Tension on A Rope of Negligible Mass

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SUMMARY

The discussion centers on the concept of tension in a massless rope, specifically how force is transmitted through a flexible cord. When a mass is suspended from a massless string, the tension remains constant throughout the length of the cord due to the absence of mass, allowing for undiminished force transmission. The net force on the cord is zero when acceleration is finite, leading to equal tension at both ends. The mathematical representation of this phenomenon can be illustrated using free-body diagrams and the principles of Newton's laws.

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  • Understanding of Newton's laws of motion
  • Familiarity with concepts of tension and force transmission
  • Basic knowledge of free-body diagrams
  • Concept of massless objects in physics
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Bashyboy
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I was a reading a little section in my physics textbook regarding tension in a flexible cord. There is a part in the paragraph that I am not sure why is true. It goes as follows: "When a flexible cord pulls on an object, the cord is said to be under tension, and the force it exerts on an object...[can be the tension]...FT. If the cord has negligible mass, the force exerted at one end is transmitted and undiminished to each adjacent piece of cord along the entire length to the other end." Could someone explain to me why the second sentence is true? Thank you.
 
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If the cord has negligible mass and acceleration is finite then the net force on the cord must be zero (ma = 0). Therefore tension on the two ends will be equal and opposite.
 
I am not sure I really understand that explanation, sorry.
 
Try looking at it this way. When a mass is hung from the string, the lowermost molecule of the string is tugged by the gravitational force acting on the suspended object. This molecule passes on the 'tug' to its adjacent upper molecule, and as it is assumed to be massless, it does this without loss of force in transmission. This tug on the second molecule is passed on to the third and so on.. Finally the uppermost molecule of the string, immediately in contact with the roof experiences this pulling force and since it has no other molecule of its own to pass on this force, it transmits this to the roof, thereby pulling the roof with the exact same force the mass was being pulled at.
 
Suppose the tension at the two ends is not the same. That would mean there is a force acting on a mass that is very close to 0. If that was the case since F = ma, the string would experience an infinite acceleration. Since this doesn't happen, the tension is either the same at both ends or the difference is so small that we can ignore it.
 
I understand that this thread is rather old, but I recently have come across this difficulty again. I understand Infinitum's argument to some extent, but I am wondering if there is some mathematical way of showing it. If the rope is massless, then molecules that constitute it are massless, and the force transmits from one molecule to the next, is there a mathematical way of showing this?
 
Bashyboy said:
I understand that this thread is rather old, but I recently have come across this difficulty again. I understand Infinitum's argument to some extent, but I am wondering if there is some mathematical way of showing it. If the rope is massless, then molecules that constitute it are massless, and the force transmits from one molecule to the next, is there a mathematical way of showing this?

You can try thinking of the rope as a chain of massless beads connected to one another by short massless springs. Now we can set up a free-body diagram for any given bead and use ##F=ma## to see what the forces on it are; the tension in the spring is the tension of the rope at that point.

Interestingly, this approach works whether the rope is massless or not. If the rope has mass you just have to assign a suitable mass to each bead.
 
Hi Nugatory

Consider a massless string going over a massless frictionless pulley with masses M1 and M2 at the end of the string.

The tension in the string would be same.

The torque equation of the pulley says (T1-T2)=Iα .Now L.H.S is zero since T1=T2 and also I=0 .

So,we have a condition 0=(0)(α) which makes α indeterminate .But we know that the pulley rotates.
So ,how is α determined ?

Is α=a/R where R is the radius of the pulley and a=[(M2-M1)g]/(M1+M2) ?
 
Tanya Sharma said:
Hi Nugatory

Consider a massless string going over a massless frictionless pulley with masses M1 and M2 at the end of the string...
that's a different problem than OP posed. Rather than hijacking his thread, it might be best to start a new one.
 
  • #10
I sincerely apolozise if you feel i have hijacked the thread.I thought that the OP's concern has been addressed .i had doubt somewhat related so posted here .

Sorry,once again .
 

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