Find tension in string between two blocks on horizontal surface

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The discussion centers on calculating the tension in a string between two blocks on a horizontal surface. The initial error was treating the two blocks as a single system without correctly assessing their individual normal forces. It is emphasized that while the blocks can be analyzed together, the total normal force must accurately reflect the weights of the individual blocks. The larger block's normal force differs from its weight, affecting the friction calculations. The conversation concludes with acknowledgment of the complexities introduced by the angle of the applied force.
I_Try_Math
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Homework Statement
Two blocks connected by a string are pulled across a horizontal surface by a force applied to one of the blocks, as shown below. The coefficient of kinetic friction between the blocks and the surface is 0.25. If each block has an acceleration of 2 m/s^2 to the right, what is the magnitude F of the applied force?
Relevant Equations
F = ma
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IMG_20240126_010206.jpg

Did I draw the free body diagram incorrectly by treating the two blocks as one system? My textbook claims the answer is 25 N.
 
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Your mistake is not exactly that you treat as one system but that you calculate the total normal force as equal to the total weight of the two bodies. Think again.

Hint : Each box has different normal force on it, in one it is equal to its weight but on the other it is NOT!!!
 
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Delta2 said:
Your mistake is not exactly that you treat as one system but that you calculate the total normal force as equal to the total weight of the two bodies. Think again.

Hint : Each box has different normal force on it, in one it is equal to its weight but on the other it is NOT!!!
If I understand correctly, the magnitude of the larger block's normal force will be different than its weight which will change the friction calculation. Thank you very much for your help on this question, I'll redo the problem soon.
 
I_Try_Math said:
the magnitude of the larger block's normal force will be different than its weight which will change the friction calculation.
That's true, but for the purpose of answering the question in post #1 you can treat the two blocks as a unit. You just have to get the total normal force right.
Answering the question in the thread title is a different matter.
 
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I_Try_Math said:
If I understand correctly, the magnitude of the larger block's normal force will be different than its weight which will change the friction calculation. Thank you very much for your help on this question, I'll redo the problem soon.
Yes. With a force of ##25N## at that angle, you might imagine that the first block is almost being pulled off the ground. It's a strange angle for the problem setter to choose.
 
Last edited:
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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