Two blocks connected by an in-extensible string passing over a light frictionless pulley are resting on two smooth inclined planes as shown in diagram. Determine the acceleration of the blocks & tension in the string. Assume string to be massless .

Acceleration = a =  g (m1sin alpha - m2sin beta) / (m1+m2)

a = 10 ( 55 x sin30 - 35 x sin60 ) / ( 55+35 )

a = 10 ( 55/2 - 35root3/2 ) / (90 )

a = 10 ( 55/2 - 60.62/2 ) / (90)

a = 10 ( -5.62/2 ) / (90)

a = -2.81/9 = - 0.31 m/s.

Hope it helps ...

 
  • 4
I did not get the logic.. Can you please elastrate a bit more..
Thanks
  • -3
esolve ok. and then do the r-mg = ma thing for the one moving downward (the heavier one obviously) with the resolved components. what i got is a = 1.38 m/s^2
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Real ans is in this pic hope it helps you

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how 90a became 9a 
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What are you looking for?