In figure below, a chain consisting of five links, each of mass 0.101 kg, it lif
ID: 1532907 • Letter: I
Question
In figure below, a chain consisting of five links, each of mass 0.101 kg, it lifted vertically with a constant acceleration of magnitude theta = 2.47 m/s^2 Find the magnitude of the force acting between link 1 and link 2 Find the magnitude of the force acting between link 2 and link 3 Find the magnitude of the force acting between link 3 and link 4 Find the magnitude of the force acting between link 4 and link 5 Find the magnitude of the force F exerted on the top link by the person lifting the chain. Find the magnitude of the net force accelerating each link.Explanation / Answer
Give that; mass m = 0.101 kg, acceleration a = 2.47 m/s^2
Force F = m*a + m*g
(a) Force on link1 from link2 = 0.101*2.47 + 0.101*9.81
= 1.24028 N
(b) Force on link2 from link3 = 1.24028 + 0.101*9.8 + 0.101*2.47
= 2.47955 N
(c) Force on link3 from link4 = 2.47955 + 0.101*9.8 + 0.101*2.47
= 3.71882 N
(d) Force on link4 from link5 = 3.71882 + 0.101*9.8 + 0.101*2.47
= 4.95809 N
(e) Force on link5 from the person = 4.95809 + 0.101*9.8 + 0.101*2.47
=6.19736 N
(f) Net force accelerating each link = 0.101*2.47
= 0.24947 N