# Question #2546b

Apr 6, 2017

See below

#### Explanation:

The question only makes sense if we take into account the mass of the string.

The small element $\mathrm{dx}$ will have mass: $\mathrm{dm} = \frac{M}{L} \mathrm{dx}$, assuming the string is of uniform density.

The FBD for the small element requires:

$\sum m a t h b f F = m m a t h b f a$

$\implies T + \mathrm{dT} - T = \mathrm{dm} \setminus a$

$\implies \mathrm{dT} = \frac{M}{L} \mathrm{dx} \setminus a$

$\implies T = \frac{M}{L} a x + C$

$T \left(0\right) = 10 \implies C = 10$

$T \left(L\right) = 20 \implies 20 = \frac{M}{L} a L + 10$

$\implies T = \frac{10}{L} x + 10$

Which is simply a linear increase along the string which is what intuitively one would expect.