Law of cosines/ Parallelogram rule

  • Thread starter fableblue
  • Start date
  • #1
17
0
I am not sure if this question should be in here but it does pertain more to these problems as compared to a general math question.

So, could someone explain why when using the parallelogram rule for obtaining the sum of 2 forces by the means of the Law of Cosines that the controller -2bc is replaced by +2bc in the equation a2=b2+c2-2bccosA
example:
The magnitude of two forces exerted on a pylon are FAB=100 and FAC=60 with angle BAC=30degrees

[tex]\Sigma[/tex]FAB+FAC=[tex]\sqrt{}[/tex](602+1002-2(600*100)cos30) = 56.637 FALSE

[tex]\Sigma[/tex]FAB+FAC=[tex]\sqrt{}[/tex](602+1002+2(600*100)cos30) = 154.895 TRUE
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
8
If the angle between FAB and FAC is less than 90 degree , cos(A) is positive. If it is more than 90 degree, cos(A) is negative.
 
  • #3
17
0
Thanks rl.bhat for you response but iI am still in question.

I am attaching 2 documents (because it is easier then trying to type them in). 1 is the proff of the law of cosins and the other is another problem that i have done using the Law of Cosins that i was taught and used in trig.
This is why i am confuessed about the operator 2ab being "-" for some and "+" for others. Both problems the angle is in Q1 but the only difference that i see is that one use the parallelogram rule and the other uses the force triangle. I am just trying to find out why sometimes it is - and sometime +. Never used the + in trig only in statics.

Thanks
 

Attachments

  • question001.jpg
    question001.jpg
    25 KB · Views: 859
  • question002.jpg
    question002.jpg
    20.2 KB · Views: 1,790
  • #4
rl.bhat
Homework Helper
4,433
8
In the law of cosine we have
a^2 = b^2 + c^2 -2bc*cos(theta) where theta is the angle between b and c and a is the opposite side of theta.
In parallelogram law, if OB and OB are b and c vectors, and theta is the angle between OB and OC, then BC is a in the above equation. But, as you can see. it is not the resultant of OB and OC.
 
  • #5
17
0
Thanks, but that does not really show why it would be different. I am thinking that it has to do with when you sum forces you are looking for a total/resultant of the two forces as apposed to when you are looking for a single force, as in the attachment question 002, you use the unmodified version of The Law of Cosines.

Would it fair to assume that when finding a single force, as with the force triangle, the triangle gets modified to be a right triangle? So, as the proof shows, x-c is substitued for side c (please refer to the diagram) in pythagorean theorem and trickles down from there? Where as using the parallelogram rule, the triangle does not get modified, so with that being said side c would stay side c?

Does this only happen when using the paraellogram method?
 
  • #6
Doc Al
Mentor
45,179
1,500
So, could someone explain why when using the parallelogram rule for obtaining the sum of 2 forces by the means of the Law of Cosines that the controller -2bc is replaced by +2bc in the equation a2=b2+c2-2bccosA
The law of cosines uses the angle within the triangle and that factor is always -2bc.
example:
The magnitude of two forces exerted on a pylon are FAB=100 and FAC=60 with angle BAC=30degrees
I presume that 30° is the angle between the two force vectors. But when you draw the sum of the two force vectors, the angle within the triangle (for applying the law of cosines) is not 30° but 180-30 = 150°. The key is that cosθ = -cos(180° - θ).
 
  • #7
17
0
Excellent:smile: I have never thought of that, now i know.

thank you
 

Related Threads on Law of cosines/ Parallelogram rule

  • Last Post
Replies
1
Views
2K
Replies
4
Views
6K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
5K
Replies
6
Views
2K
Replies
6
Views
3K
  • Last Post
Replies
3
Views
9K
Replies
28
Views
3K
  • Last Post
Replies
10
Views
8K
  • Last Post
Replies
3
Views
5K
Top