Posted by **Sarah** on Saturday, March 19, 2011 at 8:54pm.

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.

- Physics -
**Damon**, Saturday, March 19, 2011 at 9:11pm
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)

- Physics -
**Sarah**, Sunday, March 20, 2011 at 3:05am
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)

- Physics -
**Oliver**, Friday, March 25, 2011 at 5:10am
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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