1
$\begingroup$

Consider a car pulling a trailer with a tight string between the two objects. The car is moving at constant velocity. If we consider the car as a separate system from the trailer, we can state, based on Newton's third law, that the car pulls on the string, and the string pulls on the car with an equal force in the opposite direction. The force equals the tension on the string.

If we consider the string to be massless, the same tension force will pull on the trailer, and the string/trailer will form another force pair in accordance with Newton's third law.

Now let us consider the car and trailer as just one system. The tension force now becomes an internal force. In one of my textbooks, the following statement is then made for this scenario:

"The Tension force $T$, which acts on the trailer, has a reaction force $T'$ which acts (via the massless string) on the car. Both of these forces act between internal parts of the system when we choose the car and the trailer to represent just one system: they are internal forces. All internal forces occur as action-reaction pairs. According to Newton's Third Law, the forces in such an action-reaction pair are equal. As such, the net sum of the internal forces is zero".

I find this statement slightly confusing, in particular the first sentence. I have always assumed that when dealing specifically with contact forces, Newton's Third Law is valid for the two objects which are actually in contact. It therefore makes sense to me that when the forces between car/string and the forces between string/trailer are considered, Newton's Third Law applies. However, from the statement above, it seems to me that now the entire string is completely igonred as a phyhsical object, and it is now claimed that it is between the car and the trailer that Newton's Third Law applies. Even though, technically, these two objects are not directly in contact. Is this because, as we now consider the car and the trailer as just one system, we can simply ignore the massless string as a physical object, and, in a way, almost "pretend" that the car and the trailer are in direct contact?

If someone can clarify this for me, then I will greatly appreciate it!

$\endgroup$
2
  • 1
    $\begingroup$ Your are right, they are not action reaction pairs. It is an abuse of language, only useful because the cord is massless and you then can imagine it as providing a mechanism for long distance interactions. More confusing than helpful. $\endgroup$ Aug 27, 2023 at 4:39
  • 1
    $\begingroup$ @Pato Galmarini: Thank you so much for your answer. Good to hear that my criticism of the textbook language is valid! $\endgroup$
    – user12277
    Aug 27, 2023 at 7:48

1 Answer 1

1
$\begingroup$

Ideal (massless and inextensible) strings are used to "transfer" forces between objects and also to change the direction of forces.

Here is a diagram of a Car pulling a Trailer with the aid of a string which consists of three molecules 1,2, and 3.

enter image description here

Because the string is massless the net external force on it must be zero and so the magnitude of the external force $F_{1\rm T}$ ( force on molecule 1 due to trailer) must be the same as the magnitude of the external force $F_{3\rm C}$ and they must be opposite in direction.
That being said then all the forces shown in the diagram are often called "tension" and the string just having the external forces labelled $T$.

The statement in the book is confusing but is an attempt to illustrate the passive nature of an ideal string being used to connect two objects and that notion agrees with what you have written, it seems to me that now the entire string is completely ignored as a physical object.

So rather than go through the full analysis of the system, including the string, the author "pretends" that the car and the trailer are in direct contact in the interests of brevity.

$\endgroup$
1
  • $\begingroup$ Thanks a lot! This was very explanatory and helpful! $\endgroup$
    – user12277
    Sep 1, 2023 at 18:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.