Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
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
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !