A Flexible chain between two hooks (statics)

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SUMMARY

The discussion centers on the physics problem involving a flexible chain weighing 41.0 N suspended between two hooks, with each hook exerting a force at an angle of θ = 41.5° to the horizontal. The user initially calculated the force exerted by each hook as Rh = Tsinθ = Wsinθ = 27.17 N, but recognized an error in their approach. The correct method requires considering the total weight supported by both hooks, leading to the conclusion that each hook supports half the weight, resulting in a force of 20.5 N per hook.

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Homework Statement


A flexible chain weighing 41.0 N hangs between two hooks located at the same height. At each hook, the tangent to the chain makes an angle θ = 41.5° with the horizontal.
(a) Find the magnitude of the force each hook exerts on the chain.
(b) Find the tension in the chain at its midpoint.

Homework Equations


Fx = 0
Fy = 0
M = 0

The Attempt at a Solution


Let the magnitude of force each hook exerts be Rh.
Then Rh = Tsinθ = Wsinθ = 27.17 N
But this is wrong. What is my mistake in this calculation?
 

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There are two hooks and assuming the weight is distributed evenly then how much of the weight is supported by each hook?
 
I took the weight as 41/2 but the calculation is still wrong..
 

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