Calculating Tension Ratios for Equilibrium of Identical Bodies

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In the discussion about calculating tension ratios for identical bodies K, L, and M in equilibrium, the tension in rope S1 (T1) is determined to equal the weight P. The tension in rope S2 (T2) is debated, with initial thoughts suggesting it could be either 2P or 3P. Ultimately, it is concluded that T2 supports the combined weight of the two lower bodies, leading to a value of 2P. The correct ratio of T1 to T2 is established as 1/2, aligning with the textbook answer, and the equilibrium conditions at the nodes are emphasized for accurate calculations. Understanding the angles and the configuration is crucial for solving such tension problems.
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Homework Statement



Identical K,L and M bodies with a weight of P are in equilibrium . If the tension in the rope S1 is T1 and S2 is T2, then what is the ratio T1/T2? (The picture is below)

Homework Equations


The Attempt at a Solution



T1 equals P. But I am stuck with T2. I cannot decide whether it is 2P or 3P. Do we have to add the weight of K to T2? If we add, it will be 3P.
 

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S2 is the uppermost vertical string and so must support all of the weight. Its tension will be 3P.
 
Then the answer is 1/3.But according to my book the answer is 1/2.
 
the pulley is attached to a rigid support.T2 has a weight of 2P
 
Let us call A the node above K and B the node above L. Let us assume that the angle between a horizontal line and the rope connecting A to B is 45 degrees (without this value you cannot compute a solution). Then you only have to write the equilibrium conditions at nodes A (first) and B (second).
The ratio is 1/2.
 
Many Thanks
 
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