24
eureka123
·2009-09-05 10:46:05
λ=m/L
=>dm=λ.dx
Taking whole rope as system ,F=ma
Now take an element dx and distance s from where force is applied ..
m' x
=> m'=∫dm=∫λ.dx
0 0
=> m'=λx
Now see the forces on that element
=> F-T=m'a
=> F-T=(λx)(F/m)
=> T=F-mxF/Lm
=> T=F-Fx/L
=> T=F(1- x/L)
I know I made it a bit long...but that was done to elaborate the soln...I hope I am right
edited..missed a force..now its ok
1
Arshad ~Died~
·2009-09-05 10:52:07
eureka i am getting the answer as f-fx/l
24
eureka123
·2009-09-05 10:56:39
Is the rope verticla or horizontal ???
24
eureka123
·2009-09-05 10:57:06
I took it to be horizontal...
I think u guys take it to be vertical
1
Arshad ~Died~
·2009-09-05 11:07:39
eureka u hav neglected the force which is acting on the sytem....
the equation of motion should be
F-T=mxF/Lm
1
Arshad ~Died~
·2009-09-05 11:07:59
and aveek ur solution is wrong....read the question carefully
24
eureka123
·2009-09-05 11:09:10
oh yaa ..misse dit..actually I wasnt using pen or paper even..I was just typing..
1
Bicchuram Aveek
·2009-09-05 12:17:36

Taking the entire rope as one mass, F=ma; a=F/m;
Consider part AB.
F-T = Ma
Mass per unit length = m/L
Hence M=mx/L
T= F-Ma
= F-mxL.Fm
=F-FxL
The rope has been taken in a horizontal position (not a vertical one...I made a mistake in that). Cross-Secn. area has no use.
66
kaymant
·2009-09-05 13:10:05
That's right aveek.. you could have obtained that more simply if you considered the motion of the part of the rope leaving AB. Its mass is
mL(L-x)
And the acceleration F/m must be supplied by the tension. Hence,
T = mL(L-x) Fm = F(1 - xL)