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 when one block is pulled with a constant force. Initially, the user incorrectly derived the tension as T=F/3, while the correct answer is T=F/2. Participants emphasize the importance of drawing a free-body diagram to visualize the forces acting on each block. By applying Newton's second law separately to both blocks, the user eventually corrects their calculations. The final conclusion confirms that the tension in the string is indeed F/2.
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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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