Question

Parallel and normal forces Find the components of the vertical force \(\mathbf{F}=\langle 0,-10\rangle\) in the directions parallel to and normal to the following inclined planes. Show that the total force is the sum of the two component forces. A plane that makes an angle of \(\pi / 6\) with the positive \(x\) -axis

Short answer

Question: Given a force vector 𝐅=⟨0,−10⟩ and an inclined plane with an angle of 𝜋/6, show that the total force is the sum of the components of force parallel and normal to the inclined plane. Answer: The total force 𝐅 can be decomposed into parallel and normal components, 𝐅ₚ and 𝐅ₙ. We have shown that 𝐅ₚ = -10 sin(𝜋 / 6) ⟨cos(𝜋 / 6), sin(𝜋 / 6)⟩ and 𝐅ₙ = -10 cos(𝜋 / 6) ⟨-sin(𝜋 / 6), cos(𝜋 / 6)⟩. Adding the components 𝐅ₚ and 𝐅ₙ yields the total force vector 𝐅 = ⟨0, -10⟩, showing that the total force is indeed the sum of its parallel and normal components.

Step-by-step solution

Find the unit vector parallel to the inclined plane

Since the inclined plane makes an angle of \(\pi / 6\) with the positive \(x\)-axis, we can represent a vector parallel to this plane as \(\mathbf{v} = \langle \cos(\pi / 6), \sin(\pi / 6) \rangle\). Next, we need to find the unit vector in this direction. Let's call this unit vector \(\mathbf{u_{p}}\). To find the unit vector, we will divide \(\mathbf{v}\) by its magnitude: $$\mathbf{u_{p}} = \frac{\mathbf{v}}{\Vert \mathbf{v} \Vert} = \frac{\langle \cos(\pi / 6), \sin(\pi / 6) \rangle}{\sqrt{(\cos(\pi / 6))^2 + (\sin(\pi / 6))^2}} = \langle \cos(\pi / 6), \sin(\pi / 6) \rangle$$

Find the component of force parallel to the inclined plane

We can find the component of the force parallel to the inclined plane by projecting \(\mathbf{F}\) onto \(\mathbf{u_{p}}\). We can do this using the dot product: $$\mathbf{F_{p}} = (\mathbf{F} \cdot \mathbf{u_{p}}) \mathbf{u_{p}} = (\langle 0, -10 \rangle \cdot \langle \cos(\pi / 6), \sin(\pi / 6) \rangle) \langle \cos(\pi / 6), \sin(\pi / 6) \rangle$$ $$\mathbf{F_{p}} = -10 \sin(\pi / 6) \langle \cos(\pi / 6), \sin(\pi / 6) \rangle$$

Find the component of force normal to the inclined plane

To find the component of force normal to the inclined plane, let's first find the unit vector normal to the inclined plane. We can do this by rotating \(\mathbf{u_{p}}\) by \(\pi / 2\) counterclockwise, which will give us \(\mathbf{u_{n}} = \langle -\sin(\pi / 6), \cos(\pi / 6) \rangle\). Next, we can find the normal component of the force by projecting \(\mathbf{F}\) onto \(\mathbf{u_{n}}\): $$\mathbf{F_{n}} = (\mathbf{F} \cdot \mathbf{u_{n}}) \mathbf{u_{n}} = (\langle 0, -10 \rangle \cdot \langle -\sin(\pi / 6), \cos(\pi / 6) \rangle) \langle -\sin(\pi / 6), \cos(\pi / 6) \rangle$$ $$\mathbf{F_{n}} = -10 \cos(\pi / 6) \langle -\sin(\pi / 6), \cos(\pi / 6) \rangle$$

