Two masses connected in series with two massless strings are hanging from a support as shown in the figure. Find the tension in the upper string
Show Hint
To quickly find tension in a supported vertical line segment, mentally cut the line at that point and identify all the masses that would fall down . The tension force must be equal to the total weight of everything below that point. For the top string, both $m_1$ and $m_2$ are supported, so the tension is total mass ($m_1+m_2$) times $g$.
Concept:
The tension in any string in a vertical hanging system must support the total weight of all the masses attached below that specific string. This is derived from the equilibrium condition $\sum F = 0$.
Step 1: {Define the system for the upper string.}
To find the tension in the upper string, we can consider $m_1$ and $m_2$ together as a single composite system.
$$\text{Total Mass } (M) = m_1 + m_2$$
Step 2: {Set up the equilibrium equation for the total mass.}
The upward force is the tension $T_{upper}$ and the downward force is the total weight:
$$T_{upper} - Mg = 0$$
Step 3: {Solve for the tension.}
$$T_{upper} = Mg$$
Substituting $M = m_1 + m_2$:
$$T_{upper} = (m_1 + m_2)g$$
Step 4: {Alternative verification using individual FBDs.}
For $m_2$: $T_{lower} = m_2g$
For $m_1$: $T_{upper} = m_1g + T_{lower}$
$$T_{upper} = m_1g + m_2g = (m_1 + m_2)g$$
