Physics

In the figure, a chain consisting of five links, each of mass 0.100 kg, is lifted vertically with constant acceleration of magnitude a = 2.50 m/s2. Find the magnitudes of (a) the force on link 1 from link 2, (b) the force on link 2 from link 3, (c) the force on link 3 from link 4, and (d) the force on link 4 from link 5. Then find the magnitudes of (e) the force F on the top link from the person lifting the chain and (f) the net force accelerating each link.

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  1. LOL, do it backwards

    F = net force on each link
    so
    force on bottom link = F = m a
    F = .1 * 2.5 = .25 N on link 5
    that is the answer to part f

    NOW, net one up, link 4, also has net force of .25 N so force up -.25 = m a = .25
    so .25 more = .50N on link 4

    In fact the net force on EVERY link is .25N so each force up goes up by .25

    link 3 -- .5 + .25 = .75
    link 2 -- .75+.25 = 1.00
    link 1 -- 1 + .25 = 1.25 but that is the total (answer to (e)

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  2. thanks for the help but i have worked it out but for the forces on the links it is F=m(a+g)+ F(previous link)

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  3. Damon that would work if there was no friction, no gravity, no transfer of energy/work throughout the particle.

    Because they are linked, and hanging vertically, there is a few extra forces that need to be taken into account.

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  4. LOL, Damon. The only thing you did backwards was yourself! Mad kudos to Oliver and Sarah. And Damon, please refrain from being so mean... People are here to get help.

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  5. Waste

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