# Rock on a string on the ceiling?

rock on a string on the ceiling??!!

I'm stuck on this problem... any ideas??

A stone hangs by a fine thread from the ceiling and a section of the same thread dangles from the bottom of the stone. If a person gives a sharp pull on thedangling thread, where is the thread likely to break: below the stone or above it? What if the person gives a slow and steady pull? Explain your answers.

We're in a chapter about Newton's Laws

I know the rock and gravity are pulling down on the string. And the ceiling is pulling up... and without touching the string the forces of the rock on the string and the ceiling on the string create a net force of zero... but what is the effect of acceleration (you pulling on the string)??

THX

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Galileo
Homework Helper
Consider the forces (tension) in the strings when you pull on them. Suppose you pull slowly so that the system is in equilibrium at all times, until it breaks. Where do you think the string breaks, considering the tensions?

As you pull on the string, it will get slightly longer. If you pull very fast that happens faster than the block can come down. Which string would break this time?

Is it that when you pull the string very fast, the force you apply is impulsive, so the rock accelerates upwards whereas you are pulling the string downwards which results in the lower part of the string breaking?

The tensions in both halves are different.

If you pull the string slowly, the whole system is in equilibrium. But how does that effect the tensions in both parts of the string???

Galileo
Homework Helper
Is it that when you pull the string very fast, the force you apply is impulsive, so the rock accelerates upwards whereas you are pulling the string downwards which results in the lower part of the string breaking?
No, the rock doesn't accelerate upwards. It drops down a bit if you pull the string, because the string stretches. If you pull fast, the lower string will stretch faster than the block will fall.

If you pull the string slowly, the whole system is in equilibrium. But how does that effect the tensions in both parts of the string???
The string will obviously break in that part where the tension is highest. Where is the tension higher? Above or below the rock?

So if you pull slowly... the string above the rock will break because of the extra tension that the rock is putting on the string??

And if you pull faster than the rock can fall... the string below the rock will break because the string will stretch faster than the rock can fall???

Galileo
Homework Helper
So if you pull slowly... the string above the rock will break because of the extra tension that the rock is putting on the string??

And if you pull faster than the rock can fall... the string below the rock will break because the string will stretch faster than the rock can fall???
That's the idea. I`m sure the first is obvious. I hope you can see why the second one is true intuitively.

thank you! i think i understand it now!

Im sorry to bring this up again but please bear with me for a minute, there is tention 't' in the string, which is equal to m*g + f, where m is the mass of the rock, g is acceleration due to gravity and 'f' is the force you apply on the lower string.

When you do this slowly, the system is in equilibrium. As you increase the force 'f', mg being constant, at a certain point the tention becomes greater than the maximum tention the string can tolerate and hence, the upper string breaks.

When you pull the lower string quickly, the rock doesnt have time to accelerate with the string, so instantaneously, the net tension in the lower string becomes greater than the max value of the tension that the string can sustain and hence the lower string breaks.... is that right?