CBSE Class 12-science Answered
11
Asked by Harshhacker2580 | 12 Apr, 2020, 02:55: PM
Expert Answer
Figure shows the given system of 4 pulleys with masses m1, m2 and M . Tension forces acting on strings are shown in figure.
Let us assume m1 > m2 . Hence m1 is moving downward with acceleration a and m2 is moving upward with acceleration a
as shown in free body diagrams of m1 and m2 .
If we apply Newton's second law to the movement of masses m1 and m2 , we get
m1g - T1 = m1a ................(1)
T1 - m2g = m2a .................(2)
By adding eqn.(1) and (2), we get acceleration a = [ ( m1 - m2 ) / ( m1 + m2 ) ] g ...................(3)
Using equation (2) , we get T1 as, T1 = m2 ( a + g ) .................(4)
By substituting, acceleration a from eqn.(3), we get tension force T1 from eqn.(4) as
T1 = [ ( 2 m1 m2 ) / ( m1 + m2 ) ] g .........................(5)
From pully-A , we get, T2 = 2T1
Using eqn.(5), we get, T2 = [ ( 4 m1 m2 ) / ( m1 + m2 ) ] g .........................(5)
From pully-C , we get, T3 = 2T2 = 4T1
Using eqn.(5), we get, T3 = [ ( 8 m1 m2 ) / ( m1 + m2 ) ] g .........................(6)
From free body diagram of block of mass M , we get , T2 + T3 - Mg = M aM .............(7)
Where aM is acceleration of block of mass M
Using eqn.(5) and eqn.(6), we get acceleration of block of mass M from eqn.(7) as
Answered by Thiyagarajan K | 12 Apr, 2020, 08:36: PM
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