# Homework Help: Tension in string of objects connected question :o

1. Apr 21, 2012

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

In the attachment!

This is a multiple choice question where you have to choose one of the strings. Intuitively I chose Z as I would think that in the first fraction of time it would feel the force acting on it first. However, if this wasn't a multiple choice I would've gone with that the tension is equal along the strings since the acceleration has to be equal, mass is equal, and hence Force=Tension pulling it should be equal.

What i'm wondering is if this line of thought is correct? Am I in the right assuming that it's only in the first split second that the Z string experiences the biggest force? If not, would anybody care to help/explain to me the answer? Thankyou for any replies! :)

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2. Apr 21, 2012

### tms

The way to find the answer is to calculate the tension in each string. Look at the forces acting in each direction in each string.

3. Apr 21, 2012

### tiny-tim

hint: if the tensions X and Y were equal, what would be the acceleration of the mass between them?

4. Apr 23, 2012

My... Lemme think. That just opens to more confusion. I would think the tension is pulling in the same direction... God! Well. if they're pulling in opposite directions then Zero. but it still doesn't make much sense to me xd

5. Apr 23, 2012

### tiny-tim

exactly!

the equations do not lie!!

if the tension on both sides was equal, the block between would not accelerate

the only way to get each block to accelerate is for the tension in front of each block to be more than the tension behind it

(btw, you do know that the tension in each string is in opposite directions at opposite ends?

eg in string Z, the tension pulling the first block is to the left, but the tension pulling the second block is to the right)

6. Apr 23, 2012

I'm slowly latching on to the idea, I think the problem lies in that tension is a little elusive to me. I feel rather silly, but yeah. I understand that on two adjacent blocks the string tension pulling them will be in opposite directions and be equal, right? Okay.. So that would mean that for the first block, the net force acting on it would be F-the tension force acting in the opposite direction.

Then for the second block the tension pulling it to the left would have to be smaller than the tension force pulling it to the right which is = T pulling on the first block to the left?

Sorry if I'm being tedious, I'm just quite foggy on this tension idea!

7. Apr 23, 2012

### tiny-tim

try a different question first … the maths is the same, but it's static, with gravity:

suppose five weights are hanging from the ceiling, like your diagram but tilted 90° …

what is the tension in each of the pieces of rope?

which has the greatest tension?

what is the equation relating, say, T3 T4 m3 and g?​

8. Apr 23, 2012

Ok! I would think that the string with the greatest tension will be the one that is directly attached to the ceiling, intuitively, anyways!

So if i look at the bottom mass gravity would be acting on it and therefore the tension needs to counteract this force so, weight=Tension of first string.

I'd think the string above that one would then need to counteract both that tension force of the string below it=Weight of the first mass and the gravitational pull of the weight of the second mass. Therefore it'd be double that since the masses are all equal?

So then
T1=Weight of one of the identical masses.
T2= Weight of one of the identical masses x 2?

so on so forth? D: Not quite sure if i'm on the right track here, mind. Thanks for your help thus far.

9. Apr 23, 2012

### tiny-tim

yes, that's right

the tension in each string is the weight of all the masses beneath it (they needn't be equal)

so if you do F = ma for the nth block, the tension above it will be g(m1 + … mn), while the tension below it will be g(m1 + … mn-1), and of course the gravitational force on it is mng

so the total force is 0 = mng - mng, as expected

for the original, horizontal, problem, you'll get similar equations, except that there'll be no gravitational force, and you'll have "a" (the common acceleration) outside those two brackets instead of "g"

10. Apr 23, 2012