Tricky spring balance

Relevant wiki: Constraint Relation

After drawing FBD of the two blocks you will get two equations which after solving you will get acceleration and tension in the lower string. $\begin{array}{c}~10g - T = 10a \\ \space T - 5g = 5a \\ \hline \\ \boxed{a = \dfrac{g}{3}} \quad, \quad \boxed{T = \dfrac{20g}{3}} \\ \end{array}$ Assuming that the pulley is mass-less, tension in the upper string will be twice the tension in lower string, $\boxed{T_{eq} = 2T = \dfrac{40g}{3}}$ .

The reading on a Spring Balance shows the mass acting on it. But here we got tension $T_{eq}$ which is equal to weight acting on spring balance. But, we know that $W = mg \implies \dfrac{40g}{3} = mg$

$\implies \boxed{{\color{#20A900}m = \dfrac{40}{3}~kg}\quad {\color{#3D99F6} \approx 13.333~kg}} \qquad \qquad$

Call the two hanging masses $m_1, m_2$ . The standard analysis of the Atwood machine shows that they accelerate at $a_{1,2} = \pm \frac{m_1-m_2}{m_1+m_2} g.$ Thus their center of mass accelerates at a rate $a_{CM} = \frac{m_1a_1+m_2a_2}{m_1+m_2} = \left(\frac{m_1-m_2}{m_1+m_2}\right)^2\:g.$ Apply Newton's Second Law to the system of the two masses plus the pulley; the two forces acting are the upward tension from the scale $F_s$ and the downward force of gravity $F_g = (m_1+m_2)g$ , so that $F_s = F_g - F_{\text{net}} = (m_1+m_2)g - (m_1+m_2)a_{CM}$ and division by $g$ gives the reading of the scale, $\frac{F_s}g = (m_1+m_2)\left(1 - \left(\frac{m_1-m_2}{m_1+m_2}\right)^2\right) = \frac{4m_1m_2}{m_1+m_2}.$ In this case it is $\frac{4\cdot 10\cdot 5}{10+5} = \frac{200}{15} = \frac{40}{3} = \boxed{13.33}\ \text{kg}.$

Spring will measure 2T as the weight, where, T is the tension in the string attached to each weight. The equations are( g is acceleration under gravity, a is acceleration of each mass):

(1) T=5g+5a,

(2) 10g=T+10a

Solving, 2T=(40/3)g,

Answer=(40/3)=13.333.

The net force downward is only 5. So the 10kg block is falling at 5/15=1/3 of its "usual" rate. So the spring will only feel 2/3 of what it would if this weight were at rest. On the other side, the 5kg block is rising at an extra 1/3. So the spring will feel 4/3 of what it would if this block were at rest.

Total felt by the spring: 10(2/3)+5(4/3)=40/3=13.3333