How Does Friction and Acceleration Affect Tension in a Three-Block System?

In summary: My point is that the answer should be the same for a wide range of friction coefficients. Unless my intuition is failing me here, only the static case would possibly give a different result.Unless you have a specific reason to believe otherwise, you should assume that the coefficient of friction is zero.
  • #1
rudransh verma
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Homework Statement
Three blocks are connected as shown. If m1=10 kg, m2=20kg, m3=30kg, find tension T1 given that F=60N
Relevant Equations
F=ma
Since no body accelerates so net force is zero. Force on each mass is zero. T1 and T2 both are 60N.
Edit: since there is a force applied so there is acceleration on friction less surface.
 

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  • #2
rudransh verma said:
Homework Statement:: Three blocks are connected as shown. If m1=10 kg, m2=20kg, m3=30kg, find tension T1 given that F=60N
Relevant Equations:: F=ma

Since no body accelerates so net force is zero. Force on each mass is zero. T1 and T2 both are 60N.
But it’s not right!
How do you know there is no acceleration? Why have you assumed this?
 
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  • #3
Did you make free body diagrams ? :biggrin:

##\ ##
 
  • #4
BvU said:
Did you make free body diagrams ? :biggrin:

##\ ##
Yeah🤓
Steve4Physics said:
How do you know there is no acceleration? Why have you assumed this?
because it’s not mentioned in question.
 
  • #5
rudransh verma said:
Yeah🤓

because it’s not mentioned in question.
Neither is it stated that there is no acceleration. Maybe that's because you can work it out from the given facts. Let me see, there is only one external force with a horizontal component...
 
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  • #6
rudransh verma said:
BvU said:
Did you make free body diagrams ? :biggrin:

##\ ##

Yeah🤓
Did you notice the plural form? There are 3 free body diagrams to make, that will give you 3 equations with 3 unknowns.
 
  • #7
jack action said:
Did you notice the plural form? There are 3 free body diagrams to make, that will give you 3 equations with 3 unknowns.
One only needs to know the common acceleration of the masses to find ##T_1## which is the only horizontal force acting on ##m_1##.

Of course, since friction is not mentioned or implied, one has to assume that it doesn't exist. To be fair though, how can the OP, who is not familiar with problem statement conventions, deduce that acceleration (which is not mentioned) must be non-zero, yet friction (which is also not mentioned) must be zero? Given what is omitted from the statement of the problem, it is entirely possible to have zero acceleration if the coefficient of static friction is large enough, or the coefficient of kinetic friction has just the right value. Saying that the contact is frictionless costs nothing but it provides clear indication to someone, who doesn't know how to read between the lines, that the acceleration cannot be zero.

Also, if one looks at the photo posted by the OP there is a cryptic partial statement at the bottom:
"Which of the following statements abou
friction is true?
(1) coefficient of friction can be reduced t
zero."

What is that all about?

To @rudransh verma : Can you post the entire problem and not just a piece of it?
 
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  • #8
kuruman said:
deduce that acceleration (which is not mentioned) must be non-zero, yet friction (which is also not mentioned) must be zero?
what? Please look at OP again.
 
  • #9
kuruman said:
Which of the following statements abou
friction is true?
(1) coefficient of friction can be reduced t
zero."

What is that all about
Another question!
 
  • #10
Steve4Physics said:
How do you know there is no acceleration? Why have you assumed this?
You are right. I don’t know why I did that. I will try again now.
 
  • #11
kuruman said:
To be fair though, how can the OP, who is not familiar with problem statement conventions, deduce that acceleration (which is not mentioned) must be non-zero, yet friction (which is also not mentioned) must be zero?
Friction does not need to be zero.
 
  • #12
Orodruin said:
Friction does not need to be zero.
That's exactly my point. If the friction does not need to be zero, the acceleration can be zero or non-zero.
 
  • #13
kuruman said:
That's exactly my point. If the friction does not need to be zero, the acceleration can be zero or non-zero.
My point is that the answer should be the same for a wide range of friction coefficients. Unless my intuition is failing me here, only the static case would possibly give a different result.
 
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  • #14
kuruman said:
One only needs to know the common acceleration of the masses to find ##T_1## which is the only horizontal force acting on ##m_1##.
You still need at least 2 free body diagrams:
  1. One to find the acceleration;
  2. One to find ##T_1##.
That is assuming that you are experienced enough to intuitively understand how the masses relate to one another under the Force ##F##.

My advice to a beginner: Do the third free body diagram to help you find ##T_2##, even though no one asks for it. It will at the very least validate your assumption, ie. what you know intuitively.

kuruman said:
To be fair though, how can the OP, who is not familiar with problem statement conventions, deduce that acceleration (which is not mentioned) must be non-zero,
Why would one assume the acceleration is non-zero? Solving the problem doesn't imply the acceleration must be non-zero.

