Vector Components physics homework

AI Thread Summary
To find the x and y components of a 40 lb force vector at a 120-degree angle counterclockwise from the -y direction, the correct formulas are Ax = (40)(-cos(120°)) and Ay = (40)(sin(120°)). The discussion emphasizes the importance of accurately identifying the angle in relation to the positive x-axis for correct calculations. A proper diagram can help visualize the vector's position and avoid mistakes in determining the components. The conversation concludes with a reminder that understanding the relationship between different angles is crucial for solving similar problems effectively.
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Determine the x and y components. A 40 lb force vector that makes an angle of 120 degree counterclockwise from the -y direction.

I did Ax= (40)(-cos120 degree) and Ay= (40)(sin120 degree)

I got the right answer, I just want to make sure that this is the correct way of solving the problem.
 
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Susanem7389 said:
Determine the x and y components. A 40 lb force vector that makes an angle of 120 degree counterclockwise from the -y direction.

I did Ax= (40)(-cos120 degree) and Ay= (40)(sin120 degree)

I got the right answer, I just want to make sure that this is the correct way of solving the problem.

I read your earlier thread with Tiny_Tim and noticed that you wanted a rule as to when to use the formulas he suggested. When I was teaching, I used to tell my students to draw a careful diagram and label the angles the vector makes with each axis and work then use trigonometry to find the projection of the vector on to the axes. I think that way you will make fewer mistakes.
 
Okay. Thank you. Also, was the way I solved the problem correct?
 
Susanem7389 said:
Determine the x and y components. A 40 lb force vector that makes an angle of 120 degree counterclockwise from the -y direction.

I did Ax= (40)(-cos120 degree) and Ay= (40)(sin120 degree)

I got the right answer, I just want to make sure that this is the correct way of solving the problem.

Well, you said that you got the right answer, but I'm puzzled. Isn't your vector in the first quadrant making a 30 degree angle with the x axis?

Your solution is appropriate for a vector that makes an angle of 120 degrees with the Positive X axis.
 
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Yes, it is. I must have done something wrong with the equation. How would I fix it?
 
The x-component of A is \hat x \cdot \vec A (which is equal to (1)|A|\cos\theta_{\mbox{\small between $\vec A$ and $\hat x$}}).
The y-component of A is \hat y \cdot \vec A (which is equal to (1)|A|\cos\theta_{\mbox{\small between $\vec A$ and $\hat y$}}).

You can also express the components as
A_x=A\cos\theta
A_y=A\sin\theta
where \theta is the counterclockwise angle from the positive-x axis.
Note that this angle is in the range 0\leq \theta < 360^\circ (and so -1\leq \cos\theta\leq 1 and -1\leq \sin\theta\leq 1).

The above are the best facts to remember.
If you need to work with other angles, you need to draw a good picture and express the given angle in terms of the angles above [and possibly use some trig identities, especially if you want a general formula using some other choice of angles or axes].

So, for instance, if you are given a counterclockwise angle \phi with respect to the -y axis, what is the corresponding counterclockwise angle \theta from the +x-axis?
 
It would be the same numbers for both however the positive x direction, both the x and y component would be positive and for the negative y direction, both the x and y component would be negative. Thank you for your help.
 
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