Find tension in ropes connecting 3 masses being pulled on a surface

In summary: The tension in the string is the minimum necessary to maintain the length of the string. In summary, the tension in the string is the minimum necessary to maintain the length of the string.
  • #1
Mohmmad Maaitah
87
19
Homework Statement
As picture
Relevant Equations
sum of F=ma
This is the problem and my answer below it.
my friends said it's 250 N and I say it's 100 N based on what I solved what am I missing ??
1683483599280.png

my try:
1683483617696.png
 
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  • #2
Note: In my picture the weird word is in Arabic and it means **Total**.
 
  • #3
@Azim
I did analysis that and I still get same result!
what am I missing?
 
  • #4
Mohmmad Maaitah said:
@Azim
I did analysis that and I still get same result!
what am I missing?
I accidentally deleted my initial post so here's the repost:
Ask these questions to yourself:
1) How are the force ##F## and the tension ##T_1## related? (Hint: Consider the relation between ##F## and ##T_2## and then ## T_2## and ##T_1##)
2) Of the two masses involved with ##T_1## which of them is "pulling" and thus what should be the ##F=ma## analysis on that mass?

I think I made a mistake in reading your question, I thought you were the one who answered 250N. Going back over it, I think you're right, and your friends are wrong!
 
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  • #5
Mohmmad Maaitah said:
Homework Statement: As picture
Relevant Equations: sum of F=ma

This is the problem and my answer below it.
my friends said it's 250 N and I say it's 100 N based on what I solved what am I missing ??
View attachment 326103
my try:
View attachment 326104
Your answer is correct.

Although there is a more direct way to get the answer, your analysis is fine.
 
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  • #6
SammyS said:
Your answer is correct.

Although there is a more direct way to get the answer, your analysis is fine.
How could I get it in more direct way? I am unable to do it without finding all Tensions.
 
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  • #7
Mohmmad Maaitah said:
How could I get it in more direct way? I am unable to do it without finding all Tensions.
What is the Free Body Diagram on the last block and what is the only Force acting on it? What must be the value of the Force acting on it such that it moves at 10 ## m/s^2##?
 
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  • #8
Mohmmad Maaitah said:
How could I get it in more direct way? I am unable to do it without finding all Tensions.
The three bodies forming the system have the same acceleration.
As m1 and a1 are unique values, T1, which is the product of both, must be unique as well.
 
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  • #9
Okay guys I get it, here is a question in general about laws of motions?
do all connected objects on same surface with different masses (no friction) have same acceleration and velocity?
 
  • #10
Mohmmad Maaitah said:
Okay guys I get it, here is a question in general about laws of motions?
do all connected objects on same surface with different masses (no friction) have same acceleration and velocity?
In general, no. But in this question you can suppose that the strings are inextensible, and since the motion is all in one direction there is no rotation.

The remaining question is whether a string might go slack. That would require a trailing block to accelerate more than the one in front. To prove that won't happen, you would need to apply the principle that the tension in an inextensible string is the minimum necessary to maintain its length. The is essentially the same as the principle that the normal force between two bodies in contact has the minimum magnitude to prevent interpenetration.

Or you can just apply common sense.
 
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  • #11
Mohmmad Maaitah said:
Okay guys I get it, here is a question in general about laws of motions?
do all connected objects on same surface with different masses (no friction) have same acceleration and velocity?
In a system, If the bodies are rigidly connected to each other, we can consider the system as one composite body. As a result, the acceleration of the bodies in this system is the same.

Another way to find ##T_1##:
Just use Newton's second law for the body that has 10 kg mass.
The net force on the 10 kg body is ##T_1##(since there is no friction force).
##T_1=m_1a_{system}=10*10=100N##

##m_1a_{system}## is the net force that acts on the body.
 
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