What Is the Tension in the String Connecting Two Blocks of Equal Mass?

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The discussion focuses on calculating the tension in a string connecting two blocks of equal mass, each denoted as 'm', when a constant force 'F' is applied to one block. The initial incorrect approach led to the conclusion that tension 'T' equals 'F/3', while the correct calculation shows that 'T' equals 'F/2'. Participants emphasized the importance of drawing free-body diagrams and applying Newton's second law (F=ma) to each block separately to arrive at the correct solution.

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Homework Statement


Two blocks of equal mass m are tied to each other through a light string. One of the blocks is pulled along the line joining them with a constant force F. Find the tension in the string joining the blocks.

Homework Equations


I tried doing this by this method.
(I don't know how to draw free body diagram here. It is just two bodies connected by string)
I got
F+T=ma
F-T=2ma

Hence, F=3ma/2
I got T=F/3
But the answer given at the back is F/2

Can someone help me with this?
Where am I going wrong?
 
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astrophysics12 said:

Homework Statement


Two blocks of equal mass m are tied to each other through a light string. One of the blocks is pulled along the line joining them with a constant force F. Find the tension in the string joining the blocks.

Homework Equations


I tried doing this by this method.
(I don't know how to draw free body diagram here. It is just two bodies connected by string)
I got
F+T=ma
F-T=2ma

Hence, F=3ma/2
I got T=F/3
But the answer given at the back is F/2

Can someone help me with this?
Where am I going wrong?

You really need to draw a free-body diagram. You have the "front" block that has a force F pulling it in one direction and a force T pulling it in the opposite direction. You have the "rear" block that only has a force T pulling it. The accelerations are the same. So just apply F_{total} = m a for each block separately.
 
stevendaryl said:
You really need to draw a free-body diagram. You have the "front" block that has a force F pulling it in one direction and a force T pulling it in the opposite direction. You have the "rear" block that only has a force T pulling it. The accelerations are the same. So just apply F_{total} = m a for each block separately.
I did draw a free body diagram. I don't know how to draw it here. Is there some way where I can upload it?
I will do it again.
Thanks
 
astrophysics12 said:
I did draw a free body diagram. I don't know how to draw it here. Is there some way where I can upload it?
I will do it again.
Thanks

No, that's okay. Just write down "F = ma" for each block. Remember that the front block has two forces acting on it--whatever is pulling the blocks (force "F") and the tension ("T", acting in the opposite direction) The second block only has "T" acting on it. So write down your two F=ma equations and post them.
 
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stevendaryl said:
No, that's okay. Just write down "F = ma" for each block. Remember that the front block has two forces acting on it--whatever is pulling the blocks (force "F") and the tension ("T", acting in the opposite direction) The second block only has "T" acting on it. So write down your two F=ma equations and post them.
I got it.
Thank you.
I made the mistake of including F even for the second blockSystem: Second Block(behind)
T=ma
System: First Block(Front)
F-T=ma
F=ma+T
F=2ma
F/2=ma
Hence, T=F/2

Thanks, again
 

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