2
$\begingroup$

  1. Three balls of mass 10 kg, 20 kg and 10 kg are hanging by a massless string and are connected by massless springs as shown in the figure below. Initially the system is in equilibrium and all the objects are at rest.

    If the string at the top snaps suddenly, what is the acceleration of the topmost ball at that instant of time? [In the following, g denotes the acceleration due to gravity]

    (a) $0\quad$ (b) $2g\quad$ (c) $g\quad$ (d) $4g\quad$ (e) $3g$

My conceptual understanding is rather not good although I can give couple of attempts.

Attempt:

  1. Answer: $g$, since the only external force acting is gravity. But I think the spring attached to two masses might exert force.

  2. Answer: $4g$,
    taking the bottom mass, $10g = -k(x_2-x_1)$
    the mass in the middle, $20g = -kx_1+k(x_2-x_1)$
    adding both gives me $30g = -kx_1$

    at the topmost ball,
    $$T-10g-30g = -10g\\ T=0\quad g=4g$$ [$T$ tension in the string, $x_1$ displacement of the spring between topmost mass and mass in the middle, $x_2$ displacement of the spring between mass in the middle and bottom most]

Please clarify the concept that I might be mixing up and give the solution if my attempts are not correct.

$\endgroup$
  • $\begingroup$ Do you know slinky experiment ? $\endgroup$ – Creepy Creature Mar 8 '17 at 20:36
  • $\begingroup$ I only have a slight idea about it. Why do you ask? $\endgroup$ – rahul rj Mar 9 '17 at 6:30
4
$\begingroup$

Take it step by step.

The clincer is that the system is at equilibrium in the beginning.Hence all the forces on all the balls are balanced. Looking at the lowermost ball, the force exerted by the spring must be equal to weight i.e 10g.

Taking the second ball, forces acting downwards are weight and force due to the lower spring, which is 10g as obtained from previous paragraph( tension developed depends on the lower ball only). So net force acting downwards on string is 20g+10g=30g. This is balanced by force due to upper spring, which must be 30 g as well.

Now the net force acting downwards on the uppermost ball is 10g + forced due to spring(30g)=40g. This is balanced by the string and hence the body is in equilibrium.

As a result, on cutting the spring, the net force is 40g downwards and hence acceleration at that instant is 4g.

$\endgroup$
  • $\begingroup$ Thank you, Your answer was very clear. Glad my second try on it was correct. $\endgroup$ – rahul rj Mar 7 '17 at 11:40
1
$\begingroup$

The question is asking what happens to the top ball if you take away the string. The string serves to counter the downward force on the top ball. Thus when the string snaps, there will be a net downward force on the ball equal in magnitude to the tension in the string.

To find the tension in the string, it is easiest to consider everything below the string (three balls and two springs) as one system, and apply Newton's second law, which says the sum of external forces is equal to mass times acceleration. Since nothing is accelerating, the sum of external forces must be zero. Now there are only two external forces: gravity and the string tension, and they must cancel (the spring forces are internal to our system). Therefore the string tension must be the sum of gravitational forces: 40 kg * $g$.

In the first paragraph, we argued that this must be the net downward force on the ball, and since the ball has mass 10 kg, its downward acceleration must be $4g$.

$\endgroup$
0
$\begingroup$

Although the motion of the centre of mass of the system comprising the three masses must be free fall with an acceleration $g$ the motions of the individual masses will be more complex due to the extended springs exerting forces on the masses.

Draw a free body diagram for the top mass before the string is cut and label all the forces which are acting on the top mass.
Now think about what happens when the string is cut remembering that the extensions of the springs cannot change instantaneously.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.