In the arrangement shown the pulley and the strings are ideal. The acceleration of block B is
Asked by m.nilu
19th July 2018,
4:03 PM
Answered by Expert
Answer:
Tension forces acting on the strings are shown in figure. Pulleys P and R are fixed but pulley Q is moveable.
Let us get the constrain relation between P and Q. As shown in figure,
if block-A moves up by distance x, then pulley Q will be down by same distance x.
Hence 2x length of string will be free and block-B will go down by distance 2x as shown in figure.
Due to this above mentioned constarints, if a is the acceleration of block-A, then acceleration of block-B is 2×a
If we consider the forces acting on block-B that is moving with acceleration a, then we can write,
m×g - (T/2) = m×2×a ...........(1)
but from the forces acting on block-A, we have
m×g - T = m×a ...................(2)
by solving eqns.(1) and (2), we get a = g/3 , hence block-B will move with acceleration (2/3)×g
Answered by Expert
8th August 2018,
12:32 PM
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