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    I've been doing alrite till i came to this question, I am so sure my methods right but i dont get the official answer. Can someone doublecheck this to see if its my mistake or the books.

    Question 15 Page 147 Review Excercise 2

    Two particles A and B of mass 8kg and 10kg respectively, are connected by light inextensible string which passes over a light smooth pulley P. Particle B rests on a smooth horizontal table and particle A rests on a smooth plane inclined at 30degrees to the horizontal with the string taut and perpendiclar to the line of intersection of the table and plane as shown below(attachment):

    The system is released from Rest.

    (a) find the magnitude of acceleration of particle at B

    Basically diagram attached is what they give you, so I havent drawn it wrong the only thing i added was the T(tension) arrow so if anythings wrong it would only be that. Question seemed fairly straight foward to me but the answer at the back is: a=2.18ms

    My working:

    8g sin 30-T=8a
    4g-T=8a (1)
    T-10g=10a (2)

    Adding (1) and (2)

    -6g=18a
    a= -6g/18
    = -g/3
    =-3.27ms

    I really need to know if I am missing somthing here of if the answer at the backs wrong. :confused:
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    hangon illd o it now
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    Resolving in the direction of initial motion

    8gsin30-T=8a (i)

    The only force acting on particle B, in the direction of acceleration, is the tension of the string.

    T=10a (ii)

    =>8gsin30-10a=8a

    =>8gsin30=18a

    =>a=2.17rec. m(s^(-2))~2.18 m(s^(-2))

    Newton.
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    (Original post by Newton)
    Resolving in the direction of initial motion

    8gsin30-T=8a (i)

    The only force acting on particle B, in the direction of acceleration, is the tension of the string.

    T=10a (ii)

    =>8gsin30-10a=8a

    =>8gsin30=18a

    =>a=2.17rec. m(s^(-2))~2.18 m(s^(-2))

    Newton.
    Thanks Newton, Ill thank you again if a similar questions comes up in my m1 2morro
 
 
 
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