What is Tension on each part of the rope?

[Shardul Khare]

I know that when we strech the rope...all the points on that rope are under tension. But I do not understand how Newtons 3rd law of motion is applied when we work with tension.Any hep would be apreciated

 

[Nugatory]

Consider an infinitesimally small section of the rope. It is being pulled to the left by a force exerted on it by the rope to its left, and it is being pulled to the right by a force exerted on it by the rope to its right. By Newton's third law, it is also exerting equal and opposite forces on the rope to its left and to its right.

 

[Shardul Khare]

Consider an infinitesimally small section of the rope. It is being pulled to the left by a force exerted on it by the rope to its left, and it is being pulled to the right by a force exerted on it by the rope to its right. By Newton's third law, it is also exerting equal and opposite forces on the rope to its left and to its right.

Sorry,That is very difficult for me to understand...Lets consider a object suspended with a rope (Connected to a ceiling)..What would be the tension on the point of rope which is connected to the ceiling as well as the point of rpe hwich is connected to the object?

 

[Nugatory]

Sorry,That is very difficult for me to understand...Lets consider a object suspended with a rope (Connected to a ceiling)..What would be the tension on the point of rope which is connected to the ceiling as well as the point of rpe hwich is connected to the object?

Think about just the bottom of the rope, the bit that is connected directly to the object. There is a downwards force on the object from gravity - it's $mg$. However, the object is not accelerating so we know that the net force on it is zero. Thus, the little bit of rope attached to the object must be exerting an upwards force on the object, also of magnitude $mg$ so that the net force comes out zero.

Because that little bit of rope is exerting an upwards force of $mg$ on the object, we also know from Newton's third law that the object is exerting a downwards force $mg$ on the little bit of rope. But the little bit of rope is not accelerating, so the net force on it is zero - the next little bit of rope above it must be exerting an upwards force, and again Newton's third law says that there is an equal and opposite downwards force on that bit. Eventually we get to the top, where the ceiling exerts an upwards force sufficient to exactly cancel the downwards gravitational force on the rope and the object.

At every point along the rope, the tension is opposing the force of gravity on the object and all the rope below that point. If the mass of the object is $m$ and the mass of the entire length of rope is $m_R$, the tension at the bottom is $mg$ (what it takes to make the net force on the object be zero) and the tension at the top is $(m+m_R)g$ (what it takes to make the net force on the object plus the rope be zero).

 

[Battlemage!]

Think about just the bottom of the rope, the bit that is connected directly to the object. There is a downwards force on the object from gravity - it's $mg$. However, the object is not accelerating so we know that the net force on it is zero. Thus, the little bit of rope attached to the object must be exerting an upwards force on the object, also of magnitude $mg$ so that the net force comes out zero.

Because that little bit of rope is exerting an upwards force of $mg$ on the object, we also know from Newton's third law that the object is exerting a downwards force $mg$ on the little bit of rope. But the little bit of rope is not accelerating, so the net force on it is zero - the next little bit of rope above it must be exerting an upwards force, and again Newton's third law says that there is an equal and opposite downwards force on that bit. Eventually we get to the top, where the ceiling exerts an upwards force sufficient to exactly cancel the downwards gravitational force on the rope and the object.

At every point along the rope, the tension is opposing the force of gravity on the object and all the rope below that point. If the mass of the object is $m$ and the mass of the entire length of rope is $m_R$, the tension at the bottom is $mg$ (what it takes to make the net force on the object be zero) and the tension at the top is $(m+m_R)g$ (what it takes to make the net force on the object plus the rope be zero).

Stupid and off topic question, but if you wanted to integrate over the length of the rope to get the total tension, what would your integral look like? What is the dummy variable? Are the bounds are 0 to L (mg and [m+ m_R]g?)?

 

[Nugatory]

Stupid and off topic question, but if you wanted to integrate over the length of the rope to get the total tension, what would your integral look like? What is the dummy variable? Are the bounds are 0 to L (mg and [m+ m_R]g?)?

The tension at every point is equal to the weight of everything below that point. You would only need to integrate something if that weight is not obvious by inspection, which will only happen if the density of the rope is different at different points along its length. In that case you'd calculate the mass of the portion of the rope you care about the same way you calculate the mass of anything else with a non-constant density: integrate the density across the volume you're interested in.

 

[ehild]

In addition to @Nugatory's explanation that the object experiences the tension of the little piece of rope just above it, see the amazing video about "slinky drop experiment"

.Holding the slinky, it stretches and tension is built up in it. When releasing the upper end, the tension releases along the upper part, but still stays at the bottom, so the bottom part of the slinky stays stationary for a while...