Rocket + Connected Body Dynamics problem

[Mohammad Ishmas]

TL;DR Summary: A rocket has has been connected to blocks containing gas of respective masses, there are 3 such blocks, Find Tension between all the blocks 5 seconds after the launch. the main problem comes out doing the problem

There is a rocket shown in the image inserted below, it has been connected to blocks of 200 kg , 100 kg ,50 kg and 25 kg which are connected by massless strings , in those boxes/blocks gas is filled that the rocket uses for upthrust ( assume that the boundaries of boxes are massless), if the initial mass of the rocket is 1000kg and the change of mass per second that is the gas burning per second is 12.5 kg and the velocity of gases with respect to rocket is 5000 meter per second that is the velocity of upthrust of rocket or at which the rocket is travelling. Then we have to find Tension between 50 kg and 100 kg block (T1) AND tension between 100 kg and 200 kg black (T2) after 5 second of travelling of the rocket.


The question can be done but the real problem is coming after doing the question, I found the tension by first finding acceleration after 5 second which can be done by dividing u(dm/dt) by mass at 5 sec {can be found}


after this I made a free body diagram of the 50 kg block which would now weigh only 12.5 kg, after this the T1 = 833.33....... N
similarly T2 comes out to be 7499.99.... or 7450 N

However now if we look from the capability of taking force of objects of specific masses and specific acceleration then things go wrong , as
the capability of 100 kg block at 56.6... acceleration is 5666.6..... and the value of T2 comes out to be 7499.9.... that means T1 should be equal to 7499.9.... minus 5666.6...... = 1833.3...

So why isn't this working, please give a detailed explanation

 

[kuruman]

The tension is zero. First figure out how a rocket works and you will see why. You should also take into account that you cannot push with a string.

 

[Mohammad Ishmas]

The tension is zero. First figure out how a rocket works and you will see why. You should also take into account that you cannot push with a string

The tension is zero. First figure out how a rocket works and you will see why. You should also take into account that you cannot push with a string.

See this question is made by me for my students, and the string system is hypothetical I've assumed that the string has tension and have purposely added the rocket to make it a variable mass system.

 

[kuruman]

You may not assume that the string has tension because it cannot with the system you have drawn. If you don't believe me, you can simulate this as follows. Assemble four masses with strings connecting them as shown on the right and put them on a horizontal table to eliminate gravity. Push on one end as shown in the figure on the right to simulate the rocket thrust.
Questions to consider.
1. Does the string "have" tension?
2. How must you exert a force on the end so that the string acquires tension?
3. Is this force in the same direction as the thrust exerted by the gases on the rocket?

 

[Mohammad Ishmas]

View attachment 349708You may not assume that the string has tension because it cannot with the system you have drawn. If you don't believe me, you can simulate this as follows. Assemble four masses with strings connecting them as shown on the right and put them on a horizontal table to eliminate gravity. Push on one end as shown in the figure on the right to simulate the rocket thrust.
Questions to consider.
1. Does the string "have" tension?
2. How must you exert a force on the end so that the string acquires tension?
3. Is this force in the same direction as the thrust exerted by the gases on the rocket?

great point thank you, but still I have to make a question so let us consider that we apply Force by somehow pulling the system from top with the same velocity, then my question is right or wrong ?

 

[kuruman]

Yes, the system needs to be pulled by the leading module. It's the same principle as putting the horse before the cart. Can you explain what it is that you want to test with this problem? If you want to know whether your students understand internal forces in an accelerating system, why have a rocket providing the acceleration? If you want to know whether your students can handle a variable mass problem, why bother with the tensions?

 

[Mohammad Ishmas]

Yes, the system needs to be pulled by the leading module. It's the same principle as putting the horse before the cart. Can you explain what it is that you want to test with this problem? If you want to know whether your students understand internal forces in an accelerating system, why have a rocket providing the acceleration? If you want to know whether your students can handle a variable mass problem, why bother with the tensions?

