Question

A thin rod of length $l$ and mass $m$ is suspended horizontally by two vertical strings, A & B, one attached at each end of the rod. The density of the rod is given by $\rho \left(x\right)={\rho }_0\left(x/I\right)3$ corresponds to the position of string A. Give your answer in terms of $m,l,$ and $g$. What is the tension in string B?

• A. $\frac{3mg}{5}$
• B. $\frac{mg}{5}$
• C. $\frac{6mg}{5}$
• D. $\frac{4mg}{5}$

Answer 1

Answer: (D)

$\frac{4mg}{5}$

Mass of the rod, $m={\int }_0^{l}p\left(x\right)dx={\int }_0^l\frac{p_0{x}^3}{l^3}dx=\frac{p_{0}l}{4}$
Center of mass of rod$=\frac{1}{m}{\int }_0^{l}xp\left(x\right)dx=\frac{1}{m}{\int }_0^l\frac{p_0{x}^4}{l^3}dx=\frac{\frac{p_0{l}^2}{5}}{\frac{p_{0}l}{4}}=\frac{4l}{5}$
Let $T$ be the tension on the string at B, so $Tl=mg\frac{4l}{5}$
$T=\frac{4mg}{5}$