Question

A chain consisting of five links each with mass $100gm$ is lifted vertically with constant acceleration of $2.5m/s^2(\uparrow )$ as shown. Find:
$(g=10m/s^2)$
(a) the force acting between adjacent links
(b) the force $F$ exerted on the top like by the agent lifting the chain
(c) the net force on each link.

• A. $8.6N,13.8N,0.9N$
• B. $80N,48N,12N$
• C. $zero,6.15N,0.25N$
• D. $60N,42N,25N$

Answer 1

The correct option is C $zero,6.15N,0.25N$

$m=100gm$ of each link

(b) Given that whole chain is going up with an acceleration of $2.5m/s^2$

$\left(5m\right)a=F-\left(5m\right)g$

$(5\times 0.1)\times 2.5=F-(5\times 0.1)\times 9.8$

$1.25=F-4.9$

$F=6.15N$

(a) Force acting between adjacent links is zero because of newton's third law (action reaction).

means force by upper will equal in magnitude opposite in direction of force by down link.

(c) Each link is moving upward by an acceleration of $2.5m{s}^2$

So $F_{net}=ma$

$F_{net}=\left(0.1\right)\times 2.5$

$F_{net}=0.25N$