I have to make a question, such that it has a variable mass which adds to its complexity and this will make sure that the question tests the observation skills, the main thing of this question is finding the two tensions, by making Free body diagrams and calculating them. But after this a problem is occurring that is
The tension between the 200kg and 100kg block came out to be 7499.9 ... N which I found by first calculating The tension between the 100kg and 50 kg block {12.5kg after 5 secs} , So the net force in upwards direction of the 100 kg block was 7499. 99.. N - 1000N ( mg) that comes out to be 6499.99 ... N and the capability of force of 100 kg block at 56.66 ... m/s^2 is 100*56.66.66... which is 5666.66... N
Now, the 100 kg block should take should only take 5666.66... N force out of 6499.99.. N force which is correct
as that comes out to be 833.33.... N that is the tension between 50 kg {12.5 kg after 5 seconds} and 100 kg block.

In the same way I tried to find t2 by taking the top of the rocket with the pointy thing and 200 kg block
their total mass was 825 kg as the strings are massless, but after doing the same thing I did to find the tension between the 200 kg and 100 kg block by considering the top of the rocket and then by calculating the force acting on the rocket which I found to be 53125 N, So it has the net force of 53125 N - 8250 N = 44875 N

BUT the capability of taking force of this object is 825*56.66.. {56.66 .. is the acceleration of the rocket at 5 seconds) so that is coming out to be 46749.99.... N . So this is the problem that is coming

 

[jbriggs444]

I am having trouble following your work. Perhaps if you slowed down and showed us some equations along with their justification.

The tension between the 200kg and 100kg block came out to be 7499.9 ... N which I found by first calculating The tension between the 100kg and 50 kg block {12.5kg after 5 secs}

So you calculated 7499.9 Newtons... somehow.

One of the inputs to this calculation was supposedly the tension between the 100 and 50 kg block... which you do not show.

And something is 12.5 kg after 5 seconds. What, exactly?

So the net force in upwards direction of the 100 kg block was 7499. 99.. N - 1000N ( mg)

Wait, what? You counted the tension from the 200 kg block on the 100 kg block. What about the tension from the 50 kg block on the 100 kg block?

that comes out to be 6499.99 ... N and the capability of force of 100 kg block at 56.66 ... m/s^2 is 100*56.66.66... which is 5666.66... N

What is a "capability of force"?

I am completely lost. I do not even know what approach you are taking to solve the problem.

 

[kuruman]

The tension between the 200kg and 100kg block came out to be 7499.9 ... N

How did it come out to be that value? You do not explain. To get the tensions you need to find the acceleration from the free body diagram (FBD) of the entire system. Note that
All masses have the same acceleration at all times otherwise they will not move as one. See the FBD below.

Let $F$ be the net force on the 4-mass system. We will worry about a number for it later. For the strings to be under tension, this force must act on the leading mass $m_1.$ By Newton's second law, the common acceleration of the masses is $$a=\frac{F}{m_1+m_2+m_3+m_4}.$$ Now we find the tensions.
Note that the net force on the system consisting of only $m_4$ is $T_{34}$. Then $$T_{34}=m_4 a=\frac{m_4F}{m_1+m_2+m_3+m_4}.$$Note that the net force on system $(m_4+m_3)$ is $T_{23}$. Then $$T_{23}=(m_4+m_3) a=\frac{(m_4+m_3)F}{m_1+m_2+m_3+m_4}.$$Note that the net force on system $(m_4+m_3+m_2)$ is $T_{12}$. Then $$T_{12}=(m_4+m_3+m_2) a=\frac{(m_4+m_3+m_2)F}{m_1+m_2+m_3+m_4}.$$ If you want a rocket involved in all this, then $m_1$ must be the combustion chamber providing the thrust, not $m_4$. All one has to do is find $m_1$ at $t=5~$s and substitute with $F=12.5~\text{kg/s}\times 5000~\text{m/s}.$

 

[Mohammad Ishmas]

 

[Mohammad Ishmas]

This is How I found T1 and T2

 

[Mohammad Ishmas]

I am having trouble following your work. Perhaps if you slowed down and showed us some equations along with their justification.


So you calculated 7499.9 Newtons... somehow.

One of the inputs to this calculation was supposedly the tension between the 100 and 50 kg block... which you do not show.

And something is 12.5 kg after 5 seconds. What, exactly?


Wait, what? You counted the tension from the 200 kg block on the 100 kg block. What about the tension from the 50 kg block on the 100 kg block?


What is a "capability of force"?

