2
$\begingroup$

In free space an end of a rope is attached to the left side of a block, the other end is freely floating. An astronaut (on the other side of the block where the rope isn't) pulled the block by 5N force right and the block moves 5N right but the block also exerts 5N right on rope and the rope exerts 5N left on block. The total forces on the block are 5N right by astronaut and 5N left by rope. But we know that the block will move. If the net force on the block is zero, why does the block move then?

I KNOW that the action and reaction forces act on different objects but nothing feels wrong here. Let me write the action and reaction forces:

Action: Astronaut pulls the block right by 5N; Reaction: The block pulls the astronaut 5N left.

Action: The block pulls the rope by 5N force right as it is moving by 5N force right and the rope is attached to it. Reaction: The rope pulls the block by 5N force left.

If we make an FBD of the block, it is pulled 5N right by the astronaut and the rope is pulling the block 5N left. Where did I go wrong?

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ Does this answer your question? Given Newton's third law, why are things capable of moving? $\endgroup$
    – John Rennie
    Commented Jun 6 at 5:36
  • $\begingroup$ Can you clarify the experiment? Am I correct that the rope is attached to the left side of the block, but the astronaut pulls on the right side of the block, not via the rope? @JohnRennie, I'm not sure this is a duplicate. $\endgroup$
    – benrg
    Commented Jun 6 at 6:04
  • $\begingroup$ Hi benrg! Yes, you are thinking right(no pun intended), the rope is attached to the left side of the block and the astronaut pulls the block by 5N right not via the rope. $\endgroup$
    – Khushiv Batra
    Commented Jun 6 at 6:13
  • 1
    $\begingroup$ Why would you expect the free-hanging rope to pull with 5 N in the block as well? $\endgroup$
    – Steeven
    Commented Jun 6 at 6:44

5 Answers 5

Reset to default
4
$\begingroup$

The block is not pulling in the free-hanging rope with $5\,\mathrm N$. It is pulling with much less force.

That the rope "follows the motion" of the block as the block is pushed, does not mean that the rope feels the same force. "Following the motion" of another object in this context only requires that the accelerations are the same. And according to Newton's 2nd law, much less force is needed to accelerate the much lighter rope.

Improve this answer
$\endgroup$
2
  • $\begingroup$ For clarification, does it even matter that the object being pulled here consists of a "block part" and a "rope part"? We can always subdivide macroscopic objects into parts anyway, e.g. the astronaut pulls on the right half of the block, which in turn pulls on the left half of the block...? $\endgroup$
    – Marius Ladegård Meyer
    Commented Jun 6 at 6:57
  • 1
    $\begingroup$ @MariusLadegårdMeyer Sure, this is called a split system, where you are subdividing a system into smaller bits. You can do this down to each individual particle and consider how the force propagates through. You will find that the force on each particle reduces along the way. $\endgroup$
    – Steeven
    Commented Jun 6 at 7:20
2
$\begingroup$

Your mistake is thinking that the block must pull on the rope with a force of 5 N. The actual constraint is that the acceleration of the block and rope must be the same, so the net forces must be in the same ratio as the masses. Letting $F=5\text{ N}$, $f>0$ the force of the block and rope on each other, and $M$ and $m$ the masses, the constraint is $f/m=(F−f)/M$, from which you can find $f/m=F/(M+m)$, i.e. the system accelerates like a single body of mass $M+m$.

Improve this answer
$\endgroup$
2
$\begingroup$

The problem in your approach is that you are not clearly defining the systems on which you are applying the forces. Take the rope only as a system. On it, the force will be the tension force (and you cannot just assume that the tension is 5N.) 5N is the force exerted by man on the entire rope+block system, not by rope on the block

So, the rope will apply equal oppsite force on the block,(or block+man system).

Now, take only the block as a system. It has 5N force by man and tension force T by rope. Net force is 5-T towards man.

You can also take block+rope as a system. On it, there is no force other than the force of man. So, you can calculate the acceleration of entire block+rope system.

Basically, it is just like the horse-cart question. We tend to think that the horse will pull the cart with some force and the cart also pulls the horse back with the same force, therefore the net force on the "system" is 0. The "system" that we were observing was horse, but then we changed it midway to horse+cart. The ground pushes the horse+cart system ahead(by friction). Similarly, the man(analogous to ground) pulls the block+rope system forward

Read this page for better understanding.

enter image description here

Source: Concepts of Physics by HC Verma.

Improve this answer
$\endgroup$
1
$\begingroup$

Consider three separate systems, the rope, and the astronaut with the free body diagrams as shown below.

enter image description here

The net force on the block is $F-f$ to the right.

Improve this answer
$\endgroup$
0
$\begingroup$

Well if I understand the your setup correctly, here's what I've got to say.

Let's think of just the rope and the block for a moment. The block is pulling pulling the rope with a force of 5N according to you, for the moment I'll assume the this 5N force to be X (I'll further disprove that X in fact is not equal to 5N). Since the block has produced an action there must be an equal and opposite reaction (Newton's 3rd Law). So the rope must pull the block with a force X as well (towards left). Hence, there mustn't be any kind of motion between the rope and the block. In fact, there isn't. If you think about it, with respect to the rope the block is stationary and with respect to the block the rope is stationary.

But here you are forgetting something. The block is being pulled by a force of 5N towards right as well, we can say that the rope isn't aware of this. This means that even if the block was accelerating the rope wouldn't know since with respect to it the block only has the force X acting on it (towards left).
In the ground frame when you add up all the forces you must consider the equal and opposite forces X to be two different forces, not just a single one (which is where you went wrong) acting in the opposite direction to the main 5N force as they are an indirect result of it.

So as the block moves with respect to you and the rope must remain stationary (with respect to the block) the block pulls it as if it were a part of it, and hence, when you write the motion equation you must consider the block and the rope to be a single object on which you are imparting the force of 5N.
Let,
Mass of block be M.
Mass of rope be m.

Therefore the equation of motion will be,

5N = (M+m)a [where 'a' is the acceleration of the block and the rope]

Now since, you know the acceleration of the block and the rope you can work out the force X. I'll leave that up to you.
It'll shall be equal to 5m/(M+m)

Improve this answer
$\endgroup$

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.