What Determines the Maximal Elongation of a Spring in a Pulley System?

AI Thread Summary
The discussion focuses on determining the maximal elongation of a spring in a pulley system involving two bricks connected by a spring and a rope. The key point is the calculation of the amplitude (A0) of the spring's elongation, which is derived from the system's dynamics after the release of the lower brick. The acceleration of the system is analyzed using a non-inertial frame, leading to the conclusion that A0 equals mg/3k. The relationship between the forces acting on the blocks and the spring's properties is crucial for understanding the system's behavior. The explanation provided helps clarify the reasoning behind the derived formula for A0.
snate
Messages
7
Reaction score
1

Homework Statement


2 bricks of mass m are located near the edge of a smooth horizontal surface, they are connected by an ideal weightless ,unstretched spring with a length of l0 and spring constant k. The brick which is closer to the edge is connected to another brick of the same mass by an ideal unstretchable rope going through a pulley. The lowest brick is held so that the rope above it hangs vertically.
The lowest brick is released
What's the minimal time τ after which the elongation of the spring ΔL will be maximal. Find ΔL.

20645861_1928917583989680_1368339303_n.png
20631554_1928917840656321_1206854442_n.png

Homework Equations

The Attempt at a Solution


Here's the solution until the point where I'm stuck. Sorry for the Russian text, it does not contain crucial information, please ignore it
20632389_1928930010655104_1002110814_n.png

a2-a1 is the acceleration of one end of the spring from the frame of reference of another end.
Then these equations are given (y is vertical coordinate)
20631623_1928927060655399_1063382409_n.png

I understand why w2 is 3k/2m but why is the amplitude(A0) mg/3k? Can someone please explain? I only need an explanation for A0
The next equations are
20631814_1928927260655379_432063850_n.png

20632625_1928928830655222_42064552_n.png

20631703_1928928893988549_190674348_n.png

At the beginning the system is in equilibrium and then x(0)=0, that leads to B being equal to 0 and and A0+A=0 because the spring isn't stretched at the beginning.
20632455_1928929140655191_236803601_n.png

20643868_1928929220655183_226754705_n.png

20632337_1928929310655174_1726256519_n.png

Thanks.
 
Last edited:
Physics news on Phys.org
I have not tried to follow your equations, but I can suggest an easier way via a sequence of simplifications.
First, model it as all happening on a horizontal surface, with a horizontal force mg applied to the rightmost block.
Next, we can merge the two blocks connected by the inextensible string into a single block mass 2m.
Clearly the average acceleration is g/3 to the right, so take a noninertial frame accelerating that way. In this frame, the left-hand block has a pseudoforce ... to the left, and the 2m block has a pseudoforce ... to the left, leaving it with a net force ... to the right. (Fill in the blanks.)
Finally, consider the common mass centre of the two blocks. How does this move in the reference frame? How do the two blocks move in relation to it?
 
  • Like
Likes snate and phinds
haruspex said:
I have not tried to follow your equations, but I can suggest an easier way via a sequence of simplifications.
First, model it as all happening on a horizontal surface, with a horizontal force mg applied to the rightmost block.
Next, we can merge the two blocks connected by the inextensible string into a single block mass 2m.
Clearly the average acceleration is g/3 to the right, so take a noninertial frame accelerating that way. In this frame, the left-hand block has a pseudoforce ... to the left, and the 2m block has a pseudoforce ... to the left, leaving it with a net force ... to the right. (Fill in the blanks.)
Finally, consider the common mass centre of the two blocks. How does this move in the reference frame? How do the two blocks move in relation to it?
Thanks, your explanation hinted me why A0 is mg/3k. A0=amax/w2. And maximal acceleration amax after infinitesimal moment of time since the rightmost block has been released is gm/2m=g/2 because at that moment the spring is not deformed, so F=kΔx=k0=0, so we don't have to consider the spring and the leftmost block at that moment. So A0 =g*2m/(2*3*k)=mg/3k. Am I correct?
 
snate said:
Thanks, your explanation hinted me why A0 is mg/3k. A0=amax/w2. And maximal acceleration amax after infinitesimal moment of time since the rightmost block has been released is gm/2m=g/2 because at that moment the spring is not deformed, so F=kΔx=k0=0, so we don't have to consider the spring and the leftmost block at that moment. So A0 =g*2m/(2*3*k)=mg/3k. Am I correct?
Looks good.
 
  • Like
Likes snate
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top