Show that the total force is the sum of the two component forces

Lastly, we need to show that \(\mathbf{F} = \mathbf{F_{p}} + \mathbf{F_{n}}\). Summing \(\mathbf{F_{p}}\) and \(\mathbf{F_{n}}\), we have: $$\mathbf{F_{p}} + \mathbf{F_{n}} = -10 \sin(\pi / 6) \langle \cos(\pi / 6), \sin(\pi / 6) \rangle - 10 \cos(\pi / 6) \langle -\sin(\pi / 6), \cos(\pi / 6) \rangle$$ $$\mathbf{F_{p}} + \mathbf{F_{n}} = \langle -10 \cos(\pi / 6) \sin(\pi / 6) + 10 \sin^2(\pi / 6), -10 \cos^2(\pi / 6) - 10 \sin(\pi / 6) \cos(\pi / 6) \rangle$$ Recall that \(\cos^2(\pi / 6) + \sin^2(\pi / 6) = 1\), and we can simplify the above expression to: $$\mathbf{F_{p}} + \mathbf{F_{n}} = \langle 0, -10 \rangle$$ Therefore, we have shown that the total force \(\mathbf{F}\) is indeed the sum of the parallel and normal components, \(\mathbf{F_{p}}\) and \(\mathbf{F_{n}}\).

Key concepts

Inclined Plane

When objects rest on or move along surfaces that are not horizontal, they interact with inclined planes. These planes form an angle with the horizontal axis, which in this case is the positive x-axis. Understanding the angle of inclination is crucial, as it influences how forces act upon objects on the plane.
In our problem, the inclined plane has an angle of \( \pi / 6 \). Knowing this angle allows us to calculate forces that act parallel and normal (perpendicular) to the plane. An inclined plane essentially transforms how gravity affects the object, breaking the vertical force into components that interact specifically with the plane.
Key concepts include:

• Parallel Force Component: This is the portion of the total force that runs along the plane's surface.
• Normal Force Component: This part is perpendicular to the surface, preventing the object from passing through the plane.

Understanding these components is important for calculating how objects will move or remain stable on inclined surfaces.

Unit Vector

Unit vectors define direction but have a magnitude of one. They are essential in converting a given direction into a standard format.
In our exercise, we calculate the unit vector parallel to the inclined plane, \( \mathbf{u_{p}} \). To do this, we first find the vector in the desired direction, \( \mathbf{v} = \langle \cos(\pi / 6), \sin(\pi / 6) \rangle \).
Next, we normalize this vector by dividing it by its magnitude, keeping direction the same but adjusting magnitude to one:
\[ \mathbf{u_{p}} = \frac{\mathbf{v}}{\Vert \mathbf{v} \Vert} = \langle \cos(\pi / 6), \sin(\pi / 6) \rangle \]
This computation simplifies due to the properties of sine and cosine, ensuring that the magnitude is indeed one. Having unit vectors allows us to project other forces along these paths, understanding the force's implications in the given direction.

Dot Product

The dot product is a fundamental tool in vector mathematics, allowing you to project one vector onto another.
It involves multiplying corresponding components of vectors and summing the results. In our context, we compute the dot product to find how much of the force \( \mathbf{F} \) lies in the direction of another vector.
For the parallel component, the dot product is computed as:
\[ \mathbf{F} \cdot \mathbf{u_{p}} = \langle 0, -10 \rangle \cdot \langle \cos(\pi / 6), \sin(\pi / 6) \rangle \]
This results in a scalar that tells us the component of \( \mathbf{F} \) in that particular direction. The dot product is critical because it physically represents how much of the force is "effective" in moving the object along the desired path.
Without the dot product, detached from graphical or visual understanding, predicting effects would be heavily reliant on visualization rather than calculation.

Vector Projection

Vector projection is about finding out how a vector aligns or shares direction with another vector.
It relies on the dot product to obtain a scalar indicating the magnitude in the desired direction, as seen in the formula:
\[ \mathbf{F_{p}} = (\mathbf{F} \cdot \mathbf{u_{p}}) \mathbf{u_{p}} \]
This tells us how much of the force \( \mathbf{F} \) is utilized in the direction \( \mathbf{u_{p}} \). In simpler terms, projection helps us decompose one vector into parts aligning with another given vector without changing any physical properties.
This concept is practical for understanding how forces distribute themselves in restricted scenarios like inclined planes or across frictional surfaces. Essentially, it helps model and predict object behaviors accurately, showing which forces contribute to movement or resistance optimally.