How does tension in a string balance weight and cause acceleration in objects?

[Kaneki123]

Consider an example of a mass hanging by a string. The string is fixed at the other end. The mass will be at rest because weight of the object is balanced by tension in the string upwards.Consider another example in which the other end of the string is not fixed. Now if that object is coming down with some acceleration meaing it has some net force acting on it. Why is that so when weight of the object has equal reactional tension..So should'nt it be balanced by equal upward and downward forces??

 

[Simon Bridge]

Now if that object is coming down with some acceleration meaning it has some net force acting on it. Why is that so when weight of the object has equal reactional tension...

When? Please provide an example.

So should'nt it be balanced by equal upward and downward forces??

Why?

Some notes:
In real life, a string has some mass, and there is air resistance. So, when a mass with a thread attached falls, the string trails out behind it due to air resistance. Similarly the mass has an additional force, due to that air resistance, acting on it. But when a teacher sets you physics problems, there will often be unrealistic approximations in them to make the maths easier - like zero air resistance. Without air resistance, the string and the mass fall with the same acceleration and so there is no tension in the string.

You can sort out these situations by drawing an appropriate set of free body diagrams as per Newton's approach for coupled masses.

 

[Kaneki123]

When? Please provide an example.Why?

Some notes:
In real life, a string has some mass, and there is air resistance. So, when a mass with a thread attached falls, the string trails out behind it due to air resistance. Similarly the mass has an additional force, due to that air resistance, acting on it. But when a teacher sets you physics problems, there will often be unrealistic approximations in them to make the maths easier - like zero air resistance. Without air resistance, the string and the mass fall with the same acceleration and so there is no tension in the string.

You can sort out these situations by drawing an appropriate set of free body diagrams as per Newton's approach for coupled masses.

I have uploaded a diagram...In this diagram if m1 has sufficient weight, m2 will not accelerate downwards..Or you can say it as if m2's weight is 'balanced' by equal reactional tension in the string...But if m1 does not have suffiecient weight, m2 will accelerate downwards...Now, why is that so because even now m2's weight should be equal to its reactional tension in the string( Newton third law)?...If the weight has equal upwards reactional tension, should'nt it be 'balanced'?

 

[sophiecentaur]

Your model must include a mention of Friction or else M1 and M2 will always accelerate. The weight force of the hanging mass will accelerate both masses according to N2.

 

[Kaneki123]

Okay let me explain it like this...The weight of m2 will cause some tension in the string..This tension will cause some net force on m1 (of course if it is greater than its Fs )...So it will accelerate in direction of tension...My basic question is that, if, instead of that mass m1, the string was fixed to some other thing like a wall, it will not accelerate...Because in that case the weight of m2 will be equal to tension in the string...But now in THIS case, the weight of m2 dominates the tension in the string (causing it to accelerate downwards)..So why is that so when weight should still be equal to the reactional tension in the string?

 

[sophiecentaur]

You can work all this out yourself by following the following piece of logic:
1. What is the downward force on m2? (whether it's in free fall, on a table or moving down in your experiment)
2. If the string is not stretching, what total mass is being accelerated by this force?
3. What is the acceleration of the mass in 2, by the force in 1.?
4. What string Tension is needed to accelerate m1 at the rate in 3.?

But now in THIS case, the weight of m2 dominates the tension in the string

The above method will tell you the answer whatever the relative masses of m1 and m2. Stick with the Maths if you want the right answer in this sort of question. The arm waving 'reasons' are no use here.

 

[Chestermiller]

Newton's 3rd law of action-reaction says that the force that A exerts on B is equal and opposite to the force B exerts on A. But, when doing a force balance on an object, you only include the forces that are acting on that object (by other objects), not the forces with which the object exerts on other objects. So, in the force balance on A, you only include the force that B exerts on A. You don't include the force that A exerts on B.

 

[Simon Bridge]

Okay let me explain it like this...The weight of m2 will cause some tension in the string..This tension will cause some net force on m1 (of course if it is greater than its Fs )...So it will accelerate in direction of tension...My basic question is that, if, instead of that mass m1, the string was fixed to some other thing like a wall, it will not accelerate...Because in that case the weight of m2 will be equal to tension in the string...But now in THIS case, the weight of m2 dominates the tension in the string (causing it to accelerate downwards)..So why is that so when weight should still be equal to the reactional tension in the string?

