Question

A chain consisting of $5$ links each of mass $0.1kg$ is lifted vertically up with a constant acceleration of $2.5m/s^2$. The force of interaction between $1st$ and $2nd$ links as shown

• A. $5.20N$
• B. $4.92N$
• C. $9.84N$
• D. $2.46N$

Answer 1

The correct option is A $5.20N$

$5.20N$

For links of mass $0.1kg$ and upward acceleration $a=2.5m/s^2$,

$F=n\times m\times (g+a)=n\times 0.1kg\times (9.8+2.5)m/s^2=n\times 1.23N$

where n = number of links below point of interest.

For instance, the force on link 2 from link 3 has $n=3$, and

$F=3\times 1.23N=3.69N$

You have a second set of data; for that set, the force on link 3 from link 4 is

$F=(9.8+2)m/s^2\times (0.187+0.254)kg=5.20N$