Question

A chain of $10$ links each of mass $0.1kg$ is lifted vertically with a constant acceleration of $2m{s}^2$.The force of interaction between the top link and the one immediately below it is $(g=10m{s}^{-2})$:

• A. $0.2N$
• B. $2N$
• C. $10.8N$
• D. $20N$

Answer 1

The correct option is C $10.8N$

Mass of each link $m=0.1kg$

Mass of the system included in dotted box $M^{\prime }=9m=0.9kg$

Let the force of interaction between 1st link and 2nd link be $T$.

Writing equation of motion for the system ;

$9ma=T-9mg$

$\therefore$ $T=9m(a+g)$

Or $T=0.9(2+10)=10.8$ N