I am completely lost. I do not even know what approach you are taking to solve the problem.

I've posted my work, and sorry I meant capability of an object of some mass to take force at a particular acceleration, I mean if we take a two block system in which a object of 10 kg is moving with acceleration 10 m/s^2 which is also the acceleration of the whole system, the F = ma, So, 10*10=1000N and if we are applying force on this block of 2000 N then the other 1000 N would get transmitted into the string as Tension

 

[renormalize]

Please don't expect readers to follow your hard-to-read photos. Post equations in LaTeX, so they are readable and can be quoted for use in responses.

 

[kuruman]

I've posted my work, and sorry I meant capability of an object of some mass to take force at a particular acceleration, I mean if we take a two block system in which a object of 10 kg is moving with acceleration 10 m/s^2 which is also the acceleration of the whole system, the F = ma, So, 10*10=1000N and if we are applying force on this block of 2000 N then the other 1000 N would get transmitted into the string as Tension

What you say here applies to the leading block that does the pulling. You are saying that if the net force on the system is $F$ and the acceleration is $a$ then the tension between the $m_1=200$ kg block and the 100 kg block is $T=F-m_1a.$ Is this where you started your calculation of the tensions or did you start from the trailing mass of 25 kg? I cannot read your solutions.

I strongly recommend against thinking in terms of "the capability of taking force" and start thinking in terms of the acceleration of the sytem of your choice and the forces acting on it. Please read and understand how I got the tension expressions in post #9.

 

[Mohammad Ishmas]

What you say here applies to the leading block that does the pulling. You are saying that if the net force on the system is $F$ and the acceleration is $a$ then the tension between the $m_1=200$ kg block and the 100 kg block is $T=F-m_1a.$ Is this where you started your calculation of the tensions or did you start from the trailing mass of 25 kg? I cannot read your solutions.

I strongly recommend against thinking in terms of "the capability of taking force" and start thinking in terms of the acceleration of the sytem of your choice and the forces acting on it. Please read and understand how I got the tension expressions in post #9.

the 12.5 kg/s in the pdf is equals to change in mass or dm/dt

 

[jbriggs444]

As I understand it, we have a rocket assembly which totals 1000 kg initially. This includes the masses of four stages. The four stages have initial masses of 200 kg, 100 kg, 50 kg and 25 kg.

In the PDF, you calculate that after 5 seconds have elapsed, the total remaining mass of the assembly will be reduced by $5 \times 12.5 = 62.5$ kg. So the remaining mass is 937.5 kg.

You also calculate that the thrust of the rocket is given by $v \frac{dm}{dt} = 5000 \times 12.5 = 62500$ N.

Applying the calculated force to the calculated remaining mass you get $a = \frac{f}{m} = \frac{62500}{937.5} = 66.67$ m/sec²

You recognize that if this is a vertical launch in Earth gravity that this acceleration will be relative to free fall. So we need to subtract 10 m/sec² to get the acceleration of the craft relative to the launch pad. Now we are down to $56.67$ m/sec².

That much is all fine. Though the subtraction of the acceleration of free fall is misguided. Better to adopt a freely falling inertial coordinate system in which the craft is momentarily at rest. The lab frame offers no advantages over this.

You next try to calculate the tension $T_1$ that supports 1 second worth of ejecta against the downward force of gravity.

That is a pointless calculation. There is no string attached to the exhaust plume. Nor is the mass of one seconds worth of plume of any particular interest. Why one second? Why not one hour? One day? One millisecond? Two shakes of a lamb's tail? Our chosen units of measurement should not affect our calculations.

Ahh, I see. The bottom two stages started with a total of 75 kg. The 62.5 kg of ejecta came from them. So the 25 kg bottom stage is gone and all that remains of the 50 kg stage is 12.5 kg. You are calculating the downward force of gravity on that 12.5 kg stage at the 5 second mark.

It is merely an unhelpful coincidence that this is also equal to the mass of 1 second of ejecta.

Since you are working in the lab frame, this is not an unreasonable thing to do. You want a force balance on that 12.5 kg stage. You have 125 N from gravity and an unknown upward force from $T_1$. The net acceleration is 56.7 m/sec² upward. So you need a net upward force of $56.7 \times 12.5 = 708.33$ N.

