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    Two passes one of 5kg and one of 3kg attached by a string pass over a pulley. The lighter one is within a jar of gloop a experiences a resistive force. The system is released from rest after 3 seconds their speed is 6m/s.
    a)Find the value of the resistive force via Newton's second law
    b)Find the value of the resistive force via the principal of impulse
    c)If the heavier mass was in the gloop and released from after 5 seconds have a speed 15 m/s. find the resistive force. :confused: :confused:
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    (Original post by XShmalX)
    Two passes one of 5kg and one of 3kg attached by a string pass over a pulley. The lighter one is within a jar of gloop a experiences a resistive force. The system is released from rest after 3 seconds their speed is 6m/s.
    a)Find the value of the resistive force via Newton's second law
    b)Find the value of the resistive force via the principal of impulse
    c)If the heavier mass was in the gloop and released from after 5 seconds have a speed 15 m/s. find the resistive force. :confused: :confused:
    So what have you done, and where are you stuck?
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    (Original post by ghostwalker)
    So what have you done, and where are you stuck?
    I worked out the accelerations and the force of gravity down on the two masses and then tried messing about with forces and got myself in a muddle.
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    (Original post by ghostwalker)
    So what have you done, and where are you stuck?
    Why don't you just do the whole question for her? Things are far more efficient that way.
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    (Original post by XShmalX)
    I worked out the accelerations and the force of gravity down on the two masses and then tried messing about with forces and got myself in a muddle.
    OK. I presume you've drawn a diagram - always helps.

    So for part a).

    We're going to be using F=ma.

    Considering the connected masses as one mass (otherwise we need to take tension in the string into account and it gets a whole let messier)

    Mass of connected particles is?

    Acceleration is?

    Force = ??
    There are two forces here, the weight due to gravity (careful with this one).
    and the resistance of the gloop!

    Don't forget to choose one direction as positive when setting up your equation.

    So, what do you get? (Have a go, it might be right, or we can check where you've slipped up).
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    (Original post by ghostwalker)
    OK. I presume you've drawn a diagram - always helps.

    So for part a).

    We're going to be using F=ma.

    Considering the connected masses as one mass (otherwise we need to take tension in the string into account and it gets a whole let messier)

    Mass of connected particles is?

    Acceleration is?

    Force = ??
    There are two forces here, the weight due to gravity (careful with this one).
    and the resistance of the gloop!

    Don't forget to choose one direction as positive when setting up your equation.

    So, what do you get? (Have a go, it might be right, or we can check where you've slipped up).
    Mass is 8kg
    Force is 16N
    Force of gravity is (9.8*5)-(9.8*3)= 19.6 as one is going up and one down?
    So resistance of gloop is 19.6-16= 3.6N
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    (Original post by XShmalX)
    Mass is 8kg
    Force is 16N
    Force of gravity is (9.8*5)-(9.8*3)= 19.6 as one is going up and one down?
    So resistance of gloop is 19.6-16= 3.6N
    Yes. Here's how I would have worded it.


    Combined mass 8 kg.
    Acceleration = 2 ms^-2.
    Effective force due to gravity = 19.6 (derived as you did).

    If gloop has a resistance of F.

    Then 19.6-F=8x2

    F= 19.6-16 = 3.6
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    (Original post by ghostwalker)
    Yes. Here's how I would have worded it.


    Combined mass 8 kg.
    Acceleration = 2 ms^-2.
    Effective force due to gravity = 19.6 (derived as you did).

    If gloop has a resistance of F.

    Then 19.6-F=8x2

    F= 19.6-16 = 3.6
    How could I use the principle of impulse to find it though?
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    (Original post by XShmalX)
    How could I use the principle of impulse to find it though?
    Impulse = change in momentum.

    Impulse = force x time.

    Can you take it from there?
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    (Original post by ghostwalker)
    Impulse = change in momentum.

    Impulse = force x time.

    Can you take it from there?
    Yep, I think so, would c be virtually the same??
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    (Original post by XShmalX)
    Yep, I think so, would c be virtually the same??

    Yes.
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    (Original post by ghostwalker)
    Yes.
    22 N ??? Sounds alot :confused:
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    (Original post by XShmalX)
    22 N ??? Sounds alot :confused:
    Does to me. I make it -4.4 N, which is impossible. Can you check the information in the OP, and whether it is still the same situation with the 2 masses over a pulley.
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    (Original post by ghostwalker)
    Does to me. I make it -4.4 N, which is impossible. Can you check the information in the OP, and whether it is still the same situation with the 2 masses over a pulley.
    Ahhh lol sorry the masses change to 7kg and 2kg with the heavier in the gloop and reach 15m/s in 5s. Opps really sorry! :o:
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    (Original post by XShmalX)
    Ahhh lol sorry the masses change to 7kg and 2kg with the heavier in the gloop and reach 15m/s in 5s. Opps really sorry! :o:

    In that case, yes, the resisting force is 22 Newtons. Well done!
 
 
 
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