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    how would i go about doing this question??

    https://3b0a7b1bc87f5381e60f8f717510...%20Edexcel.pdf

    question 6
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    First I would draw a line on the diagram for the resultant force R which would be in the middle diagonally /. I would also label everything I knew and see where to go from there
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    (Original post by thefatone)
    how would i go about doing this question??

    https://3b0a7b1bc87f5381e60f8f717510...%20Edexcel.pdf

    question 6
    Resolve and use pythagoras to equate the magnitude to 3X then get a quadratic in X.

    Horizontal (in the direction of Q): (X-20\cos 60^{\circ})

    Vertical (upwards): 20\sin 60^{\circ}

    So (X-20\cos 60^{\circ})^2 + (20 \sin 60^{\circ})^2 = (3X)^2
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    (Original post by thefatone)
    how would i go about doing this question??

    https://3b0a7b1bc87f5381e60f8f717510...%20Edexcel.pdf

    question 6
    You can resolve vertically/horizontally, as shown by Zacken.

    Alternatively you could set up a triangle of forces and use the cosine rule. Have you used a triangle of forces before?
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    (Original post by notnek)
    You can resolve vertically/horizontally, as shown by Zacken.

    Alternatively you could set up a triangle of forces and use the cosine rule. Have you used a triangle of forces before?
    yes but i've forgotten how to do it, show me pls?

    (Original post by Zacken)
    Resolve and use pythagoras to equate the magnitude to 3X then get a quadratic in X.

    Horizontal (in the direction of Q): (X-20\cos 60^{\circ})

    Vertical (upwards): 20\sin 60^{\circ}

    So (X-20\cos 60^{\circ})^2 + (20 \sin 60^{\circ})^2 = (3X)^2
    i don't understand why X-20cos60
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    (Original post by thefatone)
    yes but i've forgotten how to do it, show me pls?



    i don't understand why X-20cos60
    There are two forces horizontally. Q:X (right) and P:20cos60 (left), since you are resolving with positive being to the right, you get X-20cos60.
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    (Original post by thefatone)
    yes but i've forgotten how to do it, show me pls?



    i don't understand why X-20cos60
    You can reposition the vectors to create a triangle of forces:



    The red vector R is the resultant force since it's equal to Q + P.

    Find angle OQP and then use the cosine rule on triangle OQP to find the magnitude of R.
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    (Original post by AlmostNotable)
    There are two forces horizontally. Q:X (right) and P:20cos60 (left), since you are resolving with positive being to the right, you get X-20cos60.

    so i have an x going right and a 20cos60 so why x-20cos60? why not add?

    (Original post by notnek)
    You can reposition the vectors to create a triangle of forces:



    The red vector R is the resultant force since it's equal to Q + P.

    Find angle OQP and then use the cosine rule on triangle OQP to find the magnitude of R.
    oh ok...
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    (Original post by thefatone)
    so i have an x going right and a 20cos60 so why x-20cos60? why not add?
    One is going left and one is going right...
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    (Original post by Zacken)
    One is going left and one is going right...
    oh i see.... derp thanks a ton as always for the help
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    (Original post by thefatone)
    oh i see.... derp thanks a ton as always for the help
    No problem.
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    Lami's theorem might be useful to know as Notnek said.
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    when it says resultant force of P and Q, is it telling me that R is (P+Q), or the 3 forces can be drawn as a closed triangle? and closed triangle means equilibrium? notnek
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    (Original post by alesha98)
    when it says resultant force of P and Q, is it telling me that R is (P+Q), or the 3 forces can be drawn as a closed triangle? and closed triangle means equilibrium? notnek
    Yes R = P + Q.

    And the forces can be drawn as a closed triangle as I showed in my post containing the diagram.

    But the forces P and Q which act on the particle are not in equlibrium.
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    (Original post by notnek)
    Yes R = P + Q.

    And the forces can be drawn as a closed triangle as I showed in my post containing the diagram.

    But the forces P and Q which act on the particle are not in equlibrium.
    What does it tell you that the forces can be drawn as a closed triangle? And also I dont know how to approach question 6 part b.

    Posted from TSR Mobile
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    (Original post by alesha98)
    What does it tell you that the forces can be drawn as a closed triangle? And also I dont know how to approach question 6 part b.

    Posted from TSR Mobile
    If you have 3 vectors a, b and c such that a = b + c then by defintion of vector addition, you can draw the vectors in a closed triangle.

    Think back to GCSE vectors.

    I'll post a diagram for 6b - give me a minute.
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    (Original post by notnek)
    If you have 3 vectors a, b and c such that a = b + c then by defintion of vector addition, you can draw the vectors in a closed triangle.

    Think back to GCSE vectors.

    I'll post a diagram for 6b - give me a minute.
    Ahh now I get it. This is very useful to me ... I completely forgot that ... thankyou so much


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    (Original post by alesha98)
    Ahh now I get it. This is very useful to me ... I completely forgot that ... thankyou so much


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    Takes your time to draw 6b. I will wait ,thanks

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    (Original post by alesha98)
    What does it tell you that the forces can be drawn as a closed triangle? And also I dont know how to approach question 6 part b.

    Posted from TSR Mobile

    The green vector is P - Q. Can you see why that is?

    And does that help?
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    (Original post by notnek)

    The green vector is P - Q. Can you see why that is?

    And does that help?
    So the green vector is the resultant force of P-Q?

    Posted from TSR Mobile
 
 
 
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