0
$\begingroup$

I am a high school student and I am a little confused about how the pulley system works here

Here in this example below, we always only shown tension on points A and B as acting upwards and then we say that the same force is being applied on the block "m" but as it's assumed that the rope is massless, the tension on them is equal in downward direction as well, it's not like the internal forces will cancel out like hey do if the rope was straight and forces were being applied as shown. So, what's going on? Can someone please explain about how the forces are acting on the rope by making a free body diagram. (assume rope is massless, pulleys are massless and there is no friction.)

this image shows the issue I am addressing

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Tension in a string becomes a force of the same magnitude at the ends of the rope (where it is attached).

You can consider the pulley acts as if the string is broken into multiple strings and is attached to the pulley at points A and B.

Because the section of the rope along the pulley is "connected" to the pulley by friction, we can imagine that the string is missing and it is just the body of the pulley that connects A and B. The tension in one section of string does not transmit through the pulley.

For a "light" pulley, the tension in the different sections will be the same. But an accelerating pulley with a moment of inertia will cause the sections to have unequal tensions.

Improve this answer
$\endgroup$
2
  • $\begingroup$ I didn't understand, like when we draw the free body diagram of rope the tensions at point A and B are also acting downwards...why don't we show it?? $\endgroup$
    – Shyam
    Commented Mar 3 at 6:36
  • $\begingroup$ Probably because the person doing so is ignoring the portion of the string between A and B. Technically the tension in that section of the string is indeterminate because you don't know how friction is acting. You can instead ignore that section and pretend that all the force from the string acts at the contact points A and B. $\endgroup$
    – BowlOfRed
    Commented Mar 3 at 7:56

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.