62consider the gravitational and tensile forces acting on an infinitesimal segment of the string.
This segment, of length dL, has weight (mass per unit length)*(gravitational acceleration)*dL -
take mass per unit length as a constant and define
q = (mass per unit length)*(g). (so that we dont have to write g all the time :)
Over the length dL, the tension on the string changes from T to T+dT.
Angle the string makes with the horizontal goes from A to A+dA.
Balance forces
In the horizontal direction only tensions on each end. Tension, by definition, is a force pointing in the same direction as the string does.
Fhoriz = 0 = -T cos(A) + (T+dT) cos(A+dA)
cos(A+dA) = cos(A) cos(dA) - sin(A) sin(dA) = cos(A) - dA sin(A)
0 = -T cos(A) + T cos(A) + dT cos(A) - T dA sin(A) - dT dA sin(A)
dT cos(A) = T dA sin(A)
I hope this gives a good start?
33Opse when i got that cos(θ+dθ) i left it.. now will try....
But how come the answer has cosh()
33Book me bhi direct mara hai "without going in detail of solving the given differential eqn....... the fn comes out to be....."
But let me reaach upto that point where my pen will stop. I'll try first.
9is final ans
Y + 1 = (eX + e-X ) /2 with bottom most point as origin ?
Y=y/c
X=x/c
9proof :
d(Tcosθ)=0
ie cosθdT=Tsinθdθ ..........1
d(Tsinθ) = kdx/cosθ
ie sinθdT + Tcosθdθ=kdx/cosθ .............2
frm 1 and 2
dT=kdY
ie
T=kY+C
now
dT=kdY
Ttanθdθ=kdY
Tdθ=kdX
(kY+C)dθ =kdX ....3
tanθ=dy/dx
sec2θdθ /dx=d2y/dx2 .......4
frm 3 and 4
sec2θ/(y+c)=d2y/dx2
put dy/dx = p
(1+p2)/(y+c) =pdP/dy
integrating and solving
p = dy/dx = √(y/c+1)2 - 1
integrate and solve
ex/c = (y/c+1) + √(y/c+1)2 - 1
observe
e-x/c = (y/c+1) - √(y/c+1)2 - 1
so
y/c + 1 = (ex/c+e-x/c)/2
