physics help....

DatEpicChick
DatEpicChick Posts: 358 Member
edited December 2024 in Health and Weight Loss
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
B) 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. >.<
«13

Replies

  • SofaKingRad
    SofaKingRad Posts: 1,592 Member
    Nevermind. I read it wrong. Magnitude won't matter.
  • DatEpicChick
    DatEpicChick Posts: 358 Member
    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?!
  • SofaKingRad
    SofaKingRad Posts: 1,592 Member
    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?)
  • SofaKingRad
    SofaKingRad Posts: 1,592 Member
    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.
  • DatEpicChick
    DatEpicChick Posts: 358 Member
    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)
  • SofaKingRad
    SofaKingRad Posts: 1,592 Member
    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.
  • DatEpicChick
    DatEpicChick Posts: 358 Member
    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.
  • SofaKingRad
    SofaKingRad Posts: 1,592 Member
    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?
  • lickmybaconcakes
    lickmybaconcakes Posts: 1,063 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
    B) 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. >.<


    resolve your resultant into the two corresponding identities ,sub in using vector addition then solve for theta (by arc of R1/R2)
  • RainRedfield
    RainRedfield Posts: 597 Member
    42
  • sixisCHANGEDjk
    sixisCHANGEDjk Posts: 243
    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/#physics
  • Changing__Christina
    Changing__Christina Posts: 245 Member
    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!
  • DatEpicChick
    DatEpicChick Posts: 358 Member
    omfg my calculator was sad to radian instead of degrees...... i hate my life right now.
  • DatEpicChick
    DatEpicChick Posts: 358 Member
    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???
  • AmandaLAcosta
    AmandaLAcosta Posts: 87
    http://faculty.rcc.edu/bhattacharya/phy4a/force_table.htm
  • DatEpicChick
    DatEpicChick Posts: 358 Member
    forcetable3.jpg

    but the two strings would be at 80 and 20, and the 3rd would be the resultant.
  • SofaKingRad
    SofaKingRad Posts: 1,592 Member
    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.
  • SofaKingRad
    SofaKingRad Posts: 1,592 Member
    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.
  • Changing__Christina
    Changing__Christina Posts: 245 Member
    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!!!!! :)
  • SofaKingRad
    SofaKingRad Posts: 1,592 Member
    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?
  • Changing__Christina
    Changing__Christina Posts: 245 Member
    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.

    I love that a drunken, stinky, bum knows physics! That is HOT! :)

    BTW, my answer was right! :laugh:
  • SofaKingRad
    SofaKingRad Posts: 1,592 Member
    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.

    Don't worry about the multiply by 10. I think in the problem you're looking at, they have the masses given to you so you can calculate the magnitude of the forces.
  • Changing__Christina
    Changing__Christina Posts: 245 Member
    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?

    we are on the same brain wave! LOL! If you weren't homeless, I would come over right now!
  • SofaKingRad
    SofaKingRad Posts: 1,592 Member
    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?

    we are on the same brain wave! LOL! If you weren't homeless, I would come over right now!

    I'm not technically homeless. My local park is my home.
  • Changing__Christina
    Changing__Christina Posts: 245 Member
    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?

    we are on the same brain wave! LOL! If you weren't homeless, I would come over right now!

    I'm not technically homeless. My local park is my home.

    Can you get to pan handling for spare change so you can buy me a drink! I will be there in a bit!
  • SofaKingRad
    SofaKingRad Posts: 1,592 Member
    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?

    we are on the same brain wave! LOL! If you weren't homeless, I would come over right now!

    I'm not technically homeless. My local park is my home.

    Can you get to pan handling for spare change so you can buy me a drink! I will be there in a bit!

    I already have my booze here. I keep it on me at all times since I'm an alcoholic.
  • DatEpicChick
    DatEpicChick Posts: 358 Member
    seriously...

    your name = so faking rad....

    i thought it was sofa king rad....


    bahahahahahahaha
  • DG_Allen
    DG_Allen Posts: 219 Member
    I think it's really an intuative problem. Where is the force going to be pulling based on the angle between the two forces? What quadrant are the forces in? You don't have all the data, so you don't really need to use an equation to figure it out.

    If this were a clock, I'm assuming a force at 0 degrees would be pulling toward 12 o'clock. Quadrant 1 is the upper right.
  • yubafarm
    yubafarm Posts: 28 Member
    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.

    I agree with this answer if the applied forces are PULLING away from the force balance at 20 and 80 degrees. The reaction force would have to be pulling back at some angle in Q3. I think that's a reasonable assumption to make even though it just says the forces are "applied" at those angles, so really it could be a pulling or pushing force (if pushing at those angles the answer would be Q1). But remembering way back to textbook type questions are written Q3 is the most likely answer.
  • SofaKingRad
    SofaKingRad Posts: 1,592 Member
    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.

    I agree with this answer if the applied forces are PULLING away from the force balance at 20 and 80 degrees. The reaction force would have to be pulling back at some angle in Q3. I think that's a reasonable assumption to make even though it just says the forces are "applied" at those angles, so really it could be a pulling or pushing force (if pushing at those angles the answer would be Q1). But remembering way back to textbook type questions are written Q3 is the most likely answer.

    Even if the forces were flipped in such that they are pushing the ring, the reactive force would sitll be in quadrant 3 since the resultant of the two forces are within quadrant one.
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