You are mistaken, in THIS case, the weight is NOT equal to the reaction tension of the string.

You can see this by drawing free body diagrams for the masses.
For simplicity, let m1=m, m2=M, M >> m, and the tension is T.
The table is frictionless, and the string does not stretch.
Then... for m1, mg-T=ma
... for m2, T=Ma
... eliminate a and solve for T in terms of m and M.

 

[sophiecentaur]

Intuition can be a real handicap in problems like this one. The OP doesn't want to do this formally but is relying on what he/she 'feels' about the situation.

 

[Kaneki123]

You are mistaken, in THIS case, the weight is NOT equal to the reaction tension of the string.

You can see this by drawing free body diagrams for the masses.
For simplicity, let m1=m, m2=M, M >> m, and the tension is T.
The table is frictionless, and the string does not stretch.
Then... for m1, mg-T=ma
... for m2, T=Ma
... eliminate a and solve for T in terms of m and M.

I know in terms of mathematical equations for tension...It's just that according to Newton's 3rd law, the weight should be equal to the reactional tension in the string...Again, my basic question is why is the weight not equal to reactional tension in string?

 

[A.T.]

according to Newton's 3rd law, the weight should be equal to the reactional tension in the string

No, that's not what Newton's 3rd law says. It just says: Force by string on mass is equal but opposite to force by mass on string. Weight doesn't appear here.

 

[Kaneki123]

No, that's not what Newton's 3rd law says. It just says: Force by string on mass is equal but opposite to force by mass on string. Weight doesn't appear here.

Right and the ''force by mass on string'' is equal to the weight of mass...( Correct me if i am wrong )...And thus the ''force by string on mass'' is reactional tension which should be equal to weight of mass...Please correct me if i am wrong

 

[sophiecentaur]

.Again, my basic question is why is the weight not equal to reactional tension in string?

Until you read and understand what Newton's Third Law actually says, you will have a problem with this. The relevance of N3, in this problem is that it tells you the tension is the same at each end of the string because the force that m2 pulls down with is the same as the force that m1 is being pulled with. Whether m1 is stuck or allowed to move, this is true. But it doesn't say that the tension is necessarily the same as the weight of m2.

 

[sophiecentaur]

( Correct me if i am wrong )...And thus the ''force by string on mass'' is reactional tension which should be equal to weight of mass...Please correct me if i am wrong

You are wrong and that's what at least three people have been telling you.

 

[A.T.]

the ''force by mass on string'' is equal to the weight of mass...

Not if the mass accelerates.

 

[Fewmet]

I think Sophiecentaur is correct in saying the OP is getting hung up on what is intuitively obvious to him or her. It’s something I routinely see in teaching introductory physics. I also routinely see students get frustrated to the point of walking away thinking physics is inaccessible or bogus.

Kaneki123: see if this line of reasoning works. Suppose, in your drawing, m1 is initially held in place and that there is no significant friction acting. Now apply the first law to m₂: it is a body at rest remaining at rest, so the net force acting on it is zero. As you correctly noted, those forces are the tension pulling up and the weight of m₂ pulling down, so the tension equals the weight. (More precisely, it has the same magnitude but is oppositely directed.)

Note that the second law leads to the same result. The acceleration of m₂ is zero s the sum of the forces acting on m₂ is zero and weight equals tension. Both approaches support your premise in your original post when m₂ is not accelerating.

What happens if you stop holding m₁? Assuming m₁ < m₂, both masses go from rest to moving. Both accelerate. If m₂ accelerates down, the net for down is no longer zero and the weight must be greater than the tension.

This leads to your initial question: doesn’t the third law say the tension and weight have to be of the same magnitude? You are left with two possibilities: the law is wrong or your interpretation is not correct. Check the latter by carefully applying the third law. I like this phrasing:

When one body applies a force to a second body, the second body applies a force of the first that is the same size but in the opposite direction.

What are the two bodies pulling on each other? (Selecting the two is probably where you are getting off course.)

 

[sophiecentaur]

You are left with two possibilities: the law is wrong or your interpretation is not correct.

You got it!

 

[Kaneki123]

Not if the mass accelerates.

''if the mass accelearates brings me to original question as to why the mass accelearates?

