Your old toy train from pre-school days consists of three cars: M 1 = 0.5 kg, M
ID: 1433199 • Letter: Y
Question
Your old toy train from pre-school days consists of three cars: M1 = 0.5 kg, M2 = 0.1 kg, and M3 = 0.4 kg. You pull the cars with a force F = 3.0 N (red arrow in the picture). The coefficient of kinetic (rolling) friction between all three cars and the ground is k = 0.11.
What is the difference in the tensions between the two ropes, FTA - FTB?
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.
Explanation / Answer
you pull at car 1, right, in the middle is car 2, and at the end car 3.
the friction force of all free cars is (m1 + m2 + m3)*g*0.1 = 1*9.81*0.1 = 0.981 N
the friction force of cars 2+3 = (m2+m3)*g*0.1 = 0.4905 N
the friction force of car 3 is m3g*0.1 = 0.3924 N.
The net pulling force is F - Ffric = 3.0 - 0.981 = 2.019 N
The acceleration of the cars is Fnet/m = 2.019/1 = 2.019 m/s^2
The force necessary to accelerate car 3 at 2.019m/s^2 is
F(3) = m3*a + friction force of m3
F3 = 0.1*2.019 + 0.3924 = 0.594 N = Tension in rope from car 2 to car 3
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the force necessary to accelerate cars 2+3 at 2.019 m/s^2 is
F(2,3) = 0.5*2.019 + 0.4905 = 1.5 N = Tension in rope from car 1 to car 2
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