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Equilibrium Question - Mechanics 1

"Three forces act upon a particle, which is in equilibrium. If the magnitudes of the forces are 4N, 5N and 6N, find the angles between the forces."

ARGGGHHH! You have no idea how long I've spent on this one question. It's driving me crazy. D:

So we're not given any other information, so how on earth is it possible to find the angles?!

I would have said that we'd use trig, something along the lines of '4cosA' (A being the angle we're trying to find). Then I'd guess that it would equal 0, due to being in equilibrium? But I don't really see how that's possible. xD

So any help would be appreciated. A nudge in the right direction? Then if I still can't do it, I'll come back. XD
Reply 1
Avatar for TenOfThem
TenOfThem
18
If the particle is in equilibrium the forces create a triangle

You can use the cosine rule to find the angles
Do you know the Cosine Rule?

The 4N, 5N and 6N forces form a triangle.
Original post by Pixie-Bee
"Three forces act upon a particle, which is in equilibrium. If the magnitudes of the forces are 4N, 5N and 6N, find the angles between the forces."

ARGGGHHH! You have no idea how long I've spent on this one question. It's driving me crazy. D:

So we're not given any other information, so how on earth is it possible to find the angles?!

I would have said that we'd use trig, something along the lines of '4cosA' (A being the angle we're trying to find). Then I'd guess that it would equal 0, due to being in equilibrium? But I don't really see how that's possible. xD

So any help would be appreciated. A nudge in the right direction? Then if I still can't do it, I'll come back. XD


what exam board is this for?
Isn't this just cosine rule?

cos⁡A=b2+c2−a22bc\cos A= \dfrac{b^2+c^2-a^2}{2bc}

So just label each magnitude a, b, c and a respective angle A, B, C. Sub in the magnitude values for each angle and you'll get the answer.
(edited 11 years ago)
Reply 5
Avatar for Pixie-Bee
Pixie-Bee
OP
2
Um, I'm with AQA I believe. :smile:

I can't believe I didn't even think of the cosine rule, I've definitely been doing these sums for too long. xD
Yet even with the cosine rule, I'm getting the wrong answer? o.O

So, say I'm trying to work out just one of the angles.
CosC = (a^2 + b^2 - c^2) / 2ab

I've rearranged that correctly right?
Then I sub in my values;

CosC = (4^2 + 5^2 - 6^2) / 2 x 4 x 5
CosC = (16 + 25 - 36) / 40
CosC = 5 / 40 (simplifies to 1/8)
C = Cos^-1 (1/8)
C = 82.82 (2dp)

Have I gone wrong there? Cause that isn't the answer my book gives. D:
Reply 6
Avatar for TenOfThem
TenOfThem
18
Original post by Pixie-Bee
Um, I'm with AQA I believe. :smile:

I can't believe I didn't even think of the cosine rule, I've definitely been doing these sums for too long. xD
Yet even with the cosine rule, I'm getting the wrong answer? o.O

So, say I'm trying to work out just one of the angles.
CosC = (a^2 + b^2 - c^2) / 2ab

I've rearranged that correctly right?
Then I sub in my values;

CosC = (4^2 + 5^2 - 6^2) / 2 x 4 x 5
CosC = (16 + 25 - 36) / 40
CosC = 5 / 40 (simplifies to 1/8)
C = Cos^-1 (1/8)
C = 82.82 (2dp)

Have I gone wrong there? Cause that isn't the answer my book gives. D:


what about 97.18
Reply 7
Avatar for Pixie-Bee
Pixie-Bee
OP
2
Yup, that's one of the right answers for the question. :biggrin:
Where did I go wrong? :s-smilie:
Reply 8
Avatar for TenOfThem
TenOfThem
18
Original post by Pixie-Bee
Yup, that's one of the right answers for the question. :biggrin:
Where did I go wrong? :s-smilie:


180-your answer
Reply 9
Avatar for Pixie-Bee
Pixie-Bee
OP
2
Original post by TenOfThem
180-your answer


..Why? (I'm having one of those really stupid moments aren't I? I blame the fact it's a holiday and I'm tired. xD) Is that for every angle, when worked out using the cosine rule?
Original post by Pixie-Bee
..Why? (I'm having one of those really stupid moments aren't I? I blame the fact it's a holiday and I'm tired. xD) Is that for every angle, when worked out using the cosine rule?



Do you have a diagram
Reply 11
Avatar for Pixie-Bee
Pixie-Bee
OP
2
Original post by TenOfThem
Do you have a diagram


...Yes.
Just a triangle. With the magnitude labelled and the angles I'm trying to find.

I don't understand. If I was asked to simply work out the angle, using this rule, I wouldn't need to subtract from 180. So why would I now? (Not denying that it works. I've found the answer to all three now, all correct. But I don't understand why it would work...)
Original post by Pixie-Bee
...Yes.
Just a triangle. With the magnitude labelled and the angles I'm trying to find.

I don't understand. If I was asked to simply work out the angle, using this rule, I wouldn't need to subtract from 180. So why would I now? (Not denying that it works. I've found the answer to all three now, all correct. But I don't understand why it would work...)


What direction have you drawn the arrows

They need to go "around" the triangle


The angles will be outside the triangle ... it is hard to explain why without being able to draw it for you with you next to me
Reply 13
Avatar for Pixie-Bee
Pixie-Bee
OP
2
Yup, I had done that.

I think I need to go search google for an online tutorial or something. I kinda get what you mean, but not certain.. :s-smilie:
Ok...

Let a = 4, b = 5 and c = 6, with respective angles A, B, C.

cos A = (b^2 + c^2 - a^2)/(2*b*c)
cos A = (25 + 36 - 16)/(60)
cos A = 0.75
A = 41.4

cos B = (a^2 + c^2 - b^2)/(2*a*c)
cos B = (16 + 36 - 25)/(48)
cos B = 0.5625
B = 55.8

cos C = (a^2 + b^2 - c^2)/(2*a*b)
cos C = (16 + 25 - 36)/(40)
cos C = 0.125
C = 82.8

Check: A + B + C - 180 ~ 41.4 + 55.8 + 82.8 = 180.

So that's the answer for the angles inside the triangle. As the other guy said, upload a picture because the angle is relative.
Original post by Pixie-Bee
Yup, I had done that.

I think I need to go search google for an online tutorial or something. I kinda get what you mean, but not certain.. :s-smilie:


In reality the forces all act at a specific point

The vectors that you have drawn are just parallel to the originals

So you have probably drawn one going right, then one going up and right from there

To get the actual angle you need to pick the up&right one up and move it to the start of the across vector ... see the angle is 90-the one you found in the triangle

If that makes any sense :smile:
Original post by Oliver.Farren



So that's the answer for the angles inside the triangle. As the other guy said, upload a picture because the angle is relative.


Gal :smile:
Original post by TenOfThem
In reality the forces all act at a specific point

The vectors that you have drawn are just parallel to the originals

So you have probably drawn one going right, then one going up and right from there

To get the actual angle you need to pick the up&right one up and move it to the start of the across vector ... see the angle is 90-the one you found in the triangle

If that makes any sense :smile:



I don't understand

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