So the first part of this question asks: A chain consisting of 5 links, each of mass .19kg, is lifted vertically with a constant acceleration of a=2.8 m/s^2. Find the magnitude of the force that link 3 exerts on link 2.(adsbygoogle = window.adsbygoogle || []).push({});

I found this answer to be 4.7 N with the following formula:

F(link 3 exerts on link 2) = ma + mg + F (link 2 exerts on link 1)

I can't seem to get the second part of this question: What is the magnitude of the force F that must be exerted on the top link to achieve this acceleration?

I used the same logic as above, namely:

F = ma + mg + F(link 5 exerts on link 4)

or, can you simply just use the formula F=ma, where m=(5)(.19kg) and a = 2.8m/s^2 ?

Thanks!

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# Homework Help: Chain consisting of five links

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