What is the tension in a rope with mass?

In summary, two blocks of mass 7kg and 5kg are connected by a heavy uniform rope of mass 4kg. An upward force of 200N is applied to the system. (a) The acceleration of the system is 2.5 m/s/s upwards. (b) The tension at the top of the rope is 87.5N upwards and the tension at the midpoint of the rope is 37.5N downwards. (c) To find the tension at the midpoint, the rope can be broken into two identical ropes, and the tension can be calculated using the equation F2 = F1 - 40N, where F1 is the tension at the top of the rope. The tension at
  • #1
Govind_Balaji
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Homework Statement


Two blocks shown in figure are connected by heavy uniform rope of mass 4kg. An upward force of 200N is applied as shown.
(a)What is the acceleration of the system?
(b)What is the tension at the top of heavy rope?
(c)What is the tension at the mid-point of the rope?

Homework Equations


I Took g=10m/s^2.

The Attempt at a Solution


See the attachment. I drew FBD.
The first question is easy.
Net gravity force=(7+4+5)g=160N downwards
Net Force in the system=200-160=40N upwards.
Acceleration of the system=40N/16kg=2.5m/s/s upwards.
This answer coincides with the Answer Key.

For third question, I should break the rope into 2 identical rope and find the tension. Right?

First I tried second question. I get stuck in the second question due to the nonm-zero net force(40N).

In the diagram, F1-70=Net force=40N. Then F1=110N upwards. Right?
F2-F1+40=40. Right?
Then F2 should be 110N downwards right?

But the answer is 137.5N

PS: I drew the FBD of three things. Did I miss any force in the FBD?
 

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  • #2
Govind_Balaji said:
In the diagram, F1-70=Net force=40N. Then F1=110N upwards. Right?

40N is the net force on the entire system. That is not the same as the net force on parts of the system.

Your equation should be "F1-70=Net-force-on-the-7kg-block=7kg*acceleration"
Govind_Balaji said:
F2-F1+40=40. Right?

Please explain this equation?
 
  • #3
Nathanael said:
40N is the net force on the entire system. That is not the same as the net force on parts of the system.

Your equation should be "F1-70=Net-force-on-the-7kg-block=7kg*acceleration"

Please explain this equation?

Thank you.

That equation was the summation of froces on the 4kg rope.

[itex]
\begin{align*} \text{Net force on 7kg block }&=7kg*2.5 m/s^2=17.5N=F_1-70N\\\Rightarrow F_1&=87.5 N \text{ upwards}\\\text{Net force on 4kg rope}&=4kg*2.5m/s^2\\10N \text{ upwards}&=F_1-F_2-40N \text{ upwards}\\10N \text{ upwards}&=-F_2+47.5N \text{ upwards}\\F_2&=37.5N \text{ downwards}\end{align*}[/itex]

??
 
  • #4
Ok a few things here, first of all (I should've said this in my first reply) your free body diagrams are wrong.

"F1" should be on the "FBD of 7kg block" and "F2" should be on the "FDB of 5kg block"Secondly, you've ignored Newton's 3rd law (for every force there is an equal and opposite force)
Because of Newton's 3rd law, each force you've drawn should have another force to complete the pair.
However, you don't need to draw the "equal and opposite force" of gravity, because it acts on the Earth (which you don't care about).
You also don't need to draw the "equal and opposite force" of the 200N-force because it acts on whatever is pulling the system up (which you don't care about).
So that leaves you with just "F1" and "F2" that need an "equal and opposite force" drawn for them.

The rope pulls down on the 5kg block with a force "F2" ... therefore the block pulls UP on the rope with a force "F2"

Similarly, the rope pulls up on the 7kg block with a force "F1" ... therefore the block pulls DOWN on the rope with a force "F1"

So the FBD of the rope needs to be changed a bit.
 
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  • #5
I corrected it. I have a small confusion over drawing Force vectors on a FBD. What does the arrowhead and the base of arrow indicate. Do they indicate the object being pushed/pulled and the object which pushes/pulls respectively?

Now I get the equation right and got the answer. Now I am going to workout the third question after having lunch.
 

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  • #6
Govind_Balaji said:
I corrected it.
...
Now I get the equation right and got the answer.

Good, good.

Govind_Balaji said:
Now I am going to workout the third question after having lunch.
I will be asleep by then (it's 1 a.m. where I am)

Govind_Balaji said:
I have a small confusion over drawing Force vectors on a FBD. What does the arrowhead and the base of arrow indicate. Do they indicate the object being pushed/pulled and the object which pushes/pulls respectively?
There is no indication of whether the object is pushed or pulled (as there is not really any real physical difference)

The arrowhead indicates the direction of the force. The base of the vector typically indicates where the force is applied, although it doesn't have to be drawn where the force is applied. Often you will see the base of the vector drawn on the center of mass, even when it is applied elsewhere.

As far as I know, there are not any strict rules for where to draw the base of your vector. You just draw it in whatever way that shows the information that needs to be shown.

(If there are strict rules for it, I certainly don't follow them :-p)
 
  • #7
I worked out my third question too. I got 112.5 N
 
  • #8
Govind_Balaji said:
I worked out my third question too. I got 112.5 N

A bit too much. Please post your working.
 

FAQ: What is the tension in a rope with mass?

What is tension in a rope with mass?

Tension in a rope with mass refers to the pulling force acting on the rope as a result of its weight and the weight of any objects attached to it.

What factors affect the tension in a rope with mass?

The tension in a rope with mass is affected by the weight of the rope itself, the weight of any objects attached to it, and the angle at which the rope is pulled.

How is the tension in a rope with mass calculated?

The tension in a rope with mass can be calculated using the formula T = mg + ma, where T is the tension, m is the mass of the rope, g is the acceleration due to gravity, and a is the acceleration caused by the pulling force.

Can the tension in a rope with mass be greater than the weight of the rope and attached objects?

Yes, the tension in a rope with mass can be greater than the combined weight of the rope and attached objects. This is because the tension is also affected by the acceleration of the pulling force.

How does tension in a rope with mass affect its strength?

The tension in a rope with mass can affect its strength by putting stress on the fibers of the rope. If the tension exceeds the breaking point of the rope, it can cause the rope to break. Therefore, it is important to consider the tension when determining the appropriate rope for a specific task.

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