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In the figure, m(1)=m(2)=10 kg. The coef...

In the figure, `m_(1)=m_(2)=10 kg`. The coefficients of friction between `A, B` and `B`, surface are `0.2`. Find the maximum value of `m_(3)` so that no block slips ( All the pullies are ideal and strings are massless).

A

`16 kg`

B

`10 kg`

C

`18 kg`

D

`14 kg`

Text Solution

Verified by Experts

The correct Answer is:
D

`(f_(2))_("lim")=*(m_(1)+m_(2))g=0.2xx20xx10=40N`
`T^(1)=m_(3)g` and `T^(1)=f_(1)+f_(2)+4T`
`rArrm_(3)(10)=20+40+4(20)rArrm_(3)=14 kg`
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