A uniform flexible rope of 60m is partly lying on a rough table and is partly just hanging across the edge of the table. If the coefficient of friction between the table and rope is 0.2,then what's the length of rope which can just hang?

Let mass of the rope =m

total length of the rope=l(=60 mtr. in this case)

length of rope(which can just hang)=x

then only friction between remaining length of rope on table will balance the hanging part's weight. 

equivalently, frictional force on rope=(coeff. of friction)*(normal reaction from table on rope)

normal reaction from table on rope=weight exerted by  rope on table=(m/l)*(l-x)*g

m/l=mass per unit length of rope(rope is uniform so this is a constant)

l-x=remaining length of rope on table

g=acceleration due to gravity

so,frictional force on rope=(0.2)*(m/l)*(l-x)*g

above is equal to weight of rope hanging=(m/l)*(x)*g

or,(0.2)*(m/l)*(l-x)*g=(m/l)*(x)*g

or,(0.2)*(l-x)=x

or,0.2(l)=1.2(x)

or,x=l/6=10 mtr.

the question use straight forward FBD(free body diagram) here.

friction=gravitational force

so 0.2 (dM)g=(dM)g

     0.2 M/60 dx=M/60 dx

integrating L.H.S from 0 to x and R.H.S from 0 to 60-x

we get,

0.2(x)=60-x

x=60\1+0.2

x=50

hence answer is 50 m