Vector Questions: Normal & Tangential Components

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The discussion revolves around expressing a vector in terms of its normal and tangential components, specifically for two diagrams involving angles and a board-like structure. The first vector points upward at a 40-degree angle, and participants suggest using trigonometric functions to express the components as cos(40)x and sin(40)y. Clarification is sought regarding the physical representation of the "board," which is described as a line from which the vector originates. The second diagram introduces an external angle, complicating the determination of components, but the approach remains similar. Overall, the focus is on using trigonometric relationships to express the vectors without needing specific lengths.
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Homework Statement



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.
 
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BeckyStar678 said:

Homework Statement



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.

Can you explain what you mean by "board"? Just a physical object like a board with a vector acting though the middle?

One way of expressing vectors is to use trig functions with the x and y components. + V Sine(40) and + V Cos(40) as they would both be positive. Is it possible they want you to express the vector of the second diagram with similar components but with a sign appropriate to positive x and y and appropriate angle?

Without further information though it is difficult to understand where you are stuck.
 
LowlyPion said:
Can you explain what you mean by "board"? Just a physical object like a board with a vector acting though the middle?

One way of expressing vectors is to use trig functions with the x and y components. + V Sine(40) and + V Cos(40) as they would both be positive. Is it possible they want you to express the vector of the second diagram with similar components but with a sign appropriate to positive x and y and appropriate angle?

Without further information though it is difficult to understand where you are stuck.

yeah the board is just the line that the vector is coming off of. like if you look at the letter "y" that's what my diagram looks like. the straight line in the y would be the board and the small line that forms the y is the vector. and the angle in the y is 40 degrees. so you think i could just have the answer be cos40(x) t + sin40(y) n, where x is the adjacent side and y is the opposite side. and then just express it in that way? because i myself don't see any way that i could actually solve for it.
 
BeckyStar678 said:
yeah the board is just the line that the vector is coming off of. like if you look at the letter "y" that's what my diagram looks like. the straight line in the y would be the board and the small line that forms the y is the vector. and the angle in the y is 40 degrees. so you think i could just have the answer be cos40(x) t + sin40(y) n, where x is the adjacent side and y is the opposite side. and then just express it in that way? because i myself don't see any way that i could actually solve for it.

There is little else from what I am given. Just remember to adjust the signs of the x component if positive x is to the right and the vector is pointing to the left. (Likewise for positive and negative normal directions.)

All I can say is good luck.
 
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