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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.

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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.

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  • $\begingroup$ when the string breaks it goes loose and there's no tension in it $\endgroup$ Dec 14, 2021 at 10:40

2 Answers 2

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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 $F_1$

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 $g$.

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  • $\begingroup$ "At the moment the string is cut, there is no longer a tension force that pulls the object." What I don't get is, that after the string is cut, just before it was cut, there was a force pulling it up, so I think that must remain as it is even after being cut. The analogy I would like to give is simply how a block which was at rest, would just remain to translate after applying a force (in an inertial frame). The force was the embodiment in this case. So why not similarly here? $\endgroup$
    – Floatoss
    Dec 14, 2021 at 11:49
  • $\begingroup$ Imagine you are pulling a cart with constant (horizontal) force, whereas friction can be neglected. There is a net external force acting on the cart in horizontal direction and the cart will accelerate which means it's velocity will be linearly increasing. The moment you stop pulling, there is no longer net external force to the object. The first Newton's law states the object will remain in equilibrium, which does not mean velocity drops to zero, only the acceleration drops to zero (instantaneously) and the velocity stops increasing and remains the same. $\endgroup$ Dec 14, 2021 at 15:15
  • $\begingroup$ Oh yes, it makes complete sense to me now. Thank you Marko. $\endgroup$
    – Floatoss
    Dec 15, 2021 at 5:19
  • $\begingroup$ @Floatoss If one of the (two) answers helped you, please consider accepting the answer to close the topic. Otherwise, the commumity will feel your question still needs attention. $\endgroup$ Dec 15, 2021 at 6:06
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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.

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