Three same-mass freight cars, why force on each is not same ?

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The discussion revolves around the forces acting on three freight cars of equal mass being pulled by a locomotive. It clarifies that the question is not about the net force on each car but rather the specific forces exerted between them. The correct forces are identified as F_A = F/3 for car A, F_B = 2F/3 for car B, and F_C = F for car C, reflecting the tension forces between the cars. There is frustration expressed regarding the ambiguity of the question, which could lead to confusion about what is being asked. Overall, the key takeaway is the distinction between net forces and the individual forces exerted between the cars.
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QUESTION
three cars of mass m are pulled with force F by a locomotive. Find forces on each cars ?

assuming the force F is exerted on car C and nearest to C is car B and remaining one is A.
\vec{F}_{A} is total sum of all (interbody-) forces on car A . similarly \vec{F}_{B} and \vec{F}_{C} are defined.

assuming the forces that the question asks is \vec{F}_{A},\vec{F}_{B} and \vec{F}_{C} and the given information is
$$\vec{F}_{CO}=\vec{F}$$

according to Newton's 2nd law
$$\vec{F}_{A}=\frac{d^{2}}{dt^{2}}m\vec{r}_{A}$$,
$$\vec{F}_{B}=\frac{d^{2}}{dt^{2}}m\vec{r}_{B}$$ and
$$\vec{F}_{C}=\frac{d^{2}}{dt^{2}}m\vec{r}_{C}$$

but since
$$\frac{d^{2}}{dt^{2}}\vec{r}_{A}=\frac{d^{2}}{dt^{2}}\vec{r}_{B}=\frac{d^{2}}{dt^{2}}\vec{r}_{C}$$, $$\frac{d}{dt}m =0$$

so above five equations would give
$$\vec{F}_{A}=\vec{F}_{B}=\vec{F}_{C}$$

according to superposition principle,
$$\vec{F}_{A}=\vec{F}_{AB}$$
because there is only one force i.e tension force due to string, exerted on A.
$$\vec{F}_{B}=\vec{F}_{BA}+\vec{F}_{BC}$$
because two tension forces (from both A and C) is acting on B.
$$\vec{F}_{C}=\vec{F}_{CO}+\vec{F}_{CB}$$
because one external force of magnitude F and one tension force from B.

according to 3rd law we also have,
$$\vec{F}_{AB}=-\vec{F}_{BA}$$ and
$$\vec{F}_{BC}=-\vec{F}_{CB}$$.

so from above five equations,
$$\vec{F}_{A}+\vec{F}_{B}+\vec{F}_{C}=\vec{F}_{CO}$$
hence
$$\vec{F}_{A}=\vec{F}_{B}=\vec{F}_{C}=\frac{\vec{F}}{3}$$
but i don't undertand why its wrong because the solution says
$$\vec{F}_{A}=\frac{\vec{F}}{3}$$,
$$\vec{F}_{B}=\frac{2\vec{F}}{3}$$ and
$$\vec{F}_{C}=\vec{F}$$

the only problem think i can think of is may be the question is asking for different forces , because there are forces that have same value for example ,
$$\vec{F}_{A}=\frac{\vec{F}}{3}$$,
$$\vec{F}_{BC}=\frac{\vec{2F}}{3}$$,
$$\vec{F}_{C0}=\vec{F}$$


thank you
 
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The question is not asking for the net force on each car, which of course must be equal. It is asking for the force that C exerts on B and B exerts on A in terms of the force F that the locomotive exerts on C.
 
so the question was indeed asking for $$\vec{F}_{CO}$$,$$\vec{F}_{BC}$$ and $$\vec{F}_A$$ .

i don't understand why not say so in the question, instead of being so short and confusing. i though the book was teaching physics not reading mind.

thanks for the help Doc Al
 
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