How to solve for the magnitude and angle in this situation?

AI Thread Summary
To solve for the resultant vector when angles are not provided, it is essential to resolve each force vector into its x and y components using the dimensions given. The angle for the 435 N vector can be calculated using arctan(200/210), resulting in approximately 43.6 degrees relative to the positive x-axis. The discussion emphasizes the importance of using similar triangles to find the x and y components, allowing for the calculation of the resultant vector by summing these components algebraically. It is noted that negative signs should be assigned to leftward and downward directions. This approach provides a systematic way to determine the resultant vector's magnitude and direction.
Vanessa Avila
Messages
94
Reaction score
1

Homework Statement



Each one of the force vectors are surrounded with dimensions. But you're not given any angles. How do you solve for the resultant vector like that?
gist.png

Homework Equations



Fr = F1 + F2 + F3

The Attempt at a Solution


I tried taking the arctan(200/210) on the 435 N vector to get the angle for that but I don't think that's right.
 
Physics news on Phys.org
What did you get for the arctan (200/210)? And what angle would that represent?
 
TomHart said:
What did you get for the arctan (200/210)? And what angle would that represent?

I got 43.6 degrees. Would that be the angle between 435 N and the positive x-axis?
 
TomHart said:
What did you get for the arctan (200/210)? And what angle would that represent?

Or would that be the entire angle between positive x and y axis?
 
If all of those numbers represent relative size of the vector, then yes, it looks like you got it right the first time - that 43.6 degrees is the angle between the 435 N vector and the positive x-axis. It's kind of a strange problem, but you are on the right track if you continue with what you are doing. It was a little confusing on the upward left pointing vector. Are the dimensions of that vector 70 and 210?
 
TomHart said:
If all of those numbers represent relative size of the vector, then yes, it looks like you got it right the first time - that 43.6 degrees is the angle between the 435 N vector and the positive x-axis. It's kind of a strange problem, but you are on the right track if you continue with what you are doing. It was a little confusing on the upward left pointing vector. Are the dimensions of that vector 70 and 210?

Oops i messed up on my drawing. The one on the left is 240 and the blue line on the top should be 70. Sorry about that!
 
TomHart said:
Are the dimensions of that vector 70 and 210?
No. From the Pythagorean Theorem it appears the magnitude of the vector, and the x and y dimensions are to a different scale or denominated in different units.

In any case the first step is to resolve each vector into its x and y components. You can calculate the hypotenuse of the right triangle formed by the x and y dimensions, and then use proportional parts of similar triangles to find the x and y components for each vector.
 
Vanessa Avila said:
Oops i messed up on my drawing. The one on the left is 240 and the blue line on the top should be 70. Sorry about that!

No, I just read it wrong. You were right. It was 70 and 240. It just looked a little confusing the way it was shown.
 
You have all the X and Y components of each vector. Are you sure you need the individual angles?
 
  • #10
Good point. It's usually easier in this situation to take advantage of the fact that corresponding sides of similar triangles are all in the same proportion.
 
  • #11
The vectors' lines of action are defined by Cartesian coordinates (you can think of the x and y dimensions as position vectors). So by similar triangles we know the magnitude of the x component of the vector divided by the magnitude of the vector equals the x dimension divided by the distance from the origin to the vector's head.
 

Attachments

  • a.jpg
    a.jpg
    18.9 KB · Views: 425
  • #12
David Lewis said:
The vectors' lines of action are defined by Cartesian coordinates (you can think of the x and y dimensions as position vectors). So by similar triangles we know the magnitude of the x component of the vector divided by the magnitude of the vector equals the x dimension divided by the distance from the origin to the vector's head.
so by doing that for each force vector, I'll be able to get the value for the resultant?
 
  • #13
Yes indeed. Just add the x force components together, and then the y force components together, as if they were scalars. Those algebraic sums will equal the x and y components of the resultant. Keep in mind the left and down directional senses are assigned a negative sign.
 
Last edited:
  • Like
Likes Vanessa Avila
  • #14
David Lewis said:
Yes indeed. Just add the x force components together, and then the y force components together, as if they were scalars. Those algebraic sums will equal the x and y components of the resultant. Keep in mind the left and down directional senses are assigned a negative sign.
Awesome. Thanks a lot!
 
Back
Top