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The friction coefficient between the tab...

The friction coefficient between the table and the block shown in figure is 0.2. Find the tension in the two strings.

Text Solution

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From the free body diagrams
`T+15a-15g=0`
`rarrT=15g-15a`..i
`Now T_1 -5g-5a=0`
`rarr T_1=5g+5a`
Again `T-(T_1+5a+mR)=0`
`rarr T-(5g+5a+5a+mR)=0` ……..ii
From equaton i and ii
`15g-15a=5g+10a+0.2(5g)`
`rarr `25a=+90` [g=10]
`rarr a=3.6m/s^2`
From equation ii, `T=5xx10+10xx3.6+02x5xx10`
`=96N` inthe left string
From equation iii,
`T=5g+5a`
`=5xx10+5xx36`
=50+18
=68N in theright stirng.
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Knowledge Check

  • An arrangement of the masses and pulleys is shown in the figure. Strings connecting masses A and B with pulley are horizontal and all pulleys and strings are light. Friction coefficient between the surface and the block B is 0.2 and between block A and B is 0.7 . The system is released from rest. (use g=10m//s^(2))

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    The magnitude of acceleration of the system is `2m//s^(2)` and there is no slipping between block `A` and block `B`.
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    The magnitude of friction force between block `A` and block` B` is `42N`
    C
    Acceleration of block `C` is `1m//s^(2)` downwards.
    D
    Tension in the string connecting block `B` and block `D` is `12N`.

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