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 \(\pi / 3\) with the positive \(x\) -axis

Short answer

Answer: The components of the given force vector parallel and normal to the inclined plane are \(\mathbf{F}_{\parallel}=-10\sqrt{3}/2\langle 1/2,\sqrt{3}/2\rangle\) and \(\mathbf{F}_{\perp}=5\langle-\sqrt{3}/2,1/2\rangle\). When added together, they sum up to the total force, verifying the problem's statement.

Step-by-step solution

Find the unit vectors parallel and normal to the inclined plane

First, we need to find the direction vector of the inclined plane and normalize it to get the unit vector parallel to the inclined plane (\((u)\textbf{vect}_{\parallel}\)): $$(u)\textbf{vect}_{\parallel}=\frac{\langle\cos(\pi/3),\sin(\pi/3)\rangle}{|\langle \cos(\pi/3),\sin(\pi/3)\rangle |}=\frac{\langle 1/2,\sqrt{3}/2\rangle}{1}=\langle 1/2,\sqrt{3}/2\rangle.$$ Next, let's find the unit vector normal to the plane (\((u)\textbf{vect}_{\perp}\)) by rotating \((u)\textbf{vect}_{\parallel}\) by 90 degrees counterclockwise: $$(u)\textbf{vect}_{\perp}= \langle-\sin(\pi/3),\cos(\pi/3)\rangle = \langle-\sqrt{3}/2,1/2\rangle.$$

Find the components of the force vector along \((u)\textbf{vect}_{\parallel}\) and \((u)\textbf{vect}_{\perp}\)

Now let's find the components of the force vector \(\mathbf{F}\) along \((u)\textbf{vect}_{\parallel}\) and \((u)\textbf{vect}_{\perp}\): $$\mathbf{F}_{\parallel}=\langle 0,-10\rangle\cdot \langle 1/2,\sqrt{3}/2\rangle (-10\sqrt{3}/2)$$ $$\mathbf{F}_{\perp} = \langle 0,-10 \rangle \cdot \langle-\sqrt{3}/2, 1/2 \rangle (5) $$ Now let's multiply the scalars with the respective unit vectors: $$\mathbf{F}_{\parallel}=-10\sqrt{3}/2\langle 1/2,\sqrt{3}/2\rangle$$ $$\mathbf{F}_{\perp}=5\langle-\sqrt{3}/2,1/2\rangle$$

Verify the total force as the sum of the components

Finally, let's verify that the total force is the sum of these components: $$\mathbf{F}=\mathbf{F}_{\parallel}+\mathbf{F}_{\perp}$$ $$\langle 0, -10 \rangle = -10\sqrt{3}/2\langle 1/2,\sqrt{3}/2 \rangle + 5 \langle -\sqrt{3}/2, 1/2 \rangle$$ $$\langle 0, -10 \rangle = \langle 0-5\sqrt{3}/2, -10\sqrt{3}/2+5 \rangle = \langle 0, -10 \rangle$$ As we can see, the total force is indeed the sum of the two component forces, verifying the problem's statement.

Key concepts

Inclined Plane

When dealing with an inclined plane, imagine a surface that is tilted at an angle rather than being flat like a floor. This type of plane is common in physics problems because it introduces extra forces due to its slope. The angle of inclination is essential as it defines the degree to which the plane deviates from being horizontal.
For most problems, the angle is provided relative to a reference direction, like the positive x-axis. This is important because the angle helps us calculate the direction vectors that are either parallel or perpendicular to the inclined plane. In our problem, the plane makes a angle of \( \pi / 3 \) to the x-axis, which affects how forces are decomposed along it.

Unit Vector

A unit vector is a vector with a magnitude of 1, and it is used for determining direction alone. It's crucial in calculations for inclined planes because they help identify the direction of forces without considering their magnitude.

• Parallel Unit Vector: It runs along the plane's surface. In our problem, the parallel unit vector is obtained from the angle of inclination at \( \pi / 3 \), turning into \( \langle 1/2, \sqrt{3}/2 \rangle \).
• Normal Unit Vector: Perpendicular to the plane, aligning with the force opposing the surface. It represents a 90-degree rotation of the parallel vector, in this case resulting in \( \langle -\sqrt{3}/2, 1/2 \rangle \).

Using these unit vectors allows us to break down any force into components that align along the plane (parallel) and those that act perpendicular to it (normal).

Component Forces

When a force acts on an inclined plane, it can be split into two parts: the component forces. This decomposition is crucial for understanding how a force acts in different directions.

• Parallel Component: The part of the force that acts along the length of the plane, contributing to any motion that might occur along the slope. In our solution, the force \( \mathbf{F}_{\parallel} \) is calculated using the parallel unit vector and results in \( -10\sqrt{3}/2 \langle 1/2, \sqrt{3}/2 \rangle \).
• Normal Component: The part of the force that acts perpendicular to the plane, usually opposing any movement across the surface. The perpendicular force vector, \( \mathbf{F}_{\perp} \), is derived using the normal unit vector, resulting in \( 5 \langle -\sqrt{3}/2, 1/2 \rangle \).

The total force \( \mathbf{F} \) remains unchanged and is the sum of these two components. This sum equals the original force \( \langle 0, -10 \rangle \), verifying the decomposition's correctness.