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Size:  128.7 KB for part c of this question I got 2 for u. However the answer should be -2. I would like some help on this please. Thanks.Attachment 519513519515
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    F is a constant retarding force so surely f would be negative in the equation. However the markscheme indicates otherwise. Its positive f in the mark scheme.
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    (Original post by coconut64)
    ...
    So you got a = -\mathbf{i} -2\mathbf{j} - good job!

    Now you know that \mathbf{F}=m\mathbf{a} so F = 2(-i-2j) = -2(i + 2j), agreeing with me so far?

    Remember that 2(-i - 2j) = 2((-1)i + (-1)2j) = 2(-1)(i + 2j) = -2(i + 2j)... so what's \mu?
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    (Original post by coconut64)
    F is a constant retarding force so surely f would be negative in the equation. However the markscheme indicates otherwise. Its positive f in the mark scheme.
    You are getting confused. The force acts in whatever direction your acceleration acts. I am not sure what you are resolving here... your acceleration and force are 2 dimensional components already resolved in the horizontal (i) and vertical (j) direction. So resolving "to the right" as you have seeming shown, makes 0 sense. If you resolve to the right, all your j components disappear...

    You are talking about F as though it is a number, it is a vector and it doesn't make sense to talk about "negative vectors" or "one vector is greater than 0" or along those lines. The technical reason for this is that the ordered pair of real numbers aren't well-ordered or don't possess a well-ordering. (lol sounds oxymoron, but oh well)
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    (Original post by Zacken)
    You are getting confused. The force acts in whatever direction your acceleration acts. I am not sure what you are resolving here... your acceleration and force are 2 dimensional components already resolved in the horizontal (i) and vertical (j) direction. So resolving "to the right" as you have seeming shown, makes 0 sense. If you resolve to the right, all your j components disappear...

    You are talking about F as though it is a number, it is a vector and it doesn't make sense to talk about "negative vectors" or "one vector is greater than 0" or along those lines. The technical reason for this is that the ordered pair of real numbers aren't well-ordered or don't possess a well-ordering. (lol sounds oxymoron, but oh well)
    I am so sorry but I am even more confused by what you just said. The equation is supposed to be f=ma. But since f is a retarding force in this question, it will act in the different direction to acceleration thus the unknown force will be negative. I don't get what you mean. Name:  IMAG0660.jpg
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    (Original post by coconut64)
    I am so sorry but I am even more confused by what you just said. The equation is supposed to be f=ma. But since f is a retarding force in this question, it will act in the different direction to acceleration thus the unknown force will be negative. I don't get what you mean. Name:  IMAG0660.jpg
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    No! F always acts in whatever direction acceleration acts. If F is a retarding force, that simply means your acceleration is a retardation. But your force still acts in the direction of acceleration.That's it. End of.
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    (Original post by Zacken)
    No! F always acts in whatever direction acceleration acts. If F is a retarding force, that simply means your acceleration is a retardation. But your force still acts in the direction of acceleration.That's it. End of.
    If force always acts in the same direction as the acceleration so why is f negative in this question? http://www.examsolutions.net/a-level...6&solution=6.3 thanks this question is similar to this question and both asks to find the retarding force.
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    (Original post by coconut64)
    If force always acts in the same direction as the acceleration so why is f negative in this question? http://www.examsolutions.net/a-level...6&solution=6.3 thanks this question is similar to this question and both asks to find the retarding force.
    The force in the question you've linked isn't the resultant force.

    Newton's second law states that F_{\text{net}} = ma and since m is a positive scalar, then F_{\text{net}} has the same direction as a.

    In the question in the video, F is just the variable given for a force, but that isn't the resultant force.

    In this question, F is the resultant retarding force.

    In any case - the video deals with one dimensional forces. Your question deals with two dimensional forces, "resolving" the way you have makes no sense.
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    (Original post by Zacken)
    The force in the question you've linked isn't the resultant force.

    Newton's second law states that F_{\text{net}} = ma and since m is a positive scalar, then F_{\text{net}} has the same direction as a.

    In the question in the video, F is just the variable given for a force, but that isn't the resultant force.

    In this question, F is the resultant retarding force.

    In any case - the video deals with one dimensional forces. Your question deals with two dimensional forces, "resolving" the way you have makes no sense.
    How do you know that this is the resultant force? It doesn't say in the question
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    (Original post by coconut64)
    How do you know that this is the resultant force? It doesn't say in the question
    Are there any other forces?
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    (Original post by Zacken)
    Are there any other forces?
    No, I have just realised that. So basically resultant force always acts in the same direction as a. But if F is only one force and there are other forces involved, it can move In the different direction to a?
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    (Original post by coconut64)
    No, I have just realised that. So basically resultant force always acts in the same direction as a. But if F is only one force and there are other forces involved, it can move In the different direction to a?
    Yes, precisely.

    But once you sum all the other forces involved, the sum is then the resultant force and moves in the direction of a.
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    (Original post by Zacken)
    Yes, precisely.

    But once you sum all the other forces involved, the sum is then the resultant force and moves in the direction of a.
    Ahh this makes more sense now. Thanks so much . I can be quite stupid at times.
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    (Original post by coconut64)
    Ahh this makes more sense now. Thanks so much . I can be quite stupid at times.
    No problem! There's no way you can be as stupid as I can be sometimes, you're not at all. :-)
 
 
 
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