A uniform chain of length L hangs partly from | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the total length of the chain be $L$ and the length hanging from the table be $\ell$. The length of the chain lying on the table is $(L - \ell

Vedclass · Accepted Answer

$\frac{\ell}{L-\ell}$

Vedclass · Answer

(C) Let the total length of the chain be $L$ and the length hanging from the table be $\ell$. The length of the chain lying on the table is $(L - \ell)$.For the chain to be in equilibrium,the force of friction must balance the weight of the hanging part.The weight of the hanging part is $W = \lambda \ell g$,where $\lambda$ is the linear mass density of the chain.The normal force exerted by the table on the chain is $N = \lambda (L - \ell) g$.The maximum static friction force is $f_{max} = \mu N = \mu \lambda (L - \ell) g$.Equating the forces for the limiting case: $\lambda \ell g = \mu \lambda (L - \ell) g$.Solving for $\mu$: $\mu = \frac{\ell}{L - \ell}$.

$A$ uniform chain of length $L$ hangs partly from a table and is kept in equilibrium by friction. If the maximum length that can hang without slipping is $\ell$,then the coefficient of friction between the table and the chain is:

Similar Questions

An object takes $n$ times as much time to slide down a $45^{\circ}$ rough inclined plane as it takes to slide down a perfectly smooth inclined plane of the same inclination. The coefficient of kinetic friction between the object and the rough incline is given by:

$A$ block of weight $1 \,N$ rests on an inclined plane of inclination $\theta$ with the horizontal. The coefficient of friction is $\mu$. The minimum force that has to be applied parallel to the inclined plane to make the body just move up the plane is

$A$ given object takes $n$ times the time to slide down a $45^{\circ}$ rough inclined plane as it takes to slide down an identical perfectly smooth $45^{\circ}$ inclined plane. The coefficient of kinetic friction between the object and the surface of the inclined plane is:

$A$ body is moving up an inclined plane of angle $\theta$ with an initial kinetic energy $E$. The coefficient of friction between the plane and the body is $\mu$. The work done against friction before the body comes to rest is

$A$ block slides down on an inclined surface of inclination $30^{\circ}$ with the horizontal. Starting from rest,it covers $8 \ m$ in the first two seconds. Find the coefficient of friction. (Take $g = 10 \ m/s^2$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module