Physics, 15.04.2020 03:45 snowww5492
A crate with a mass of 187.5 kg is suspended from the end of a uniform boom. The upper end of the boom is supported by the tension of 3206 N in a cable attached to the wall. The lower end of the boom pivots at the location marked X on the same wall. Calculate the mass of the boom.
Answers: 3
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![T * sin (\alpha + \beta ) * L -[\frac{m_bg }{2} +mg ]* L cos \beta = 0](/tpl/images/0601/1667/d4be2.png)
![T * sin (\alpha + \beta ) * L = [\frac{m_bg }{2} +mg ]* L cos \beta](/tpl/images/0601/1667/51548.png)
the mass of the boom the subject of the formula ![m_b = \frac{2 *[\frac{T sin(\alpha +\beta ) }{cos \beta } -mg]}{g}](/tpl/images/0601/1667/f4d8b.png)
![m_b = \frac{2 *[\frac{3206 * sin (19.65 + 15.95)}{cos 15.95} - 187.5*9.8]}{9.8}](/tpl/images/0601/1667/46d73.png)
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A crate with a mass of 187.5 kg is suspended from the end of a uniform boom. The upper end of the bo...
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