Determine magnitude of the components of these forces

Click For Summary
SUMMARY

The discussion focuses on resolving the force F2 into components along the u and v axes, specifically calculating F2u and F2v. The user initially calculated F2v as 482 N using the Law of Sines but struggled with F2u, mistakenly obtaining values between 600-1000 N. After clarification, it was established that F2u should be calculated from the tip of F2v to the tip of F2, confirming the angles in the triangle formed by these vectors. The correct approach involves recognizing the parallel nature of the triangle to the u axis, simplifying the calculations.

PREREQUISITES
  • Understanding of vector resolution and component forces
  • Familiarity with the Law of Sines
  • Basic knowledge of trigonometric functions (sine, cosine)
  • Ability to interpret force diagrams
NEXT STEPS
  • Study vector resolution techniques in physics
  • Learn advanced applications of the Law of Sines in force problems
  • Explore trigonometric identities and their applications in physics
  • Practice solving problems involving concurrent forces and their components
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and vector analysis, as well as educators looking for examples of force resolution techniques.

pka
Messages
11
Reaction score
0

Homework Statement


Resolve the force F2 into components acting along the u and v axes and determine the magnitudes of the components.

So! I was able to finally determine F2u. I cannot for the live of me get F2v. I am having a very hard time solving this problem. I have included the diagram of the problem. All of it is what is given.

Here is the link to the image:
http://i16.tinypic.com/6je9m60.jpg


Alright! So in order to first solve for F2v I constructed a triangle where the left side was the F2 force of 500 N and the right side was the F2v vector. They're both connected by the same point. It looks exactly as in the diagram I included. And via the diagram included I determined that the angles inside the triangle are 45, 65 and 70 counterclockwise. So! I used the Law of Sines. F2v is then 482 N.

Ok, so in order to calculate F2u I tried something similar but my numbers are far too large. I get numbers in the range of 600-1000 N. And...F2v should be 376 N. I tried the same thing I did with F2v but I can't seem to determine the base angles so that I can use Law of Sines again. I get 115 degrees as the angle between the concurrent forces (F2 and F2u). But I don't know how to get the other two angles. I'm just stumped. I thought maybe a different approach would be better. So I tried just solving for the opposite side of the u line to get F2u...In other words I said that F2u is (500 N)*cos(65 deg). 65 because that's the angle between 70 and 45. And...well I get 211 N. But that's not it either. =/

If anyone could tell me the way I'm supposed to be thinking about this problem I would greatly appreciate it because there are more problems like there I'm having trouble with but I really need some help. hahaha. Many thanks in advance!

--pka
 
Physics news on Phys.org
So! I was able to finally determine F2u. I cannot for the live of me get F2v.
.
.
F2v is then 482 N.

Ok, so in order to calculate F2u I tried something similar but my numbers are far too large. I get numbers in the range of 600-1000 N. And...F2v should be 376 N.

You mean you can't get F2u? You just said that F2v was 482 N and then you said it was 376 N. I'm not sure I'm following...

Where you are taking your F2u? It should be from the tip of F2v (so along the v axis) to the tip of F2 and it will be parallel to the u axis. Those angles you stated are correct for the triangle those vectors form. You should just be able to use the sine law to get the answer.
 
Last edited:
OOOOOHHHH! Thank you so much. I hadn't realized that the triangle I constructed was parallel to u. And yes, it was F2u is the one I needed to find. And of course it parallel right? Because otherwise how could I possibly have determined the angles of the triangle. I made this problem more difficult for myself than it needed to be. lol. In any case, thank you for the help!
 
You're welcome.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
15K
  • · Replies 30 ·
2
Replies
30
Views
2K
Replies
3
Views
1K
  • · Replies 21 ·
Replies
21
Views
3K
Replies
2
Views
1K
Replies
17
Views
3K
Replies
4
Views
2K
  • · Replies 31 ·
2
Replies
31
Views
8K
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K