A uniform rope of mass M and length L is fixe | vedclass

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Question

(b)

For the lower part

Mass of the lower part is $m^{\prime}=$ Mass per unit length $	imes$ length of lower part

$=\frac{M}{L}(L-I)$

So,

Vedclass · Accepted Answer

$\frac{M g}{L}(L-l)$

Vedclass · Answer

(b)

For the lower part

Mass of the lower part is $m^{\prime}=$ Mass per unit length $	imes$ length of lower part

$=\frac{M}{L}(L-I)$

So, Using $\vec{F}_{	ext {net }}=m \vec{a}$

here $\vec{a}=0$

$T-\frac{M}{L}(L-I) g=0$

$T=\frac{M}{L}(L-I) g$

A uniform rope of mass $M$ and length $L$ is fixed at its upper end vertically from a rigid support. Then the tension in the rope at the distance $l$ from the rigid support is $x$

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