Calculating Acceleration of Particles on a Sloping Roof

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To calculate the acceleration of two particles connected by a string over a pulley, the initial equations set up were incorrect. The tension in the string and the forces acting on both particles must be accurately represented, particularly considering the angle of the slope. Adjusting the equations led to the realization that the correct acceleration is 4/5g, which aligns with the book's answer of 7.84 m/s². The discussion highlights the importance of correctly applying forces and tension in physics problems involving inclined planes. Clarifying the setup and equations resolves the discrepancy in the calculated acceleration.
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Two particles, A (mass 0.2 kg) and B (mass 0.3 kg), are attached to an inexstensible string which passes over a smooth pulley. A initially rests on a sloping roof that makes a 30 degree angle with the horizontal and B hangs freely below the roof. Find the acceleration, a, of the particles when released from rest.
(I tried to draw a diagram but it came out looking a mess).
My workings (T is the tension in the string):
(1) 0.3g - T = 0.3a
(2) T - 0.2g sin 30 deg = 0.2a

(1) T = 0.3g - 0.3a
(2) 0.3g - 0.3a - 0.1g = 0.2a
(2) 0.5a = 0.2g
(2) a = 0.4g
taking g to be 9.8 m/s/s:
a = 3.92 m/s/s
My book says 7.84 m/s/s.
What did I do wrong?

Thanks for any help.
Jimi
 
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hendrix7 said:
My book says 7.84 m/s/s.
What did I do wrong?

Thanks for any help.
Jimi
Try sloping your roof in the other direction. Looks like that would just about give the quoted answer, but I did not calculate it
 
According to OlderDan's tip you should replace eq [2] by

T + 0.2gsin30^{\circ},

which should give you 4/5g.
 
Yep, you're both right, I see it. Thanks.
 
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