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  4. A chain consisting of 5 links of mass 0.1 kg each is lifted vertically upwards with a constant acceleration 5 m / s 2 as shown in figure. The force of interaction between the top link and the link immediately below it will be: ( g =10 m / s 2)

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Q. A chain consisting of $5$ links of mass $0.1\, kg$ each is lifted vertically upwards with a constant acceleration $5 \,m / s ^{2}$ as shown in figure. The force of interaction between the top link and the link immediately below it will be : $\left( g =10 \,m / s ^{2}\right)$Physics Question Image

Laws of Motion

Solution:

$F = M ( g + a )=5 \times 0.1(10+5)=7.5 \,N$
for top link $F-F_{1}=m(g+a)$
$\Rightarrow F _{1}=7.5-0.1(15)=6 \,N$
FBD of top link
image