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Size:  131.3 KB for ii) of this question why isn't R = 2mg-3/5t?Attachment 507807507809 this is my diagram of it and tension acts in the same direction as r buy the mark scheme suggests that tension acts in the opposite direction. Why?? Thanks
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    Also for 3ii) of this question, why is the acceleration for mkg the same as a for 2mkg? Thanks Name:  Screenshot_2016-02-26-12-31-04.png
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    (Original post by coconut64)
    Also for 3ii) of this question, why is the acceleration for mkg the same as a for 2mkg? Thanks Name:  Screenshot_2016-02-26-12-31-04.png
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    Independent of mass.
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    Ref the Tension, you have not marked a direction for tension in your diagram so far as I can see.

    At the Top Left Ring, Tension acts down to the Right whilst Friction acts to the left. Possibly you are looking at force on the rough pole as you have Friction towards the Right on the LH ring and towards the Left on the RH ring

    Please try to have one post for one question.
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    (Original post by Zacken)
    Independent of mass.
    "said the young man from Mauritius. The professor shifted uncomfortably in his seat, and sighed. He looked round at the elegantly oak-lined walls of his study, nestling in the ancient surroundings of Slaughterhouse College, Camford, and, briefly giving thanks that he had rejected the offer from that northern place (how he hated black pudding!), he mumbled the time-honoured incantation, as he so often did when faced with a promising but unformed proto-mathematician:

    "Could you expand on that, please?"

    He settled back in his seat, puffed on his pipe, and took a sip of port. The fire crackled; the silence deepened; the tension mounted."
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    (Original post by atsruser)
    "said the young man from Mauritius. The professor shifted uncomfortably in his seat, and sighed. He looked round at the elegantly oak-lined walls of his study, nestling in the ancient surroundings of Slaughterhouse College, Camford, and, briefly giving thanks that he had rejected the offer from that northern place (how he hated black pudding!), he mumbled the time-honoured incantation, as he so often did when faced with a promising but unformed proto-mathematician:

    "Could you expand on that, please?"

    He settled back in his seat, puffed on his pipe, and took a sip of port. The fire crackled; the silence deepened; the tension mounted."
    Oh dear, I enjoyed that acutely. It's brilliant, cracked me up at 'proto-mathematician'. :lol:
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    am I too late?
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    (Original post by Zacken)
    Oh dear, I enjoyed that acutely. It's brilliant, cracked me up at 'proto-mathematician'. :lol:
    I'm glad that you liked it, but I was really trying to point out that you didn't fully answer the question, and that further clarification was needed.
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    (Original post by coconut64)
    Also for 3ii) of this question, why is the acceleration for mkg the same as a for 2mkg? Thanks Name:  Screenshot_2016-02-26-12-31-04.png
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    Consider both sides of F=ma for this particle. All the forces are proportional to the mass, and so is the RHS, so the mass can be divided away, showing that the acceleration is independent of the mass.
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    (Original post by coconut64)
    Also for 3ii) of this question, why is the acceleration for mkg the same as a for 2mkg? Thanks Name:  Screenshot_2016-02-26-12-31-04.png
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    astruster has already explained this part but, more intuitively, think about the proportionality of acceleration in the equation a=\dfrac{F}{m}

    Clearly, the acceleration of an object under an applied force is proportional to the force acting on it and is inversely proportional to it's mass, meaning a greater mass will accelerate at a lower rate compared to a less mass if the applied force is the same. However, for an object in a gravitational field, the force on an object is its weight, which is proportional to its mass according to Newton's law of gravitation. This proportionality indicates the gravitational attraction of a mass itself towards other masses (or effectively towards our earth for our proposes in the question), but this proportionality is equivalent to that of the force required with respect to the resulting acceleration which is due to the equivalency of the gravitational and inertial mass. This fact comes from an experimental observation demonstrated by Galileo rather than a priori assumption and can be written as a ratio of the masses. a=\dfrac{F}{m}=\dfrac{m_g}{m_i}g
 
 
 
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