A uniform rod of length $L$ and mass $M$ has been placed on a rough horizontal surface. The horizontal force $F$ applied on the rod is such that the rod is just in the state of rest. If the coefficient of friction varies according to the relation $\mu = Kx$ where $K$ is a $+$ ve constant. Then the tension at mid point of rod is
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$Let\,us\,consider\,an\,element\,dx\,a\,distance\,x$
$from\,one\,of\,the\,ends\,of\,the\,ends\,of\,the\,rod.$
$dF = \mu (x) \times \frac{{Mg}}{L} \times dx = \frac{{Mg}}{L}kxdx$
$F = \frac{{Mgk}}{L}\left( {\frac{{{x^2}}}{2}} \right)L - 0 = \frac{{MgkL}}{2}$
$k = \frac{{2F}}{{MgL}}$
$Tx = \frac{F}{4}$
$Thus\,the\,tension\,at\,midpo\operatorname{int} \,of\,rod\,is\,Tx = \frac{F}{4}.$
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