A Flexible chain between two hooks (statics)

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The discussion focuses on a problem involving a flexible chain weighing 41.0 N suspended between two hooks at the same height, with each hook making an angle of 41.5° with the horizontal. The user attempts to calculate the force exerted by each hook, initially assuming that the weight is evenly distributed and using the equation Rh = Tsinθ. However, the calculation of Rh as 27.17 N is identified as incorrect, prompting questions about the distribution of weight and the correct approach to the problem. The user seeks clarification on how to properly account for the forces and tensions involved in the scenario. The thread highlights the importance of correctly applying static equilibrium equations to solve for forces in systems with multiple supports.
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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..
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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