The problem involves a uniform fine chain of length l that is suspended with its lower end just touching a horizontal table. The objective is to find the pressure on the table when a length x of the chain has reached the table. The discussion revolves around the concepts of force, pressure, and momentum in the context of the chain's motion.
The discussion is active, with various interpretations and approaches being explored. Some participants have offered insights into the momentum of the chain and its implications for the forces acting on the table. There is an ongoing examination of the assumptions in the problem, particularly regarding the chain's release and the forces involved.
Some participants note the lack of explicit details in the problem statement, particularly regarding the area of contact and the nature of the chain's release. There is also mention of differing interpretations of the term "pressure" in the context of the problem.
I assume the question says, or is supposed to say, that the chain is released. So there is not only the weight of the chain already on the table but some force, perhaps, from the new links arriving.Suyash Singh said:Homework Statement
a uniform fine chain of length l is suspended with lower end just touching a horizontal table. Find the pressure on the table, when a length x has reached the table..
Homework Equations
Pressure = force/area
The Attempt at a Solution
let mass density, m= mass/l (this gives the mass per unit length)
force= mgx
How do i proceed now?
so how do i do it?haruspex said:I assume the question says, or is supposed to say, that the chain is released. So there is not only the weight of the chain already on the table but some force, perhaps, from the new links arriving.
Consider momentum.
How the unit of pressure can be ##Nm## (Newtonxmeter) ?Suyash Singh said:the answer is 3mgx and no more details are given. From one site i found that m here represents mass density but i am not sure about that.
Sort of.Suyash Singh said:something like mass x velocity?
Yes, that m would be the mass per unit length.Suyash Singh said:found that m here represents mass density
haruspex said:Sort of.
Consider a short period of time Δt.
At time 0, the end of the chain is just touching the table and x=0.
When there is a length x on the table, what is the velocity of the chain?
How much chain hits the table in time Δt?
What is the change in its momentum ?Yes, that m would be the mass per unit length.
I meant the change in momentum of the piece that hits the table.Vriska said:change in momentum is... m*dv/dt+ vdm/dt
haruspex said:I meant the change in momentum of the piece that hits the table.
You can write the mass of that piece in terms of your other variables.
Good, but since you did not originate the thread you probably should have held off a while before posting the solution.Vriska said:Okay momentum of piece is mass *change in velocity. but the mass that'd have fallen in time dt is Lvdt =L(sqrt(2gx))dt but the change in velocity is sqrt(2gx) . dp = 2Lgx dt. 2Lgx, force due to change in momentum is 2Lgx. force on the table due to weight already on it = Lgx, which is 3Lgx. Thanks!