 

[Kaneki123]

I think Sophiecentaur is correct in saying the OP is getting hung up on what is intuitively obvious to him or her. It’s something I routinely see in teaching introductory physics. I also routinely see students get frustrated to the point of walking away thinking physics is inaccessible or bogus.

Kaneki123: see if this line of reasoning works. Suppose, in your drawing, m1 is initially held in place and that there is no significant friction acting. Now apply the first law to m₂: it is a body at rest remaining at rest, so the net force acting on it is zero. As you correctly noted, those forces are the tension pulling up and the weight of m₂ pulling down, so the tension equals the weight. (More precisely, it has the same magnitude but is oppositely directed.)

Note that the second law leads to the same result. The acceleration of m₂ is zero s the sum of the forces acting on m₂ is zero and weight equals tension. Both approaches support your premise in your original post when m₂ is not accelerating.

What happens if you stop holding m₁? Assuming m₁ < m₂, both masses go from rest to moving. Both accelerate. If m₂ accelerates down, the net for down is no longer zero and the weight must be greater than the tension.

This leads to your initial question: doesn’t the third law say the tension and weight have to be of the same magnitude? You are left with two possibilities: the law is wrong or your interpretation is not correct. Check the latter by carefully applying the third law. I like this phrasing:

When one body applies a force to a second body, the second body applies a force of the first that is the same size but in the opposite direction.

What are the two bodies pulling on each other? (Selecting the two is probably where you are getting off course.)

Okay...thanks for long explanation...The two bodies pulling on each other are...the mass m1 pulling on string in downward direction and the reactional tension in string pulling m1 in upwards direction..(correct me if i am wrong)...Maybe its better if i put my question in this way that, What is the 'relation' of the other end of string (whether it is fixed or free) with the acceleration of m1 in downwards direction?

 

[Kaneki123]

I think Sophiecentaur is correct in saying the OP is getting hung up on what is intuitively obvious to him or her. It’s something I routinely see in teaching introductory physics. I also routinely see students get frustrated to the point of walking away thinking physics is inaccessible or bogus.

Kaneki123: see if this line of reasoning works. Suppose, in your drawing, m1 is initially held in place and that there is no significant friction acting. Now apply the first law to m₂: it is a body at rest remaining at rest, so the net force acting on it is zero. As you correctly noted, those forces are the tension pulling up and the weight of m₂ pulling down, so the tension equals the weight. (More precisely, it has the same magnitude but is oppositely directed.)

Note that the second law leads to the same result. The acceleration of m₂ is zero s the sum of the forces acting on m₂ is zero and weight equals tension. Both approaches support your premise in your original post when m₂ is not accelerating.

What happens if you stop holding m₁? Assuming m₁ < m₂, both masses go from rest to moving. Both accelerate. If m₂ accelerates down, the net for down is no longer zero and the weight must be greater than the tension.

This leads to your initial question: doesn’t the third law say the tension and weight have to be of the same magnitude? You are left with two possibilities: the law is wrong or your interpretation is not correct. Check the latter by carefully applying the third law. I like this phrasing:

When one body applies a force to a second body, the second body applies a force of the first that is the same size but in the opposite direction.

What are the two bodies pulling on each other? (Selecting the two is probably where you are getting off course.)

Okay i have uploaded another diagram...According to my understanding, ALL of the forces (represented by arrows) in the diagram are equal in magnitude, whether the body M is accelerating or not...Correct me with some explanation...

 

[jbriggs444]

According to my understanding, ALL of the forces (represented by arrows) in the diagram are equal in magnitude, whether the body M is accelerating or not...Correct me with some explanation...

That is not correct. Newton's second law applies. If the block is accelerating then the sum of the external forces on the block is

a. Zero
b. Non-zero.

Pick one.

 

[Fewmet]

The two bodies pulling on each other are...the mass m1 pulling on string in downward direction and the reactional tension in string pulling m1 in upwards direction..(correct me if i am wrong)...

You are correct. The third law tells you those two pulls are equal in size and opposite in direction.

Note you could also say the two bodies are the Earth pulling down on m₂ gravitationally and m₂ pulling up on the Earth gravitationally. Or the string pulling on m₁ and m₁ pulling back on the string. Or ignore the physicality of the string and say m₁ pulls on m₂ and m₂ pulls equally hard back on m₁. (It is not intuitively obvious, but it is important that the tension is uniform in the string. The string pulls equally hard on m₁ and m₂)

Maybe its better if i put my question in this way that, What is the 'relation' of the other end of string (whether it is fixed or free) with the acceleration of m1 in downwards direction?

