Question

A child pulls on a toy locomotive of mass 0.979 kg with a force of 3.25 N at an angle of 42º off the ground. The locomotive is connected to two train cars by cables. Friction in the axles results in an effective coefficient of kinetic friction between the floor and the train which is 0.110. One car has a mass of 0.952 kg and the other has a mass of 0.419 kg. (a) What is the acceleration of the train? (b) What is the tension in the cable between the locomotive and the car connected to the locomotive?

Answer 1

@Michele_Laino

Answer 2

here we can compute the total friction force, which is:
\[\large \begin{gathered}
R = \mu \left\{ {\left( {{m_1} + {m_2} + {m_3}} \right)g - F\sin 42} \right\} = \hfill \\
\hfill \\
= 0.11\left\{ {\left( {0.979 + 0.952 + 0.419} \right) \cdot 9.81 - 3.25\sin 42} \right\} = ...? \hfill \\
\end{gathered} \]

Answer 3

please try to do such computation

Answer 4

@Michele_Laino so I'll just calculate it and get the answer

Answer 5

hint:
the requested acceleration, is given by the subsequent formula:
\[\Large a = \frac{{\left( {F \cdot \cos 42} \right) - R}}{{{m_1} + {m_2} + {m_3}}}\]

Answer 6

a=(F⋅cos42)−R/m 1 +m 2 +m 3
[a=(-3.25*cos42)-2.86/ 0.979+0.952+0.419\]

Answer 7

@Michele_Laino so now I just solve it like that?

Answer 8

I got \(R=2.29\), please check such value. So, we have to compute this:
\[\Large \begin{gathered}
a = \frac{{\left( {F \cdot \cos 42} \right) - R}}{{{m_1} + {m_2} + {m_3}}} = \hfill \\
\hfill \\
= \frac{{\left( {3.25 \cdot \cos 42} \right) - 2.29}}{{0.979 + 0.952 + 0.419}} = ...? \hfill \\
\end{gathered} \]