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.
Waste
To solve these problems, we need to apply Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration (F = ma).
(a) The force on link 1 from link 2:
Since link 1 is being pulled upward by link 2, the force on link 1 will be equal to the mass of link 1 multiplied by its acceleration. The mass of link 1 is 0.100 kg, and the acceleration is 2.50 m/s^2. So, the force on link 1 from link 2 is:
F1_2 = m1 * a = (0.100 kg) * (2.50 m/s^2) = 0.250 N.
(b) The force on link 2 from link 3:
Similarly, the force on link 2 from link 3 will also be equal to the mass of link 2 multiplied by its acceleration. The mass of link 2 is also 0.100 kg, and the acceleration is still 2.50 m/s^2. So, the force on link 2 from link 3 is:
F2_3 = m2 * a = (0.100 kg) * (2.50 m/s^2) = 0.250 N.
(c) The force on link 3 from link 4:
Following the same logic, the force on link 3 from link 4 will be:
F3_4 = m3 * a = (0.100 kg) * (2.50 m/s^2) = 0.250 N.
(d) The force on link 4 from link 5:
Once again, the force on link 4 from link 5 will be:
F4_5 = m4 * a = (0.100 kg) * (2.50 m/s^2) = 0.250 N.
(e) The force F on the top link from the person lifting the chain:
Since all the links are connected in a chain, the force applied by the person will be equal to the sum of all the forces between the links. In this case, the top link is being pulled up by four other links, so the force on the top link will be equal to the combined forces from links 1 to 5:
F = F1_2 + F2_3 + F3_4 + F4_5
F = 0.250 N + 0.250 N + 0.250 N + 0.250 N
F = 1.0 N.
(f) The net force accelerating each link:
The net force accelerating each link can be found by subtracting the force on the link above it from the force it exerts on the link below it. Since all the forces between the links are equal, the net force on each link will be zero. Therefore, the net force accelerating each link is zero Newtons (0 N).
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.
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.
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)
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)