Are you asking about the tension in the other end of the string? The acceleration of the other end of the string? Does "fixed or free" mean "attached to an immovable wall or attached to a movable mass", or perhaps attached to nothing?

Whatever your intent, it can be answered by running the same kind of analysis I gave above. What I sketched out in post #16 let's you see if the tension in the string equals the weight of m₂ or if it is less than the weight. Does that let you move forward?

 

[Kaneki123]

That is not correct. Newton's second law applies. If the block is accelerating then the sum of the external forces on the block is

a. Zero
b. Non-zero.

Pick one.

of course it is non-zero...meaning the block has weight which ''dominates'' the opposite direction tension in string...Why does the weight dominates the tension or why does acceleration occur in the block?

 

[Kaneki123]

You are correct. The third law tells you those two pulls are equal in size and opposite in direction.

Note you could also say the two bodies are the Earth pulling down on m₂ gravitationally and m₂ pulling up on the Earth gravitationally. Or the string pulling on m₁ and m₁ pulling back on the string. Or ignore the physicality of the string and say m₁ pulls on m₂ and m₂ pulls equally hard back on m₁. (It is not intuitively obvious, but it is important that the tension is uniform in the string. The string pulls equally hard on m₁ and m₂)Are you asking about the tension in the other end of the string? The acceleration of the other end of the string? Does "fixed or free" mean "attached to an immovable wall or attached to a movable mass", or perhaps attached to nothing?

Whatever your intent, it can be answered by running the same kind of analysis I gave above. What I sketched out in post #16 let's you see if the tension in the string equals the weight of m₂ or if it is less than the weight. Does that let you move forward?

Okay let me state my question like this...The B1 in diagram is at rest because its weight is ''balanced'' by tension in string...The B2 is not balanced because its weight somehow dominates the tension in the string...If this is related to the ''fixed'' end (as in B1) and the ''fixed to body with lower mass''(as in B2) then please explain?If not then please explain how the weight can ''dominate'' the tension in string when it clearly does not do so in B1?

 

[jbriggs444]

of course it is non-zero...

[referring to the net external force on dangling mass M in the diagram posted in #20]
Good. Now you have also claimed that...

According to my understanding, ALL of the forces (represented by arrows) in the diagram are equal in magnitude, whether the body M is accelerating or not

So let us back up. The net external force on dangling mass M is non-zero. What forces on the diagram contribute to that net external force?

 

[sophiecentaur]

Okay i have uploaded another diagram...According to my understanding, ALL of the forces (represented by arrows) in the diagram are equal in magnitude, whether the body M is accelerating or not...Correct me with some explanation.

This is becoming very hard work.
That diagram is of no use if you don't say what's on the other end of the string. Without any more information, we can only say that there is no reason to suggest that T is the same magnitude as the Weight force of M. I realize that this is what you firmly believe but it is just wrong and makes no sense, in fact.
I suggest that you read Newton;s First and Second laws of motion (modern / textbook versions will make it easier to understand). Then apply what you have read to your suggested situation. Notice I have not suggested that you should look at Newton's Third Law, which has nothing to do with this diagram.

 

[sophiecentaur]

I have a feeling that the OP believes that, if he re-states his ideas in an appropriate way, we will all say 'Yes, you are right and we are all wrong".
This is just not going to happen because what he is saying dos not belong in conventional Physics. I wish he would start to find how he is wrong and not find reasons to show he is in fact, right and the rest of Science is out of step. (This is not an uncommon phenomenon in PF).

 

[Kaneki123]

[referring to the net external force on dangling mass M in the diagram posted in #20]
Good. Now you have also claimed that...So let us back up. The net external force on dangling mass M is non-zero. What forces on the diagram contribute to that net external force?

The weight of mass M...

 

[jbriggs444]

The weight of mass M...

What other force contributes to the total external force on the dangling mass M?

 

[A.T.]

''if the mass accelearates brings me to original question as to why the mass accelearates?

Mechanical systems aren't analysed by "why" questions, but by applying Newtons Law and solving the resulting equations.