1. Tardigrade
  2. Question
  3. Physics
  4. A chain of 5 links each of mass 0.1 kg is lifted vertically with a constant acceleration 1.2 m s -2 . The force of interaction between the top link and the one immediately below it is

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

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

Laws of Motion

Solution:

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