# Tension in a light inextensible string

1. Sep 26, 2012

### hms.tech

1. The problem statement, all variables and given/known data

The attachment shows the string which is light and inextensible passing over two pulleys which are NOT massless. Hence they have some mass and some Moment of inertia.

There is no friction in the pulley thus the pulleys are free to rotate.

K ≥ 1

we know that the system will accelerate and the heavier pink object would accelerate downwards.

The red table is frictionless.

The thing that i am unable of grasp is finding the Tension in each part of the string.

I am not sure, if i have labelled them correctly.
Would someone please tell me the correct values.
Although i am quite sure that T1 and T2 are correctly labelled (ie showing that the tension in both of these parts of the string would be DIFFERENT)

What is the tension T3 and T4 equal to in terms of magnitude (is it the same as T1 or T2) and direction (will they be in the same direction as which i have drawn?)

2. Relevant equations

none

3. The attempt at a solution

the attachment is an attempt (and might contain errors )

#### Attached Files:

• ###### Untitled.png
File size:
9.2 KB
Views:
214
2. Sep 27, 2012

### hms.tech

Basically i have to find the tension[magnitude and direction] (sown in orange) in the diagram.

The directions of T3 and T4 are just arbitrary, they are Assumed to be acting in that sense, but it's not necessary they are (as i mentioned before the directions might be wrong) therefore i need you guys to help me determine the direction of T3 and T4

Note : the yellow circles are Pulleys and the pink ovals are masses. The mass on the right is heavier than that on the left. The red line is the table.

3. Sep 27, 2012

### Staff: Mentor

You have 3 rope segments and thus three different tensions, not four. Label the tension in the horizontal segment T3.

When drawing forces it's good to keep in mind what those forces are acting on. It's usually best to draw the forces acting on the massive bodies.

Your drawing of the forces acting on the two pink bodies is correct. But you have to fix the drawing of the forces acting on the pulleys. The vertical rope segments pull down on the pulleys--you have that correct. Now fix the tension forces due to the horizontal rope segment, remembering that ropes can only pull.

4. Sep 27, 2012

### Staff: Mentor

You'll need to know the masses and moments of inertia to find the tension forces.
See my last post. The ropes pull on the pulley. That should tell you the directions of the tension forces on the pulley.

5. Sep 27, 2012

### hms.tech

You are implying that if i reverse the directions of $T_{3}$ and $T_{4}$, that would give me the correct directions of the Tensions.

Also that $T_{3}$ = -$T_{4}$

Another thing which confuses me : Why is it not that $T_{1}$ = $T_{2}$ ?
Ans : Is it because the pulley has some mass (and is not mass-less or "light") ?

6. Sep 27, 2012

### Staff: Mentor

That would give you the correct directions of the tension forces exerted by the middle string segment on the pulleys.

I would let $T_{3}$ represent the magnitude of the tension force.

Exactly. A continuous segment of massless rope has the same tension throughout, but not when it's divided over a massive pulley.

Don't know of anything off hand. I suggest getting a review book and solving as many problems as possible. You'll also find many, many problems discussed in great detail right here on PF. But you'll need to poke around to find them.

7. Sep 27, 2012

### hms.tech

I have done some more reading on this topic and found out (another reason) :

"If the pulley is smooth, the tension in the string is same throughout its length while on the other hand if the pulley is rough the tension in the strings on either side of the pulley is different"

Does this reasons add up to the one mentioned before ?

8. Sep 28, 2012

### Staff: Mentor

I presume they mean that the pulley is so smooth that the string slides over it without rotating it. In that case, sure, the pulley would not change the tension in the string.

But usually it is assumed that the pulley does turn with the movement of the string. So the tension would be the same on both sides only if the pulley is massless and frictionless (on its axle). When mass (and thus moment of inertia) is involved or there is friction, there must be a difference in tension from one side to the other to create the net torque needed to rotate the pulley (at least when there is acceleration involved).