Question

A chain consisting of five links each with mass 100gm is lifted vertically with a constant acceleration of 2m/sec sq. as shown find the forces acting between adjacent links

Answer 1

Answer: Force exerted by lifting the chain on the top is 4.92N .

Given mass = 0.100Kg and with a acceleration of $2m/s^2$

Let as consider the force on each link is same. Hence the net force,

$Force_{net} = ma$

(0.100Kg) x (2m/s2) = 0.250Kg.m/s2 (where 1N = 1Kg.m/s2)

=0.250N

We know that W =mg

$W\quad =\quad 100Kg\times 9.91m/s^2$

By simplifying we get, W= 0.981N

Consider in each link, there is a weight W and upward force U and downward force D,

So, $F_{net} = U-D-W$

Therefore, $U=F_{net} + W + D$

=(0.250 N) + (0.981 N) + D

=1.23 N  

For bottom link D=0;

U=1.23 N + 0

=1.23 N  

For above link

U= 1.23 N + 1.23 N

U= 2.46 N

For next link above,

U = 1.23 N + 2.46 N = 3.69 N

For next link above

U = 1.23 N + 3.68 N = 4.92 N

Hence the force exerted by lifting the chain on the top is 4.92 N.