A uniform rod of length L and mass M has been | vedclass

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Question

$Let\,us\,consider\,an\,element\,dx\,a\,distance\,x$

$from\,one\,of\,the\,ends\,of\,the\,ends\,of\,the\,rod.$

$dF = \mu (x) 	imes \frac{{Mg}}{L

Vedclass · Accepted Answer

$F/4$

Vedclass · Answer

$Let\,us\,consider\,an\,element\,dx\,a\,distance\,x$

$from\,one\,of\,the\,ends\,of\,the\,ends\,of\,the\,rod.$

$dF = \mu (x) 	imes \frac{{Mg}}{L} 	imes dx = \frac{{Mg}}{L}kxdx$

$F = \frac{{Mgk}}{L}\left( {\frac{{{x^2}}}{2}} ight)L - 0 = \frac{{MgkL}}{2}$

$k = \frac{{2F}}{{MgL}}$

$Tx = \frac{F}{4}$

$Thus\,the\,tension\,at\,midpo\operatorname{int} \,of\,rod\,is\,Tx = \frac{F}{4}.$

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