physics help....
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DatEpicChick
Posts: 358 Member
yeah, obviously, this has nothing to do with weight loss hahahaha. But i CANNOT find help on google, and everyone on my facebook has no idea how to work this problem....
okay...
Question:
Two forces are applied to a ring of a force table, one at an angle of 20 degrees, and the other at an angle of 80 degrees. Regardless of the magnitudes of the forces, will the equilibrant be in the
A) first quadrant
second quadrant
C) Third Quadrant
D) Fourth Quadrant
E) You cannot tell which quadrant from the available information....
i dont want the answer, i want to know how to work this out...
so far what i have is this...
i believe this equation goes into this...
Direction Fe opposite to Fr so direction Fr= direction Fe- 180 degree.
Fe = Equilibrant
Fr= Resultant
And i'm pretty sure the magnitude doesnt matter, but somehow i am supposed to use 'sin' to figure out what the 'newtons' for the pull would be, gravity is 9.80 ms/2 but i have no idea how to set this up. I wouldnt ask for help, but i have looked everywhere, i have been on this problem for over an hour and my physics partners are not texting me back. >.<
okay...
Question:
Two forces are applied to a ring of a force table, one at an angle of 20 degrees, and the other at an angle of 80 degrees. Regardless of the magnitudes of the forces, will the equilibrant be in the
A) first quadrant
![B) B)](https://community.myfitnesspal.com/en/resources/emoji/sunglasses.png)
C) Third Quadrant
D) Fourth Quadrant
E) You cannot tell which quadrant from the available information....
i dont want the answer, i want to know how to work this out...
so far what i have is this...
i believe this equation goes into this...
Direction Fe opposite to Fr so direction Fr= direction Fe- 180 degree.
Fe = Equilibrant
Fr= Resultant
And i'm pretty sure the magnitude doesnt matter, but somehow i am supposed to use 'sin' to figure out what the 'newtons' for the pull would be, gravity is 9.80 ms/2 but i have no idea how to set this up. I wouldnt ask for help, but i have looked everywhere, i have been on this problem for over an hour and my physics partners are not texting me back. >.<
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Replies
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Nevermind. I read it wrong. Magnitude won't matter.0
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my biggest issue is like how to find force? do i automatically multiply is by 10? and if so why? I mean force= mass*acceleration, but i dont know the mass, so does that mean i cant figure it out?!0
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I think a picture would be more beneficial to see where the actual forces are acting on the ring. It all depends on where the forces are being applied to the ring (internal or external), and what the angle is relative to (horizontal? Vertical?)0
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my biggest issue is like how to find force? do i automatically multiply is by 10? and if so why? I mean force= mass*acceleration, but i dont know the mass, so does that mean i cant figure it out?!
That doesn't matter. What matters is the ring is in equillibirium, so you know that whatever the two forces are, there is a reactive force that equals the sum of the other two forces. To figure out where the reactive force is at, you need to know where the other two forces are being applied.0 -
the angle is horizontal, and these are being used on a force table. and the ring is simply holding the masses up, (which we dont know the masses)0
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the angle is horizontal, and these are being used on a force table. and the ring is simply holding the masses up, (which we dont know the masses)
This is where a picture is worth a thousand words. A sketch would be helpful so that you can draw your free body diagram.0 -
like i tried this:
Fg= mg
= (10) (9.81)
Fg= 98.1 N
Sin= opp/hyp
sinO = 1/2 Fg / Ft
Ft= (1/2 Fg) / sinO
Ft= 49.05 / sin(20)
But that isnt right when i checked it against an example
idk why i would multiply by 10
or WTF
this was a step by step example i found, but it doesnt some out right.0 -
like i tried this:
Fg= mg
= (10) (9.81)
Fg= 98.1 N
Sin= opp/hyp
sinO = 1/2 Fg / Ft
Ft= (1/2 Fg) / sinO
Ft= 49.05 / sin(20)
But that isnt right when i checked it against an example
idk why i would multiply by 10
or WTF
this was a step by step example i found, but it doesnt some out right.
Can you post a picture?0 -
yeah, obviously, this has nothing to do with weight loss hahahaha. But i CANNOT find help on google, and everyone on my facebook has no idea how to work this problem....
okay...
Question:
Two forces are applied to a ring of a force table, one at an angle of 20 degrees, and the other at an angle of 80 degrees. Regardless of the magnitudes of the forces, will the equilibrant be in the
A) first quadrantsecond quadrant
C) Third Quadrant
D) Fourth Quadrant
E) You cannot tell which quadrant from the available information....
i dont want the answer, i want to know how to work this out...
so far what i have is this...
i believe this equation goes into this...
Direction Fe opposite to Fr so direction Fr= direction Fe- 180 degree.
Fe = Equilibrant
Fr= Resultant
And i'm pretty sure the magnitude doesnt matter, but somehow i am supposed to use 'sin' to figure out what the 'newtons' for the pull would be, gravity is 9.80 ms/2 but i have no idea how to set this up. I wouldnt ask for help, but i have looked everywhere, i have been on this problem for over an hour and my physics partners are not texting me back. >.<
resolve your resultant into the two corresponding identities ,sub in using vector addition then solve for theta (by arc of R1/R2)0 -
420
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How I wish I could help but I can't. Thought this might be helpful though if you've not seen it.
http://www.khanacademy.org/#physics0 -
ummm, just pick C. I did that on all my tests and sailed through school!!!
whew, I am soooo glad that I do not have to figure out physics! that *kitten* seems hard!
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omfg my calculator was sad to radian instead of degrees...... i hate my life right now.0
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there isnt a picture to post... i could like draw it out and kinda show you what i THINK it is asking... or do you need a picture of the force table???0
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but the two strings would be at 80 and 20, and the 3rd would be the resultant.0 -
there isnt a picture to post... i could like draw it out and kinda show you what i THINK it is asking... or do you need a picture of the force table???
No, I don't need a picture of hte force table now. I think I know what it's asking. I think it's in Quadrant 3, just intuitively because the two forces that you know are there are in Quadrant 1. Equal and opposite reaction shows that the opposit reaction would be roughly in quadrant 3. But, you need to prove that.0 -
Actually, you can only prove that mathematically if you know the magnitude of the forces. So, my answer would be the third quadrant, Only becaus the two applied forces are in the first quadrant, your reactive force would be in the third.0
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there isnt a picture to post... i could like draw it out and kinda show you what i THINK it is asking... or do you need a picture of the force table???
No, I don't need a picture of hte force table now. I think I know what it's asking. I think it's in Quadrant 3, just intuitively because the two forces that you know are there are in Quadrant 1. Equal and opposite reaction shows that the opposit reaction would be roughly in quadrant 3. But, you need to prove that.
SEE I WAS RIGHT!!!!! YAY FOR ME!!!!!0 -
there isnt a picture to post... i could like draw it out and kinda show you what i THINK it is asking... or do you need a picture of the force table???
No, I don't need a picture of hte force table now. I think I know what it's asking. I think it's in Quadrant 3, just intuitively because the two forces that you know are there are in Quadrant 1. Equal and opposite reaction shows that the opposit reaction would be roughly in quadrant 3. But, you need to prove that.
SEE I WAS RIGHT!!!!! YAY FOR ME!!!!!
I bet you never thought a homeless hobo who smells of booze and cigarettes would discuss physics with you, huh?0
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