A chain of 5 links each of mass 0.1 kg is lifted vertically with a constant acceleration 1.2 `ms^2`. The force of interaction between the top link and the one immediately below it is

Options

A. 5.5 N
B. 4.4 N
C. 3.04 N
D. 7.6 N

Answer

B

Explanation

If F is the upward force, a is the net acceleration upward and m is the mass of each link, then
`(5m)a - F- (5mg or F = 5m(a + g) `
If R is the force of interaction between the top link and the one immediately below it, then
`ma=F-mg-R or R `
or `= F-m(a + g) = 4m(a + g) `(Using (i))
Substituting the given values, we get
` R = 4 xx 0.1(1.2 + 9.8) = 4.4 N`