Question

A chain of $5$ links each of mass $0.1kg$ is lifted vertically with a constant acceleration $1.2m{s}^{-2}.$ The force of interaction between the top link and the one immediately below it is

• A. 5.5 N
• B. 4.4 N
• C. 3.04 N
• D. 7.6 N

Answer 1

Answer: (C)

If $F$ is the upward force, $a$ is the net acceleration upward and $m$ is the mass of each link, then
$(5m)a=F-(5m)g\text{ or }F=5m(a+g)$
If $R$ is the force of interaction between the top link and the one immediately below it, then
$ma=F-mg-R$
or $R=F-m(a+g)=4m(a+g)(Using(i))$
Substituting the given values, we get
$R=4\times 0.1(1.2+9.8)=4.4N$