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Physics Question, I can't understand why that answer is correct. Watch

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    The answer to the question was D. However, I don't understand why :confused:
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    (Original post by MSB47)
    The answer to the question was D. However, I don't understand why :confused:
    Hello and welcome to TSR.

    This is a question about vectors and resolving vertical and horizontal forces.

    The force EQ acts in a horizontal direction, mg acts vertically.

    The definition of

    tan \theta = \frac{opposite}{adjacent}

    therefore

    tan \theta = \frac{EQ}{mg}


    EQ = mg.tan \theta

    D is incorrect.
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    (Original post by uberteknik)
    Hello and welcome to TSR.

    This is a question about vectors and resolving vertical and horizontal forces.

    The force EQ acts in a horizontal direction, mg acts vertically.

    The definition of

    tan \theta = \frac{opposite}{adjacent}

    therefore

    tan \theta = \frac{EQ}{mg}


    EQ = mg.tan \theta

    D is incorrect.
    That makes a lot of sense however, isn't B also incorrect I haven't been able prove that expression is correct and it is bugging me out of curiosity on how to derive that expression. Could you please tell me how answer B is a correct expression?

    Many thanks
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    (Original post by MSB47)
    That makes a lot of sense however, isn't B also incorrect I haven't been able prove that expression is correct and it is bugging me out of curiosity on how to derive that expression. Could you please tell me how answer B is a correct expression?

    Many thanks
    EDIT: See post 6 for correct answer



    Yes. it looks like an error.

    Not sure what they meant to put.

    T = \frac{mg}{Sin \theta} \mathrm \ or \ T = \frac{EQ}{Cos \theta}
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    (Original post by uberteknik)
    Yes. it looks like an error.

    Not sure what they meant to put.

    T = \frac{mg}{Sin \theta} \mathrm \ or \ T = \frac{EQ}{Cos \theta}
    Yeah I got something like T=mg/cos0 + EQ/Sin0



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    (Original post by MSB47)
    Yeah I got something like T=mg/cos0 + EQ/Sin0



    This bugged me too, so I had a cup of tea and came back to it and solved it. It's not obvious at first but quite simple when the penny drops:

    mg = T.Cos \theta..........eq1

    EQ = T.Sin \theta...........eq2

    T^2 = mg^2 + EQ^2

    T^2 = mg.mg + EQ.EQ............eq3

    sub eq1 and eq2 into eq3

    T^2 = mgT.Cos \theta + EQT.Sin \theta

    divide through by T

    T = mg.Cos \theta + EQ.Sin \theta

    Answer B is correct after all.

    QED.
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    (Original post by uberteknik)
    This bugged me too, so I had a cup of tea and came back to it and solved it. It's not obvious at first but quite simple when the penny drops:

    mg = T.Cos \theta..........eq1

    EQ = T.Sin \theta...........eq2

    T^2 = mg^2 + EQ^2

    T^2 = mg.mg + EQ.EQ............eq3

    sub eq1 and eq2 into eq3

    T^2 = mgT.Cos \theta + EQT.Sin \theta

    divide through by T

    T = mg.Cos \theta + EQ.Sin \theta

    Answer B is correct after all.

    QED.
    That has eased my brain so much. I was starting to get annoyed that it wouldn't make sense, I guess it just required thinking outside the box. Thanks for the help, really appreciated.
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    (Original post by MSB47)
    That has eased my brain so much. I was starting to get annoyed that it wouldn't make sense, I guess it just required thinking outside the box. Thanks for the help, really appreciated.
    You are very welcome.
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    (Original post by MSB47)
    That has eased my brain so much. I was starting to get annoyed that it wouldn't make sense, I guess it just required thinking outside the box. Thanks for the help, really appreciated.
    Having slept on it, there is another way this expression could be derived from the trig identity:

    Sin^2 \theta + Cos^2 \theta = 1

    multiply by T and expand

    (TSin \theta)Sin \theta + (TCos \theta)Cos \theta = T

    then from

    TSin \theta = EQ \  \mathrm {\   and \   } \ TCos \theta = mg

    T = mgCos \theta + EQSin \theta


    I'm not a closet OCD! LOL No really I'm not!


    Moral of the story: Getting some good rest is just as important as getting the work done. Do both!
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    (Original post by uberteknik)
    Having slept on it, there is another way this expression could be derived from the trig identity:

    Sin^2 \theta + Cos^2 \theta = 1

    multiply by T and expand

    (TSin \theta)Sin \theta + (TCos \theta)Cos \theta = T

    then from

    TSin \theta = EQ \  \mathrm {\   and \   } \ TCos \theta = mg

    T = mgCos \theta + EQSin \theta


    I'm not a closet OCD! LOL No really I'm not!


    Moral of the story: Getting some good rest is just as important as getting the work done. Do both!
    That makes sense too. You make it seem easy
 
 
 
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