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

Explanation

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.