# Question #4acab

Aug 6, 2017

See below.

#### Explanation:

I'm not sure what your current knowledge of trigonometry is so I will include a more in-depth explanation below.

Trigonometry and Forces

Suppose we have some object on an incline, and we need to take inventory of all the forces acting on the object.

Consider this "frictionless" inclined plane: Because the force of gravity acts vertically on the object, it occurs at an angle relative to the plane. The diagram breaks the force of gravity up into its parallel (x, horizontal) and perpendicular (y, vertical) components. Let's examine how this is done. As shown above, we can imagine that a triangle is formed by the force of gravity and the vertical. We can imagine the normal force like the positive y-axis of an x-y plane and set up our coordinate system accordingly.

You mentioned using the Pythagorean theorem, which, in combination with trigonometry, is how the components of a force are found.

${\vec{F}}_{n e t} = \sqrt{{\left({F}_{x}\right)}^{2} + {\left({F}_{y}\right)}^{2}}$

Which looks very similar to $c = \sqrt{{a}^{2} + {b}^{2}}$

The net force is the hypotenuse of the right triangle. The parallel (x) and perpendicular (y) components then make up the horizontal and vertical sides of the triangle, respectively.

For an inclined plane like this, the angle between the net force and the vertical is equal to the angle of incline. There is a fairly simple geometrical proof of this, but it is not necessary to know and I will not include it to avoid confusion, though I would be happy to explain upon request.

If we then imagine connecting the sides to form a right triangle as shown in the above diagram, what we are left with is something like this: And this is a basic right triangle just like the ones we dealt with before.

So, time for an example problem. Say the angle of incline is ${60}^{o}$ and the mass of the object is $2 k g$.

This means that the force of gravity is

${\vec{F}}_{G} = m g = \left(2 k g\right) \left(9.81 \frac{m}{s} ^ 2\right) = 19.62 N$ downward.

This is our hypotenuse.

Now we want to find the parallel and perpendicular components of the force of gravity.

We can see from the triangle that the parallel (x) component is opposite the angle $\theta$, so we'll want to use the sine function.

$\sin \left(\theta\right) = \left(\text{opposite")/("hypotenuse}\right)$

$\implies \sin \left(\theta\right) = \frac{{\left({F}_{G}\right)}_{x}}{{F}_{G}}$

We want to find ${\left({F}_{G}\right)}_{x}$, so we can rearrange the above equation to solve for it.

$\implies {\left({F}_{G}\right)}_{x} = {F}_{G} \sin \left(\theta\right)$

$= m g \sin \theta$

$= \left(19.62 N\right) \sin \left({60}^{o}\right)$

$= 16.99 \approx 17 N$

So the parallel component of gravity is about 17 newtons.

Onto the perpendicular component. We can see that it is adjacent to the angle, so we want to use the cosine function.

$\cos \left(\theta\right) = \left(\text{adjacent")/("hypotenuse}\right)$

$\implies \cos \left(\theta\right) = \frac{{\left({F}_{G}\right)}_{y}}{{F}_{G}}$

$\implies {\left({F}_{G}\right)}_{y} = {F}_{G} \cos \left(\theta\right)$

$= m g \cos \left(\theta\right)$

$= \left(19.62 N\right) \cos \left({60}^{o}\right)$

$= 9.8 N$

So the perpendicular component of gravity is about 9.8 newtons.

We can verify this result using the Pythagorean theorem.

${a}^{2} + {b}^{2} = {c}^{2}$

$\implies c = \sqrt{{a}^{2} + {b}^{2}}$

$\implies {F}_{G} = \sqrt{{\left(17\right)}^{2} + {\left(9.8\right)}^{2}}$

$= 19.62 N$

Just as we calculated above for the magnitude of the force of gravity.

You can apply this method in a similar way to find resultant forces. Break the forces up into their respective components, be sure to keep track of what is positive and what is negative (be consistent), add all of those that are horizontal and add all of those that are vertical, and then use the pythagorean theorem to find the magnitude as shown above.

$- - - - -$

Trigonometry is a branch of mathematics that deals with the relationship between the sides and angles of triangles and the functions of the angles. These functions are called trigonometric ("trig") functions . You will likely only need to worry about three of them: the sine, the cosine, and the tangent, as you are mostly likely to be dealing with right triangles (triangles with a ${90}^{o}$ angle).

Right triangle: A right triangle with angles A, B, C and side lengths a,b,c.

The trigonometric functions I mentioned above are functions of an angle. You would see $\cos \left(\theta\right) , \sin \left(\theta\right) ,$and $\tan \left(\theta\right)$ as the representations of these functions.

Cosine

The cosine of an angle is given as a ratio (as in a fraction) of the length of the adjacent side to the length to the hypotenuse of the triangle. The adjacent side length is the one closest to the angle, and the hypotenuse is the longest side of the triangle, labeled $c$ in the above diagram.

$\cos \left(\theta\right) = \left(\text{adjacent")/("hypotenuse}\right)$

So from the triangle above, for the angle A we would write:

$\cos \left(A\right) = \frac{b}{c}$

For angle B:

$\cos \left(B\right) = \frac{a}{c}$

We don't take the cosine of angle C as the cosine of ${90}^{o}$ is zero, but I will mention that a bit later.

Sine

The sine of an angle is given as the length of the opposite side to the length of the hypotenuse of the triangle.

$\sin \left(\theta\right) = \left(\text{opposite")/("hypotenuse}\right)$

$\sin \left(A\right) = \frac{a}{c}$

$\sin \left(B\right) = \frac{b}{c}$

Tangent

The tangent of an angle is given as the length of the opposite side to the length of the adjacent side.

$\tan \left(\theta\right) = \left(\text{opposite")/("adjacent}\right)$

$\tan \left(A\right) = \frac{a}{b}$

$\tan \left(B\right) = \frac{b}{a}$

A good way to remember these functions is SOH CAH TOA.

$\textcolor{b l u e}{\text{S}}$ine: $\textcolor{b l u e}{\text{O}}$pposite over $\textcolor{b l u e}{\text{H}}$ypotenuse, $\textcolor{b l u e}{\text{C}}$osine: $\textcolor{b l u e}{\text{A}}$djacent over $\textcolor{b l u e}{\text{H}}$ypotenuse, $\textcolor{b l u e}{\text{T}}$angent: $\textcolor{b l u e}{\text{O}}$pposite over $\textcolor{b l u e}{\text{A}}$djacent