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# m1 dynamics watch

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1. 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)

-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.
Attached Images

2. hangon illd o it now
3. 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.
4. (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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