Simple mechanics question: will reading of the scale change?

AI Thread Summary
The discussion centers on whether the reading of a scale will change when a marble descends in a beaker after a string is cut. Initially, the net force on the beaker is equal to the weight of the marble minus the buoyant force, leading to a stable reading. When the string is cut, there is a temporary decrease in weight as the marble accelerates, followed by a return to the original reading once it reaches constant velocity. The final reading remains unchanged because the total mass of the system does not alter, despite transient fluctuations during the marble's descent. Overall, the weight reading on the scale will be the same once the marble comes to rest at the bottom of the beaker.
tonnijames
Messages
2
Reaction score
1
Homework Statement
A beaker filled with water is placed on the scale.
A weight denser than water is tied to one end of a string; the other end of the string is tied to a stick. The weight is then placed in water with the stick supported over the rim of the beaker. Total weight at this time is recorded. If the string is suddenly cut and the weight sinks at a constant velocity, will the reading of the scale change? Assuming stick and string are massless.

Any help is appreciated. Thank you!
Relevant Equations
Ft=Fg - F(buoyant) = F(stick on beaker)
Fnet final = 0 on the weight.
My reasoning is that
1)at initial condition, net force of stick on the beaker is equal to the weight of the weight minus the buoyant force applied by the water to the weight, which is the tension force in the string. The recorded weight should be equal to mass of beaker + water + net force of stick on beaker.
2)when string is cut, as it sinks at constant velocity Fnet=0 for the weight. Thus the weight is "canceled out" by the buoyant force and net forced applied by stick on the beaker is 0. I suspect that the final reading will be less, but I could not be sure.
Below I have drew a diagram.
242428
 
Physics news on Phys.org
The final reading - after the marble has come to rest on the bottom of the beaker - must be the same as before cutting the string, because the mass of the beaker + water + marble + string + stick has not changed, and all parts are at rest in the end, as they were in the beginning. Since the weight of a stationary object depends only on its mass, the weight must also be the same.

There will be a slight reduction in weight immediately after the string is cut, as the marble accelerates to its constant descent velocity, returning to the normal weight as soon as that velocity is reached. That is mirrored by a slight increase in weight between when the marble first contacts the floor and when it comes to rest. That increase in weight relates to the deceleration.

The 'blips' of decreased and increased weight are similar in some ways to the lightness you feel in a lift (US elevator) when it starts to descend and the increased weight when as it slows to stop at a lower floor.

Often in physics problems it is assumed that the descent or ascent velocity is infinitesimally slow, in order to make these blip variations as the item accelerates to and decelerates from the ascension/descent speed small enough to ignore.
 
andrewkirk said:
The final reading - after the marble has come to rest on the bottom of the beaker - must be the same as before cutting the string, because the mass of the beaker + water + marble + string + stick has not changed, and all parts are at rest in the end, as they were in the beginning. Since the weight of a stationary object depends only on its mass, the weight must also be the same.

There will be a slight reduction in weight immediately after the string is cut, as the marble accelerates to its constant descent velocity, returning to the normal weight as soon as that velocity is reached. That is mirrored by a slight increase in weight between when the marble first contacts the floor and when it comes to rest. That increase in weight relates to the deceleration.

The 'blips' of decreased and increased weight are similar in some ways to the lightness you feel in a lift (US elevator) when it starts to descend and the increased weight when as it slows to stop at a lower floor.

Often in physics problems it is assumed that the descent or ascent velocity is infinitesimally slow, in order to make these blip variations as the item accelerates to and decelerates from the ascension/descent speed small enough to ignore.
Thank you for the explanation. In my original post I made a mistake that the weight of the marble will be canceled by the buoyancy of water, due to marble descending at constant velocity. Now I see that regardless of the motion of marble, as long as it's stationary and the mass does not change, reading would not change.
 
There is another effect which may be worth considering.

The object (a marble or whatever) is denser than water.

What is true of the center of mass of the system as the object sinks?
 
SammyS said:
There is another effect which may be worth considering.

The object (a marble or whatever) is denser than water.

What is true of the center of mass of the system as the object sinks?
That effect is already contemplated in the response by @andrewkirk. Yes, the center of mass of the system is moving downward. But what can be said about its downward velocity?
 
jbriggs444 said:
That effect is already contemplated in the response by @andrewkirk. Yes, the center of mass of the system is moving downward. But what can be said about its downward velocity?
Oh my!

Well yes, it is constant. Damn you Sir I. Newton !
 
  • Like
Likes jbriggs444 and PeroK
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top