- #1

Kanda ryu

- 14

- 1

## Homework Statement

A uniform chain of mass 'm' and length 'l' rests on a rough incline (inclination is angle 'Q') with its part hanging vertically. The chain (inclined) starts moving up the incline (and the vertical part moving down) provided the hanging (vertical) part equals to 'n' times the total chain length. Find the work done by friction as it completely slides off the incline.

## Homework Equations

Newton's 2nd law

## The Attempt at a Solution

I assumed friction coefficient to be 'u'.

Frictional force would be N u.

I took the inclined part a system to draw the free body diagram.

N = m'gCosQ where m' is the mass of chain on the incline.

Work (fric) = N u s (displacement) x Cos 180° = - m'gCosQ x s

As mass m' will decrease as chain slides up the incline. I considered a small part of chain 'x' of mass dm. As chain is uniform density,

m/l = dm/x => dm = m x /l

For a small work dW

= -dm gCosQ dx

= - (m/l) gCosQ xdx

Let the length of vertically hanging part be 'a'

a/l = n => a=nl

So length of inclined part initially is l-nl

So I integrated xdx putting limits (l-ln) to 0 and got W= mlgCosQ u / 2 but the problem is I'm unable to find a way to write 'u' in terms of given variables.

I was hoping for an idea to find 'u' or any other better method to solve this (within the scope of basic calculus and mechanics).