Net force on a mass when a string (massless) is cut

Question

In the below given figure, $m_1 = 5$ kg and $m_2 =2$ kg and $F=1$ N. We have to find the acceleration of the either block and also find with what acceleration will $m_1$ fall after the string breaks but the force "F" still acts on the mass $m_1.$ Given the rope and the pulley are massless and the friction between the rope and pulley is negligible.

I solved for the acceleration with which the blocks move by applying Newton's second law. But I'm confused about the part where we have to solve for after the string is cut. I believe that, after the string is cut (breaks), we have to take the force of tension the rope is applying on the block $m_1$ into consideration too, i.e, $\sum{F_{m_1}} = m_1g+1-T_{\text{by the above rope}}$. But the solution does not consider this ($T_{\text{by the above rope}}$) into account and just accounts for $m_1g$ and the $F$. The rope is indeed pulling the block before the string got cut and hence I think we have to consider it.

Answer 1

This is one-dimensional problem so I will not write vectors but only magnitudes with the appropriate direction (sign).

We write equations for the second Newton's law for the system as follows:

$$m_1 a_1 = w_1 + F_1 - T_1 \quad \text{and} \quad m_2 a_2 = -w_2 - F_2 + T_2$$

where $w = mg$ is the object weight, $F$ is an external force that acts on the object in the same direction as the weight, $T$ is the force with which rope pulls the objects in the direction opposite to the weight, and accelerations $a_1$ and $a_2$ act in the direction of $F_1$ and $T_2$, respectively. Note that the acceleration for both objects is the same $a_1 = a_2$.

From $F_1 = F_2$ which is given in the OP and from $T_1 = T_2$ which follows from the fact that the rope and the pulley are massless, the above equations are combined into

$$a \cdot (m_1 + m_2) = w_1 - w_2 = g \cdot (m_1 - m_2)$$

Finally, the acceleration is

$$a = g \cdot \frac{m_1 - m_2}{m_1 + m_2}$$

When:

• $m_1 = m_2$ then $a = 0$ which means there is no net force and the system is in equilibrium
• $m_1 > m_2$ then $a > 0$ which means the resultant force acts in the same direction as $F_1$
• $m_1 < m_2$ then $a < 0$ which means the resultant force acts in the opposite direction of $F_1$

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At the moment the string is cut, the rope becomes loose and there is no longer a tension force that pulls the objects, hence:

$$T_1 = 0 \quad \text{and} \quad T_2 = 0$$

and the system is reduced to:

$$a_1 = g+ \frac{ F_1}{m_1} \quad \text{and} \quad a_2 = g+\frac{ F_2}{m_2}$$

where $a_1$ and $a_2$ act in the direction of $g$.

Answer 2

I believe that, after the string is cut (breaks), we have to take the force of tension the rope is applying on the block $m_1$ into consideration too

No you don't.

When the string is cut there is no tension, and since it is massless, it is no longer present as a force or as a mass to be pulled. The two blocks then fall independently of one another per Newton's second law

$$a=\frac{F_{net}}{m}$$

Thus, for each block:

$$a_{1}=\frac{F+m_{1}g}{m_{1}}=\frac{F}{m_{1}}+g$$

$$a_{2}=\frac{F+m_{2}g}{m_{2}}=\frac{F}{m_{2}}+g$$

Hope this helps.