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    Name:  Picture2.png
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Size:  230.9 KBWhy is the tensions in both strings the same?
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    (Original post by ps1265A)
    Name:  Picture2.png
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Size:  230.9 KBWhy is the tensions in both strings the same?
    If you mean "the tension in both chunks of the string": it must be, because the string is inextensible. Suppose the tension were different at two different points on the string. Then there would be a net force on a point in between those points, so that in-between point would have to be moving (because things move when a force is applied to them). But in this problem, the string is stationary.

    The idea is that "if we look at a perpendicular cross-section of the string, we can't determine how far along the string we took the cross-section". Is that idea intuitively clear to you?
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    (Original post by Smaug123)
    If you mean "the tension in both chunks of the string": it must be, because the string is inextensible. Suppose the tension were different at two different points on the string. Then there would be a net force on a point in between those points, so that in-between point would have to be moving (because things move when a force is applied to them). But in this problem, the string is stationary.

    The idea is that "if we look at a perpendicular cross-section of the string, we can't determine how far along the string we took the cross-section". Is that idea intuitively clear to you?
    Thanks for that! But I have done questions in which the string is inextensible and the tensions are different?
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    (Original post by ps1265A)
    Thanks for that! But I have done questions in which the string is inextensible and the tensions are different?
    The ring is smooth (key point), so there is no additional force acting along the string, and tension is constant throughout.
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    (Original post by ghostwalker)
    The ring is smooth (key point), so there is no additional force acting along the string, and tension is constant throughout.
    Thanks! If I have 2 supports, and the plank is on the point of tilting about support A, this means reaction of support at B is 0 right? And normal reaction changes at the point where it is about to tilt right?
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    (Original post by ps1265A)
    Thanks! If I have 2 supports, and the plank is on the point of tilting about support A, this means reaction of support at B is 0 right? And normal reaction changes at the point where it is about to tilt right?
    If it's tending to lift off from B, then yes the reaction will be zero there.

    I don't understand your last sentence - needs a lot more context.
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    (Original post by ghostwalker)
    If it's tending to lift off from B, then yes the reaction will be zero there.

    I don't understand your last sentence - needs a lot more context.
    Thanks!

    For Q1, why is there no normal reaction?
    https://3b0a7b1bc87f5381e60f8f717510...%20Edexcel.pdf

    And do I ALWAYS have to take moments about a support?
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    (Original post by ps1265A)
    Thanks!

    For Q1, why is there no normal reaction?
    https://3b0a7b1bc87f5381e60f8f717510...%20Edexcel.pdf

    And do I ALWAYS have to take moments about a support?
    Normal reaction where?

    You can take moments about wherever you want. Taking it about a support is usually more useful, as eliminates the forces acting at the support from your equations.
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    (Original post by ghostwalker)
    Normal reaction where?

    You can take moments about wherever you want. Taking it about a support is usually more useful, as eliminates the forces acting at the support from your equations.
    On P
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    (Original post by ps1265A)
    On P
    Well it's just a particle on a string. You could resolve the 12N force into components parallel to and normal to OP if you wish. Or do you mean normal to something else?

    You have three forces acting, tension in the string, the weight of P, and the 12N force.
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    (Original post by ghostwalker)
    Well it's just a particle on a string. You could resolve the 12N force into components parallel to and normal to OP if you wish. Or do you mean normal to something else?

    You have three forces acting, tension in the string, the weight of P, and the 12N force.
    Thanks!

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

    For Q8c), I've done everything apart from the last bit
    Why can I not use S=+1.175 u =+0.7 a=+9.8???
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    (Original post by ps1265A)
    Thanks!

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

    For Q8c), I've done everything apart from the last bit
    Why can I not use S=+1.175 u =+0.7 a=+9.8???
    a = +ve implies you're taking downwards as positive. In which case u will be -ve, as it's going up when the string breaks.
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    (Original post by ghostwalker)
    a = +ve implies you're taking downwards as positive. In which case u will be -ve, as it's going up when the string breaks.
    Why is it going up? It's falling down???

    Another Q, say if I have a plank and a support; there will be a normal reaction at the support. Is this normal reaction a force that is exerted on to the plank?
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    (Original post by ps1265A)
    Why is it going up? It's falling down???
    The initial velocity of the mass when the string breaks is upwards. It's the lighter of the two masses.

    Another Q, say if I have a plank and a support; there will be a normal reaction at the support. Is this normal reaction a force that is exerted on to the plank?
    Yes.
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    (Original post by ps1265A)
    Why is it going up? It's falling down???

    Another Q, say if I have a plank and a support; there will be a normal reaction at the support. Is this normal reaction a force that is exerted on to the plank?
    When the string breaks, B is on it's way up and A is on it's way down. B the travels freely under gravity until it hits the floor again. If you're having problems picturing it, imagine giving something on a string a tug upwards - it doesn't stop moving up when you stop pulling, it continues to move with a slack string.

    For your second question, yes.
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    (Original post by ghostwalker)
    The initial velocity of the mass when the string breaks is upwards. It's the lighter of the two masses.



    Yes.
    Thanks! When I'm considering vectors, when a vector is north of another, i components are the same. When something is north-west, why is it -i=j
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    (Original post by Mr T Pities You)
    When the string breaks, B is on it's way up and A is on it's way down. B the travels freely under gravity until it hits the floor again. If you're having problems picturing it, imagine giving something on a string a tug upwards - it doesn't stop moving up when you stop pulling, it continues to move with a slack string.

    For your second question, yes.
    Thanks!
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    (Original post by ps1265A)
    Thanks! When I'm considering vectors, when a vector is north of another, i components are the same. When something is north-west, why is it -i=j
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    Not a great explanation but I hope it helps
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    (Original post by Mr T Pities You)
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    Not a great explanation but I hope it helps
    Thanksss sooooooo much!

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

    For Q5a), why is a= NEGATIVE 9.8???? What about the positive 9.8 when it's going up?
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    (Original post by ps1265A)
    Thanksss sooooooo much!

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

    For Q5a), why is a= NEGATIVE 9.8???? What about the positive 9.8 when it's going up?
    Acceleration due to gravity always acts down, and if you always pick up as being positive, g is always negative.
 
 
 
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