Your old toy train from pre-school days consists of three cars: M 1 = 0.2 kg, M
ID: 1432863 • Letter: Y
Question
Your old toy train from pre-school days consists of three cars: M1 = 0.2 kg, M2 = 0.4 kg, and M3 = 0.1 kg. You pull the cars with a force F = 4.5 N (red arrow in the picture). The coefficient of kinetic (rolling) friction between all three cars and the ground is ?k = 0.13.
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.
M3 M2Explanation / Answer
The data given in the question is,
mases M1 = 0.2 kg, M2 = 0.4 kg, and M3 = 0.1 kg
Total mass = 0.7kg.
force F = 4.5 N,The coefficient of kinetic (rolling) friction between all three cars and the ground is k = 0.13
What is the difference in the tensions between the two ropes, FTA - FTB=?
the friction force of all free cars is (m1 + m2 + m3)*g*0.1 = 0.7*9.81*0.13 = 0.8927 N
the friction force of cars 2+3 = (m2+m3)*g*0.13 = 0.6376 N
the friction force of car 3 is m3g*0.13 = 0.1275 N.
The net pulling force is F - Ffric = 4.5 - 0.8927 = 3.607N
The acceleration of the cars is Fnet/m = 3.607/0.7 = 5.15 m/s^2
The force necessary to accelerate car 3 at 5.15 m/s2is
F(3) = m3*a + friction force of m3
F3 = 0.1*5.15 + 0.1275 = 0.642 N = Tension in rope from car 2 to car 3
the force necessary to accelerate cars 2+3 at 5.13 m/s2 is
F(2,3) = 0.5*5.15 +0.6376= 3.212 N = Tension in rope from car 1 to car 2
FTA - FTB =3.212-0.642=2.570N
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