# Silly Parallelogram Question

by mpaige1
Tags: parallelogram, silly
 P: 24 1. The problem statement, all variables and given/known data Determine resultanat force of vectors P (40N) and Q (60N). P is 20 degrees from reference plane and Q is 45 degrees from reference plane. 2. Relevant equations When using the parallelogram rule, should you always start with the lowest angle vector? (like in the bottom diagram of the attached image) 3. The attempt at a solution I solved the question using the parallelogram in two different ways and ended up with different answers. Is there a right way to do this? Solution 1: P vector, then Q vector with resultant vector starting at tail of P to tip of Q. Law of Cosines yields, R = 97.726N Law of Sines yields, Angle A = ∠15.039 alpha = ∠20 + ∠15.039 = ∠35.04 Solution 2: Q vector, then P vector with resultant vector starting at tail of Q to tip of P. Law of Cosines yields, R = 98.542N Law of Sines yields, A = ∠7.9803 alpha = ∠45 - ∠7.9803 = ∠37 Diagram attached.
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 Quote by mpaige1 1. The problem statement, all variables and given/known data Determine resultanat force of vectors P (40N) and Q (60N). P is 20 degrees from reference plane and Q is 45 degrees from reference plane.
You say "reference plane" which implies three dimensions. But in that case, there are an infinite numbers of candidates for P and an infinite number of candidates for Q. I will assume this is a two dimensional problem and you mean "reference line".

 2. Relevant equations When using the parallelogram rule, should you always start with the lowest angle vector? (like in the bottom diagram of the attached image)
Do you understand why it is called the "parallelogram" rule? A parallelogram has four sides, two of which represent P and two which reresent Q- it doesn't matter which you "start with".

 3. The attempt at a solution I solved the question using the parallelogram in two different ways and ended up with different answers. Is there a right way to do this? Solution 1: P vector, then Q vector with resultant vector starting at tail of P to tip of Q. Law of Cosines yields, R = 97.726N
It's impossible to say what you might have done if you don't say what you have done! Here, you have a triangle with two sides of lengths 40 and 60 and angle between them of 140 degrees. Applying the cosine law I do NOT get 97.726.

 Law of Sines yields, Angle A = ∠15.039 alpha = ∠20 + ∠15.039 = ∠35.04 Solution 2: Q vector, then P vector with resultant vector starting at tail of Q to tip of P. Law of Cosines yields, R = 98.542N Law of Sines yields, A = ∠7.9803 alpha = ∠45 - ∠7.9803 = ∠37 Diagram attached.
All I can say is I get a completely different answer. I can't say what you are doing wrong. I presume you know that the cosine law says that $c^2= a^2+ b^2- 2abcos(C)$. The fact that that is "symmetric" in a and b should tell you that it doesn't matter which side you call "a" and which side you call "b". What angle do you have between the two vectors?
 P: 24 Sorry, I used the wrong value for the angle in cosine. It should have been 155 and I used 160. This should be correct, the textbook I'm working from has the same as their answer. It could be wrong but I doubt it. Here's the complete walkthrough of the problem: R^2=〖60〗^2+〖40〗^2-2*40*60cos(155) R = 97.726N sin(A)/40 = sin(155)/97.726 A = 9.9612 alpha = 35.039
P: 24

## Silly Parallelogram Question

I'm not sure if the diagram I gave is showing up but it should have everything you need on there. No worries guys, it was a stupid mistake.

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