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.

Answer 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)

Answer 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)

Answer 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.

Answer 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.

Answer 5

Waste