Problem involving Torque and a Derrick from Feynman Exercises

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SUMMARY

The discussion focuses on solving a torque problem involving a derrick as presented in Feynman Exercises. The user initially derived the tension formula as $$T = \frac L x Wtan\theta$$ using torque principles. However, the book's answer includes an additional term, $$\frac W 2$$, which accounts for the weight of the boom in addition to the block's weight. The user successfully clarified the source of this term with assistance from other forum members.

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Homework Statement
2.17 A derrick is made of a uniform boom of length L and weight w, pivoted at its lower end, as shown in Fig. 2-15. It is supported at an angle with the vertical by a horizontal cable attached at a point a distance x from the pivot, and a weight W is slung from its upper end. Find the tension T in the horizontal cable.
Relevant Equations
##\tau = F_{\perp}r##
Work = Fd
I attempted to solve this problem by considering the torque caused by the perpendicular components of the tension and weight with respect to the derrick. $$ Tcos\theta x = Wsin\theta L$$ $$T = \frac L x Wtan\theta$$ Using the principle of virtual work I also arrived at the same answer by considering the derrick falling a distance y. $$Wsin\theta y = Tcos\theta \frac x L y $$The answer listed in the book is ##T = \frac L x (W + \frac W 2 )tan\theta ## and I can't figure out where the ##\frac W 2 ## term comes from. Thank you in advance for helping me.
 

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In addition to the weight W of the block, the boom has weight w.
 
TSny said:
In addition to the weight W of the block, the boom has weight w.
Ok I see now. Thanks for pointing that out.
 
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