In the figure below, a chain consisting of five links, each of mass 0.105 kg, is
ID: 1965272 • Letter: I
Question
In the figure below, a chain consisting of five links, each of mass 0.105 kg, is lifted vertically with a constant acceleration of magnitude a = 2.59 m/s2.figure: http://www.webassign.net/hrw/hrw7_5-43.gif
(a) Find the magnitude of the force acting between link 1 and link 2.
___N
(b) Find the magnitude of the force acting between link 2 and link 3.
___N
(c) Find the magnitude of the force acting between link 3 and link 4.
___N
(d) Find the magnitude of the force acting between link 4 and link 5.
___N
(e) Find the magnitude of the force F exerted on the top link by the person lifting the chain.
___N
(f) Find the magnitude of the net force accelerating each link.
___N
Explanation / Answer
According to newton's 2nd law, F = ma ------------------ (1) a) F2-1= m(a +g)= (0.105 kg)(2.59m/s2 + 9.8 m/s2)
=1.300 N
b)
F3-2= m(a + g) + F2-1 =(0.105 kg)(2.59 m/s2 + 9.8 m/s2) + 1.300N =2.60 N c) F4-3 =m(a + g) + F3-2 =(0.105 kg)(2.59 m/s2 + 9.8 m/s2) + 2.60 N =3.90 N =3.90 N d) F5-4 =m(a + g) + F4-3 =(0.105 kg)(2.59 m/s2 + 9.8 m/s2) + 3.90 N = 5.20N e) F - F4-5 - mg =ma F = ma + mg +F4-5 = m(a + g) + F4-5 = (0.105 kg)(2.59m/s2 + 9.8m/s2) + 5.20N
= 6.50 N
f)
Fnet= mg =(0.105 kg)(9.8m/s2) =(0.105 kg)(9.8m/s2) =1.029N