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    Particles P and Q of masses 2m and m are attatched to ends of a light inextensible string which passes over a smooth fixed pulley. They both hang at a distance of 2m above horizontal ground, the system is released from rest.

    Find the magnitude of acceleration of the system.

    When resolving to do this, why do you use mg and 2mg for the mass of the particles? I thought you just used the mass? or is it weight (as w = mg ) in f = ma?
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    (Original post by lollage123)

    When resolving to do this, why do you use mg and 2mg for the mass of the particles?
    you don't
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    I think you'd just have:

    (1) 2mg - T = 2ma
    (2) T - mg = ma

    But I haven't done mechanics in a while
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    Find net force then use combined mass
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    Ignore wat i said
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    (Original post by L'Evil Fish)
    But I haven't done mechanics in a while
    .... then it's been too long.......

    The force that's making the 2m mass accelerate is gravity
    The force that's opposing the movement of P is T
    the effective force = mass x acceleration
    2mg - T = 2ma

    The force that's making Q accelerate comes from the tension in the string
    The force that's opposing that is gravity
    T - mg = ma
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    Pok im not sure anymore is is (2m-m)g/(2m+m)= g/3 assuming gravity = 10 it is 3.3333333333333333 ect
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    (Original post by gdunne42)
    .... then it's been too long.......

    The force that's making the 2m mass accelerate is gravity
    The force that's opposing the movement of P is T
    the effective force = mass x acceleration
    2mg - T = 2ma

    The force that's making Q accelerate comes from the tension in the string
    The force that's opposing that is gravity
    T - mg = ma
    I left the g out

    My apologies... I'll edit
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    (Original post by lollage123)
    Particles P and Q of masses 2m and m are attatched to ends of a light inextensible string which passes over a smooth fixed pulley. They both hang at a distance of 2m above horizontal ground, the system is released from rest.

    Find the magnitude of acceleration of the system.

    When resolving to do this, why do you use mg and 2mg for the mass of the particles? I thought you just used the mass? or is it weight (as w = mg ) in f = ma?
    The mass of each of the masses is 2m and m, right?
    You do use F=ma, and in fact you have both, F and m, and you're looking for a.
    the F is 2mg and mg, respectively.
    Remember that the weight of a particle is:
    • The force that particle has towards the Earth
    • The force caused by acceleration due to gravity (9.8ms^-2)
    • In Newtons


     F = ma
    Sub in your values of F.

    1)  2mg = ma
    2)  mg = ma

    Hopefully you can see a common factor that you can eliminate, then equate the equations to eachother and find a solution.



    edit:
    sorry I forgot about T and i'm far too tired to even be attempting to help anyone...
    I hope this helps in any case
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    Ive seem to have done it in a much simpler way but got same answer as gdunne g/3 would i receive less marks
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    The tension bit is not need in my solution as it is inextensible making the system essentially a combined fixed object being moved
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    (Original post by chapmouse)
    The mass of each of the masses is 2m and m, right?
    You do use F=ma, and in fact you have both, F and m, and you're looking for a.
    the F is 2mg and mg, respectively.
    Remember that the weight of a particle is:
    • The force that particle has towards the Earth
    • The force caused by acceleration due to gravity (9.8ms^-2)
    • In Newtons


     F = ma
    Sub in your values of F.

    1)  2mg - T = ma
    2)  T - mg = ma

    Spoiler:
    Show
    T = ma + mg

    2mg - ma + mg = ma
    3mg - 2ma = 0
    m(3g-2a) = 0
    So I get a to be 3g/2
    ^

    Where have I gone wrong?

    Posted from TSR Mobile
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    Okay thanks guys.
    Knowing that the speed of P as it hits the ground is 3.6ms-1, and given that the particle q does not reach the pulley, how do you work out the height q reaches above the ground?
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    Use F=ma to calculate the acceleration of Q. Now you should be able to use suvat to calculate the height q reaches. Remember that the final velocity of q will be 0 because it will be momentarily stationary.
 
 
 
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