How would one do a free body diagram where there are no fixed points without including the ##ma## vector somewhere?
kuruman said:
yet friction (which is also not mentioned) must be zero?
[SARCASM]Have you ever thought that maybe there is friction, just that there is no gravity?[/SARCASM]

It doesn't mention aerodynamic drag, or that the masses are made of ferrous material in the presence of a magnetic field either. I tend to assume they're all inexistent as well.

My point is: There is no need to make the question more complicated than it is.
 
  • #15
jack action said:
You still need at least 2 free body diagrams:
  1. One to find the acceleration;
  2. One to find ##T_1##.
That is assuming that you are experienced enough to intuitively understand how the masses relate to one another under the Force ##F##.
I guess our opinions differ about the need for a free body diagram in this particular case. The way I see it, a single external force acting on a system is also the net force. In that case, there is no need for an FBD the purpose of which is to sort out force components, find the resultant net force and write down Newton's second law often in two dimensions. With a single external force, one can directly write down ##\vec F=\vec F_{\text{net}}=m_{\text{sys }}\vec a## and be done with it. Of course here the equation needs to be written twice, once for the system of three masses and then again for ##m_1##. As you say, there is no reason to make this more complicated than it is.
jack action said:
Why would one assume the acceleration is non-zero? Solving the problem doesn't imply the acceleration must be non-zero.
Perhaps I was not clear about what I meant. I tried to look at this from a beginner's point of view: Acceleration is not mentioned. Lack of friction is not mentioned. Air resistance is not mentioned. Gravity is not mentioned. A whole lot of other things are not mentioned. Which am I justified in including as relevant to this question and which can I safely ignore?
jack action said:
[SARCASM]Have you ever thought that maybe there is friction, just that there is no gravity?[/SARCASM]
[STRAIGHT FACE]I have not.[/STRAIGHT FACE]

If there is (are) response(s) to this post, please use PM to keep the focus on assisting the OP.
 
  • #16
rudransh verma said:
Another question!
I suppose you are right. But the continued indentation is suggesting otherwise. What textbook (solutions manual?) is this ?

So: no friction in this thread ?

##\ ##
 
  • #17
BvU said:
indentation is suggesting otherwise. What textbook (solutions manual?) is this ?

So: no friction in this thread ?
T1 = 10N
 
  • #18
As @Orodruin has previously pointed out, the tension, T1 does not depend upon the coefficient of friction provided that this is not the static case.
(It should be noted that the only other requirement for this to be true is that the coefficient of friction be the same for all of the blocks.)

The coefficient can be zero. It can be small. It can be large. It can be such that the acceleration is zero. In any case, if the blocks are sliding on the surface, any of these scenarios gives the same value for T1 .
 
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  • #19
The point of this problem as I see it is to be able to make assumptions and test if those assumptions make sense or not. Is there friction or not and what kind of friction? Is there acceleration or not and under what circumstances? One has to make an assumption and test if it is consistent and then make another if the first doesn't lead to a consistent answer.

I see five simple cases to consider.

1) No friction. This is easily solvable.
2) Static friction.
i) The static friction on each block is less than the maximum allowable by the materials involved.
ii) The static friction is at the maximum possible value.
3) Kinetic Friction.
i) The acceleration is zero but the velocity is not.
ii) Both the acceleration and velocity are non zero.

Each case is either solvable with the given information or it is unsolvable. All solvable cases lead to the same answer for the tension of ##T_1## but under different assumptions. The unsolvable cases just tell you what you would need to solve it in that case but since that information is not provided you cannot solve it but can assume the problem is not one of those cases. In other words, you still learn something. I would encourage you to try and work each case one by one to fully understand how to approach such a problem.

BTW, It would be nice to see the whole problem.
 
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1. What is the concept of "Three blocks and a force"?

The concept of "Three blocks and a force" refers to a physics problem in which three blocks are connected by strings and a force is applied to one of the blocks. The problem involves calculating the acceleration and tension in the strings.

2. How do you approach solving a "Three blocks and a force" problem?

The first step in solving a "Three blocks and a force" problem is to draw a free body diagram for each block, showing all the forces acting on them. Then, write down the equations of motion for each block and solve them simultaneously to find the acceleration and tension in the strings.

3. What are the key equations used in solving a "Three blocks and a force" problem?

The key equations used in solving a "Three blocks and a force" problem are Newton's second law of motion (F=ma), the equations of motion for each block (ΣF=ma), and the equations for tension in the strings (T=ma).

4. How does the mass of the blocks affect the solution to a "Three blocks and a force" problem?

The mass of the blocks affects the solution to a "Three blocks and a force" problem by changing the acceleration and tension in the strings. The larger the mass of the blocks, the greater the force needed to accelerate them and the greater the tension in the strings.

5. Can the "Three blocks and a force" problem be solved using only algebra?

Yes, the "Three blocks and a force" problem can be solved using only algebra. By writing down the equations of motion for each block and solving them simultaneously, the values for acceleration and tension can be found without the need for calculus or other advanced mathematical techniques.

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