How can I find the resultant force using only the law of cosines and sines?

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Homework Help Overview

The discussion revolves around finding the resultant force using only the law of cosines and sines. The original poster expresses difficulty in identifying a usable angle after drawing a parallelogram to represent the vectors involved.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss isolating triangles and the importance of identifying angles between vectors. There is mention of using head-to-tail vector addition and the law of cosines to find the resultant. Questions arise regarding the angles in the parallelogram and their relationship to the vectors.

Discussion Status

Participants are exploring various interpretations of the angles involved and discussing how to manipulate the vectors to gain more information. Some guidance has been offered regarding the identification of angles and the application of the law of cosines.

Contextual Notes

The original poster is constrained by the requirement to use only the law of cosines and sines, which adds complexity to their problem-solving approach. There is also a mention of the original poster's struggle with similar triangles that are not right triangles.

Saladsamurai
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Homework Statement


Okay, I am only allowed to use law of cosines and sones to find the resultant force F_r

I am having a hell of a time finding a usable angle after drawing my parellelogram..so obviously I am in need of sleep.

What am I missing here? (pic is clickable)
th_Photo2.jpg



Casey
 
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I know I need to isolate a triangle, but I really suck at similar triangles when they are not necessarily right triangles.
 
Draw yourself a diagram where you add the vectors (head to tail addition). The two vectors are the two sides of a triangle. You should be able to identify the angle between them and apply the law of cosines to find the third side of the triangle, which will be the resultant.
 
The 120 degree angle should be the angle corresponding to your resultant vector in the triangle. For the other angles:

HINT: What is the angle between the 80lb force and the -x axis? Does this help you find another angle in your parallelogram?
 
Is the angle between them by any chance 60 degrees? In the parellepgram 2(120)=240 leaving 120/2 to give me four angles that add to 360
 
Saladsamurai said:
Is the angle between them by any chance 60 degrees?
Yes. (Assuming you're talking about the triangle I referred to.)
 
I'll be back in 20 minutes. All of the theological students just showed up at StarBucks and I can't take their banter...I'm going home.

Casey

Doc Al said:
Yes. (Assuming you're talking about the triangle I referred to.)

Angle between 60 and 80 with tail of 60 at tip of 80.
 
Run for it, man! :wink:
 
I made it home!

Thanks Doc and G01! I am taking this Statics course over X-mass break :points gun into mouth and fake blows brains out:

So is the general approach to these to use the fact that vectors can be moved around to redraw the scenario in a manner that helps to generate more information from the given info?

Casey
 
  • #10
Yes. When adding vectors, you can move them around at will as long as you keep the magnitude and direction the same.
 

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