The Weak Link

Classical Mechanics Level 1

A chain made of identical links is hanging between two points (indicated by the hooks). We can increase the length of the chain by adding more links to it until the chain becomes very heavy and breaks.

At which point is the chain most likely to break?

A, at the top B, in the middle C, at the bottom It is equally likely to break from any point

2 solutions

Jesse Nieminen
Mar 5, 2017

The chain will break because it weighs too much.
Thus, it will most likely break at point $\text{A}$ because more weight is focused on the point $\text{A}$ than on the other named points.

Hence, the answer is $\boxed{\text{A, at the top}}$ .

can you prove tensiom at A is more than that at C ???

space sizzlers - 4 years, 3 months ago

Yes. Notice that at C, the tension is balanced by the chains on both sides

Agnishom Chattopadhyay - 4 years, 3 months ago

Isn't the tension at C acting in the horizontal direction due to symmetry .....?.

space sizzlers - 4 years, 3 months ago

@Space Sizzlers – Yes, the tension at $C$ is not zero, rather it acts horizontally. Tension force always acts tangentially to the rope/chain.

Rohit Gupta - 4 years, 3 months ago

@Rohit Gupta – @Rohit Gupta yes , please explain me the equilibrium of the half chain from A to C . If tension at A is vertical and that at C is horizontal , the weight of half chain also is vertical , then how is the half chain in equilibrium . I mean which other force balances the horizontal tension at C ?

space sizzlers - 4 years, 3 months ago

@Space Sizzlers – I think the diagram here is misleading. The link at $A$ should not be perfectly vertical rather inclined at some angle so that the tension at $A$ is inclined as well. The horizontal component of tension at point $A$ will balance the tension at point $C$ for the half chain.

Rohit Gupta - 4 years, 3 months ago

@Agnishom Chattopadhyay i need to add a picture here how do i do it ? And if not here where should i add it ? thanks in advance

space sizzlers - 4 years, 3 months ago

You use ![image link](image description)

Agnishom Chattopadhyay - 3 years ago

we can actually do it very easily by using shear force and bending moment of the chain in mind assuming that the chain is a simply supported beam and its self weight as a uniformly distributed load the bending moment is maximum at the middle of the chain and shear force is zero at the middle of the chain as it is evident by looking at it the bending action is highest at C the shear forces are maximum at the two ends of the chain AKA beam so the chain will break where the maximum shear stresses are acting thus it will break at the end of the chain AKA answer is A, at the top

Maruthi Nandan - 4 years, 3 months ago

I agree with you for the part that the shear stress is maximum at the top and it will break from there. But, how could you say that the shear force is zero at the middle of the chain, I agree that it should be minimum but not zero.

Rohit Gupta - 4 years, 3 months ago

for a uniformly distributed load(that load being the self weight in this case) for a simply supported beam like the chain above the shear force from right to left or left to right depending on your choice changes from negative to positive(or positive ti negative depending on your sign convention) there will definitely be point where the shear force zero where the shear is force is transforming from negative direction to positive ( that point is also called some name i can't remember its point of conflucture or something like that i will find out) for a uniformly distributed load it is at the middle at middle the shear force is zero and the bending moment is maximum at that point

Maruthi Nandan - 4 years, 3 months ago

I though like that too. A is holding the most weight

Half-god Dragon - 3 years ago

Lance Fernando
Mar 6, 2017

Imagine this: the point A is the one who carries the rest of the points in a stack-up, including point B and C. With this, the probability of breaking the stack is increasing exponentially as more points stack up. At any point, A will be stressed too much and will fall down, causing the break of the stack made by the points. The concept is the same as the chain problem here - only inversely.

This is a very interesting way to visualize it. Why does the same thing work when we invert the stack upside down?

Agnishom Chattopadhyay - 4 years, 3 months ago

This is because the lower the chain point, the more loose it is - with it, the point doesn't carry much weight, as it can sway around within what can the top point reach. On the other hand, the top point is locked in its movement, because it carries a lot of points within itself - with heave, it can't move well and can break if the pressure applied is too much.

Lance Fernando - 4 years, 3 months ago

Center of gravity of chain is more near the point C than at any point A. Thus point C should experience more tension.

ANKUR KURMI - 4 years, 3 months ago

This is unclear to me. I am not sure how the location of the center of gravity will affect the tension in the links. Can you elaborate it further?

Rohit Gupta - 4 years, 3 months ago

The comment about the center of gravity isn't relevant.
If you wish, draw a force diagram for each link. The higher the link, the more force on it pulling down.

Richard Desper - 4 years, 3 months ago

The Weak Link

2 solutions

0 pending reports