A uniform chain of length L & weight W is hanging vertically from it's ends A & B which are close together at A given instant the end B is released. Find the tension at A when B has fallen a distance X (X<L)
Answers
A uniform chain of length L & weight W is hanging vertically from it's ends A & B which are close together at A given instant the end B is released. Find the tension at A when B has fallen a distance X (X<L)
Explanation:
We first start by letting the chain fall through distance ′ dn ′ in the time ′ Δt ′
initial momentum of dx/2 part going towards end A of the chain
=( dx/2 )( W/Lg )v
Because the chain has started falling freely
v= √ 2g( dx /2 )
the final momentum as the =0 ( the chain stops waning as it tied to the A end)
the change in momentum will be:
=( dx/2 )( W/Lg )v
This happens in Δt time
force =( W / Lg )( dx/2Δt )v
= Lg W (v 2 )
This in the force due to change in momentum at the end of the chain attached at point A.
force due to weight of
dx/2 = W/L dx/2
Total force due to ( dx/2 ) length = W /L dx/2 + W/Lg v²
= W/L dx/2 + W/ Lg dx 3/2 W / L dx
The weight due to initial hanging will be:
L/2 length of chain = W /L ( L/2 )
Total force =Total weight
W/L ( L/2 )+ 3/2 W /L dx
for ′ x ′ length of fall f total = W/L ( L/2 )+ 3/2 3/2 W/L x
→f total = 2 W (1+ L 3x )