0
$\begingroup$

I've attached a question that we went through I class. Could someone please help me understand why the tension in between the mass and the rope is 2K. Its also 2k in other pulley systems that I've seen where a pulley is attached with a rope from its centre to a fixed celling and two masses hanging on each side with tension k on each rope. I must be missing something very obvious but please help or provide a resource where I can lean more about such systems.enter image description here

$\endgroup$
1
$\begingroup$

This isn't always true. In your case, it probably is.

There are pieces of information that haven't been explicitly stated.

1) the mass of the rope and the pulley are negligible. (We assume the masses are zero).

2) the two ropes on either side of the pulley are parallel (vertical).

As a consequence of (1), the tension on the rope on either side of the pulley is equal, and in your problem, you call this tension $K$.

Also as a consequence of (1) the total force on the (mass-less) pulley is zero (recall $F=ma$, and if $m=0$, then $F=0$). Since the two ropes are both pulling vertically upward on the pulley, each with force $K$, this must be balanced by the tension of the rope fixed to the axle of the pulley. So the tension in this rope is pulling down on the pulley must be $2K$. I hope this answers your question.

If the two ropes were not directly vertical, but more in a V shape, then you would have to deal with the forces by breaking them into the vector components and then add. In the case of a V shaped configuration, the tension would be less than $2K$.

$\endgroup$
  • $\begingroup$ Thats very helpful but I think I am confused because i don't understand where the tension in the rope attached to the axel is coming from. Could you give a description of what it means for the rope to be attached to the axel. I thought it was just hanging. so the tension should be equal to the force of gravity on the mass m1. $\endgroup$ – Christopher Moskovitz Sep 24 '17 at 13:52
  • $\begingroup$ Hmm, these types of mechanics problems take a bit of practice. Maybe it helps if I walk you though a few pieces. So we showed that since the net force on the pulley is zero, the rope connected to the axle must be pulling down on the pulley with force $2K$. Newton's third law tells us that the pulley is pulling up on the rope with force $2K$. But if the rope is massless, then this means that net force on it is zero, and the mass $m_1$ must be pulling down on the rope with force $2K$. Newton's third law tells us the rope must be pulling up on $m_1$ with force $2K$. $\endgroup$ – AlbertB Sep 24 '17 at 20:15
  • $\begingroup$ Now we look at the total forces acting on $m_1$ and those are $2K$ from the rope and $-m_{1}g$ from gravity. This gives us the expression $m_{1}a_{1}=F=2K-m_{1}g$. (If we use the sign convention that up is positive.) $\endgroup$ – AlbertB Sep 24 '17 at 20:17
  • $\begingroup$ Note that the tension on the rope attached to the axle is not equal to the weight because the mass is accelerating. $\endgroup$ – AlbertB Sep 24 '17 at 20:19
  • $\begingroup$ That was a very clear explanation thank you! AlbertB $\endgroup$ – Christopher Moskovitz Sep 26 '17 at 18:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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