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
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.
Physics - Anonymous, Monday, September 28, 2015 at 5:22pm
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.