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In the figure m(A) = m(B) = m(C) = 60 kg...

In the figure `m_(A) = m_(B) = m_(C) = 60 kg`. The coefficient of friction between `C` and ground is `0.5 B` and ground is `0.3, A` and `B` is `0.4. C` is pulling the string with the maximum possible force without moving. Then the tension in the string connected to `A` will be

A

`120 N`

B

`60 N`

C

`100 N`

D

zero

Text Solution

Verified by Experts

The correct Answer is:
D

Maximum force of friction between `c` and ground is
`(f_(c))_(max) = (0.5) (60) (10) = 300 N`
Since it is pulling the blocks by the maximum force (without moving). Therefore the applied force is `F = 300 N`

`(f_(AB))_(max) = 0.4 xx 60 xx 120 xx 10 = 240 N`
`(f_(BG))_(max) = 0.3 xx 120 xx 10 = 360 N`
Since `(f_(BG))_(max)` is greater than `300 N`, blocks will not move. Free body diagrames of block are as show below.
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