56.3^\circ

So, the angle is 56.3° below the +x axis, or: θ=303.7∘ counterclockwise from +x axis\boxed{\theta = 303.7^\circ \text{ counterclockwise from +x axis}}

Explanation

This problem involves Newton’s second law in vector form. We’re told that a 2.00 kg box experiences two forces, one of which is given. The goal is to find the second force that causes a specific acceleration. First, we use Newton’s second law: F⃗net=ma⃗\vec{F}_{\text{net}} = m \vec{a}. Since the mass is 2.00 kg and acceleration is 12.0 m/s², the net force must be 24.0 N.

The given force F⃗1\vec{F}_1 has a magnitude of 20.0 N at 30° to the x-axis, which we resolve into x and y components using trigonometry: F1x=20.0cos⁡(30∘)=17.32 N,F1y=20.0sin⁡(30∘)=10.0 NF_{1x} = 20.0 \cos(30^\circ) = 17.32\, \text{N}, \quad F_{1y} = 20.0 \sin(30^\circ) = 10.0\, \text{N}

To find the second force F⃗2\vec{F}_2, we subtract F⃗1\vec{F}_1 from the net force vector. Assuming the net force is along the x-axis (a common simplification unless a direction is specified), the net force is 24.0 i^24.0 \, \hat{i}. Subtracting F⃗1\vec{F}_1 gives: F⃗2=(24.0 i^)−(17.32 i^+10.0 j^)=6.68 i^−10.0 j^\vec{F}_2 = (24.0 \, \hat{i}) – (17.32 \, \hat{i} + 10.0 \, \hat{j}) = 6.68 \, \hat{i} – 10.0 \, \hat{j}

This result shows the second force acts partly forward and partly downward. We compute its magnitude with the Pythagorean theorem and direction using inverse tangent. The final direction is either as a negative angle from the x-axis or converted to a positive counterclockwise angle for standard notation.