There are two forces on the 2.00 kg box in the overhead view of Figure, but only one is shown. For F1 = 20.0 N, a = 12.0 m/s², and ? = 30.0°, 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:
We are given the following:
- Mass of the box: m=2.00 kgm = 2.00\, \text{kg}
- Acceleration: a⃗\vec{a}, magnitude a=12.0 m/s2a = 12.0\, \text{m/s}^2, direction θ=30.0∘\theta = 30.0^\circ below the +x-axis
- Force F⃗1=20.0 Ni^\vec{F}_1 = 20.0\, \text{N} \hat{i}
- We are to find the second force F⃗2\vec{F}_2
Newton’s Second Law:
F⃗net=ma⃗=F⃗1+F⃗2\vec{F}_{\text{net}} = m\vec{a} = \vec{F}_1 + \vec{F}_2
So we rearrange to solve for F⃗2\vec{F}_2: F⃗2=ma⃗−F⃗1\vec{F}_2 = m\vec{a} – \vec{F}_1
Step 1: Find ma⃗m\vec{a} in unit-vector form
Given:
- a=12.0 m/s2a = 12.0\, \text{m/s}^2
- θ=30∘\theta = 30^\circ below the +x-axis
Break into components: ax=acos(30∘)=12.0×32=10.39 m/s2a_x = a \cos(30^\circ) = 12.0 \times \frac{\sqrt{3}}{2} = 10.39\, \text{m/s}^2 ay=−asin(30∘)=−12.0×12=−6.00 m/s2a_y = -a \sin(30^\circ) = -12.0 \times \frac{1}{2} = -6.00\, \text{m/s}^2
Now multiply by mass: ma⃗=2.00 kg×(10.39i^−6.00j^)=(20.78i^−12.0j^) Nm\vec{a} = 2.00\, \text{kg} \times (10.39 \hat{i} – 6.00 \hat{j}) = (20.78 \hat{i} – 12.0 \hat{j})\, \text{N}
Step 2: Subtract F⃗1\vec{F}_1 to get F⃗2\vec{F}_2
F⃗2=(20.78i^−12.0j^)−(20.0i^)=(0.78i^−12.0j^) N\vec{F}_2 = (20.78 \hat{i} – 12.0 \hat{j}) – (20.0 \hat{i}) = (0.78 \hat{i} – 12.0 \hat{j})\, \text{N}
Final Answers:
(a) Unit vector notation: F⃗2=0.78i^−12.0j^ N\boxed{\vec{F}_2 = 0.78 \hat{i} – 12.0 \hat{j}\, \text{N}}
(b) Magnitude of F⃗2\vec{F}_2: ∣F⃗2∣=0.782+12.02=0.61+144≈12.03 N|\vec{F}_2| = \sqrt{0.78^2 + 12.0^2} = \sqrt{0.61 + 144} \approx \boxed{12.03\, \text{N}}
(c) Direction relative to +x-axis: tanθ=∣−12.0∣0.78⇒θ=tan−1(12.00.78)≈86.3∘\tan\theta = \frac{|-12.0|}{0.78} \Rightarrow \theta = \tan^{-1}\left(\frac{12.0}{0.78}\right) \approx 86.3^\circ
Since it’s below the +x-axis: θ=86.3∘ below the +x-axis\boxed{\theta = 86.3^\circ\ \text{below the +x-axis}}
Explanation
This problem involves determining an unknown force acting on a 2.00 kg box, given its acceleration and one known force. According to Newton’s second law, the net force acting on an object is equal to the product of its mass and acceleration (F⃗net=ma⃗\vec{F}_{\text{net}} = m\vec{a}). The box is subject to two forces: one known force, F⃗1\vec{F}_1, acting directly along the +x-axis, and a second unknown force, F⃗2\vec{F}_2, which we are solving for.
We start by calculating the net force required to accelerate the 2.00 kg box at 12.0 m/s212.0\, \text{m/s}^2 at a 30∘30^\circ angle below the x-axis. To do this, we decompose the acceleration vector into its x and y components using trigonometric functions. The x-component is found using cosine, and the y-component using sine—keeping in mind the direction (negative y because it’s below the x-axis).
Multiplying these acceleration components by the mass yields the net force in vector form. From this net force, we subtract the given force F⃗1\vec{F}_1 (which has only an x-component) to isolate the second force F⃗2\vec{F}_2. This gives us the x and y components of F⃗2\vec{F}_2.
Finally, we compute the magnitude of F⃗2\vec{F}_2 using the Pythagorean theorem and determine the direction using the arctangent of the ratio of the y-component to the x-component. Because the force points downward, the angle is measured below the x-axis.
This approach combines Newton’s laws with vector analysis to find a missing force in a dynamic system.
