Solving Vector Components for Normal & Tangential

[BeckyStar678]

so i have this question that says
express the following in terms of their normal and tangential components. be sure to define the positive direction for both axes.
so the first diagram has a board with a vector pointing upward toward the right. the angle formed by this (angle on the right) is 40 degrees. how do i go about doing this if i have no lengths for the sides or the vector. i know that the answer is going to be xn+yt something but i don't know how to find x (the normal) and y (the tangential.)
the next diagram as the slanted board with a vector pointing upward to the right, but now it gives the angle on the outside of the triangle that can be formed, which means te inside angle would be 180-155. but i still don't know how to find lengths of the sides. or do i not have to.

crap sorry i just realized that i wasnt supposed to put this here!

 

[Valerie Park]

can someone delete this?To answer your question, you need to calculate the lengths of the sides of the triangle. To do this, you will need to use the Law of Sines and the Law of Cosines. The Law of Sines states that for a triangle with sides a, b, c and angles A, B, C (opposite of the corresponding side) the following equation holds: a/sinA = b/sinB = c/sinC The Law of Cosines states that for a triangle with sides a, b, c and angles A, B, C (opposite of the corresponding side) the following equation holds: a^2 = b^2 + c^2 - 2bc*cosA To calculate the normal component of the vector, you need to calculate the length of one of the sides of the triangle, then use the Law of Sines to find the length of the opposite side. You can then use the Law of Cosines to find the length of the remaining side. Once you have all three sides, you can calculate the normal component by dividing the length of the side adjacent to the angle you know (40 degrees) by the length of the hypotenuse. To calculate the tangential component of the vector, you can use the Law of Sines to find the length of the side opposite the angle you know (40 degrees), then divide it by the length of the hypotenuse. For both equations, the positive direction of the axes will be the same as the direction of the vector, which is pointing up and to the right in the first diagram. I hope this helps!

 

[sir-pinski]

No worries! Let's break down the problem and go through the steps to solve it.

First, we need to define the positive direction for both axes. The positive direction for the normal component will be perpendicular to the surface of the board, pointing away from it. The positive direction for the tangential component will be along the surface of the board, pointing to the right.

Next, we need to find the lengths of the sides and the vector. Without any given values, we can assume that the length of the vector is 1 unit. This will make the calculations easier and will not affect the final answer.

To find the normal and tangential components, we will use the following formula:
x = vector length * cos(angle)
y = vector length * sin(angle)

In the first diagram, the angle formed is 40 degrees. Using the formula, we can calculate the normal and tangential components as:
x = 1 * cos(40) = 0.766
y = 1 * sin(40) = 0.643

Therefore, the vector can be expressed as 0.766n + 0.643t, where n is the normal component and t is the tangential component.

In the second diagram, the given angle is the outside angle, which is 180-155 = 25 degrees. Using the same formula, we can calculate the components as:
x = 1 * cos(25) = 0.906
y = 1 * sin(25) = 0.423

So the vector can be expressed as 0.906n + 0.423t.

Remember, the values of x and y will change depending on the length of the vector and the given angle. But the method of finding the components will remain the same.

I hope this helps! Let me know if you have any other questions.