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 intersection between the top link and the one immediately below it is

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

Answer 1

Answer: (B)

$4.4N$

Here,

$F=5mg+5ma$

$=5m(g+a)$

If the force of interaction between top link and the second is $T$. Then,

$ma=F-mg-T$

$T=F-mg-ma$

$=5mg+5ma-mg-ma$

$4mg+4ma=4m(g+a)$

Therefore,

$T=4\times 0.1(9.8+1.2)=4.4N$