A system comprising blocks, a light frictionless pulley, a frictionless incline,

Question

A system comprising blocks, a light frictionless pulley, a frictionless incline, and connecting ropes is shown in the figure above. The 9-kg block accelerates downward when the system is released from rest. Find the acceleration of the system. *Please use the axis I used so I can compare your solutions to mine. I need to understand what I did wrong..Thank you! :) X-axis: towards the pulley. on the incline plane.
Y-axis: upwards
T1:Between 6 kg block and 4 kg block T2:Between 4 kg block and pulley = T2:Between pulley and 9 kg block
(It would be great if you can explain why these two tensions are equal..is it because they are connected to the same string, and the "tautness" is same throughout the string?)
I got...
4 kg block (4*-9.8=-58.8)
ma= -58.8sin30+T1= -6a
6 kg block (6*-9.8=-39.2)
ma= -T1-39.2sin30+T2= -4a
9 kg block (9*-9.8=88.2)
ma= T2-88.2 = -9a Where do I go from here..? Thanks

Explanation / Answer

I think this will help you Given Mass of the blocks on the iinclined plane are m1 = 6 kg, m2 = 4 kg And the mass hanging is m3 = 9 kg Angle of inclination, ? = 30^0 Net force on m1 is m1 a = T1 - m1g sin30 T1 = 6 kg( a + 9.8 m/s^2 *0.5) T1 = 6 kg ( a + 4.9 m/s^2)-------(1) Net force on m2 is M a = T2 - Mg sin30   [where M = m1 +m2 = 10 kg] T2 = M( a + gsin30)     = 10 kg ( a + 9.8 m/s^2 *0.5) T2 = 10 kg (a +4.9 )----------(2) Net force on m3 is m3 a = m3 g - T2 T2 = 9 kg (g -a)------(3) Equating eq 1 and 3 we get 10 kg (a +4.9 ) = 9 kg (g -a) 10 (a +4.9 ) = 9 (9.8 m/s^2 - a) on solving we get a = 2.1 m/s^2 I think this will help you Given Mass of the blocks on the iinclined plane are m1 = 6 kg, m2 = 4 kg And the mass hanging is m3 = 9 kg Angle of inclination, ? = 30^0 Net force on m1 is m1 a = T1 - m1g sin30 T1 = 6 kg( a + 9.8 m/s^2 *0.5) T1 = 6 kg ( a + 4.9 m/s^2)-------(1) Net force on m2 is M a = T2 - Mg sin30   [where M = m1 +m2 = 10 kg] T2 = M( a + gsin30)     = 10 kg ( a + 9.8 m/s^2 *0.5) T2 = 10 kg (a +4.9 )----------(2) Net force on m3 is m3 a = m3 g - T2 T2 = 9 kg (g -a)------(3) Equating eq 1 and 3 we get 10 kg (a +4.9 ) = 9 kg (g -a) 10 (a +4.9 ) = 9 (9.8 m/s^2 - a) on solving we get a = 2.1 m/s^2

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