Question

Your old toy train from pre-school days consists of three cars: M_1 = 0.4 kg. M_2 = 0.4 kg, and M_3 = 0.1 kg. You pull the cars with a force F = 5.0 N (red arrow in the picture). The coefficient of kinetic (rolling) friction between all three cars and the ground is mu_k = 0.14. What is the difference in the tensions between the two ropes, F_TA - F_Tb? Give your answer in Newtons to at least three significant digits to avoid rounding errors. Your answer will not be graded on the number of digits you provide.

Answer 1

as all the cars are attached so the they have comman acceleration ( lets say a)

as Fnet = ma

5 - fr = (m1+m2+m3) a      

{ fr = friction force = mu * mg = 0.14 * ( 0.4+0.4+0.1) *9.81 = 1.23 N}

5 - 1.23 = ( 0.4+0.4+0.1) a

a = 4.18 m/s2

1) now take m1 only

F - F_TA - Fr1 = m1 a

{ Fr1 = mu * 0.4 *9.81 =0.55 N}

5 - F_TA- 0.55 = 0.4 * 4.18

F_TA = 2.78 N                 ..................(1)

2) now on m3

F_TB - Fr = m3 *a

{ Fr = mu m3 g = 0.14 * 0.1 *9.81 = 0.137 N}

so F_TB -0.137 = 0.1 * 4.18

F_TB = 0.55 N

so    F_TA - F_TB = 2.78 -0.55

= 2.22 N answer