I agree that this requires a tension of 708.33 + 125 = 833.33 N.

 

[Mohammad Ishmas]

As I understand it, we have a rocket assembly which totals 1000 kg initially. This includes the masses of four stages. The four stages have initial masses of 200 kg, 100 kg, 50 kg and 25 kg.

In the PDF, you calculate that after 5 seconds have elapsed, the total remaining mass of the assembly will be reduced by $5 \times 12.5 = 62.5$ kg. So the remaining mass is 937.5 kg.

You also calculate that the thrust of the rocket is given by $v \frac{dm}{dt} = 5000 \times 12.5 = 62500$ N.

Applying the calculated force to the calculated remaining mass you get $a = \frac{f}{m} = \frac{62500}{937.5} = 66.67$ m/sec²

You recognize that if this is a vertical launch in Earth gravity that this acceleration will be relative to free fall. So we need to subtract 10 m/sec² to get the acceleration of the craft relative to the launch pad. Now we are down to $56.67$ m/sec².

That much is all fine. Though the subtraction of the acceleration of free fall is misguided. Better to adopt a freely falling inertial coordinate system in which the craft is momentarily at rest. The lab frame offers no advantages over this.

You next try to calculate the tension $T_1$ that supports 1 second worth of ejecta against the downward force of gravity.

That is a pointless calculation. There is no string attached to the exhaust plume. Nor is the mass of one seconds worth of plume of any particular interest. Why one second? Why not one hour? One day? One millisecond? Two shakes of a lamb's tail? Our chosen units of measurement should not affect our calculations.

 

[Mohammad Ishmas]

How does T1 support the system for only 1 Second ?? , T1 is not between 25kg and 50kg if that was the case then what you are saying would happen

 

[kuruman]

As I understand it, we have a rocket assembly which totals 1000 kg initially. This includes the masses of four stages. The four stages have initial masses of 200 kg, 100 kg, 50 kg and 25 kg.

Ahem.
200 + 100 + 50 + 25 = 375 kg, not 1000 kg.
That is also wrong in the pdf attachment.

I thought that this calculation was in free space. Since it is near the surface of the Earth, quantity $F$ in the post #9 equations should be replaced by $\left[u\dfrac{dm_1}{dt}-(m_1+m_2+m_3+m_4)\right]$ where $u$ is the velocity of the expelled gases relative to the rocket.

 

[Mohammad Ishmas]

Ahem.
200 + 100 + 50 + 25 = 375 kg, not 1000 kg.
That is also wrong in the pdf attachment.

I thought that this calculation was in free space. Since it is near the surface of the Earth, quantity $F$ in the post #9 equations should be replaced by $\left[u\dfrac{dm_1}{dt}-(m_1+m_2+m_3+m_4)\right]$ where $u$ is the velocity of the expelled gases relative to the rocket.

Well the blocks are of 200,100,50 and 25 kg respectively and the top of the rocket that pointy thing would have the rest of the mass

 

[Mohammad Ishmas]

Well the blocks are of 200,100,50 and 25 kg respectively and the top of the rocket that pointy thing would have the rest of the mass

Ahem.
200 + 100 + 50 + 25 = 375 kg, not 1000 kg.
That is also wrong in the pdf attachment.

I thought that this calculation was in free space. Since it is near the surface of the Earth, quantity $F$ in the post #9 equations should be replaced by $\left[u\dfrac{dm_1}{dt}-(m_1+m_2+m_3+m_4)\right]$ where $u$ is the velocity of the expelled gases relative to the rocket.

well considering this my method is not wrong tho right ??

 

[jbriggs444]

well considering this my method is not wrong tho right ??

Not wrong. Adopting launch-pad relative coordinates just means that you are adding or subtracting $g$ from everything. It cancels out in the end.

 

[jbriggs444]

How does T1 support the system for only 1 Second ?? , T1 is not between 25kg and 50kg if that was the case then what you are saying would happen

You took pain to point out that the 12.5 kg/sec in the PDF was the mass flow rate. I then misread the drawing to indicate that the 12.5 kg in the PDF was the mass flow for one second rather than the remaining mass of the next-to-bottom stage.

 

[Mohammad Ishmas]

Not wrong. Adopting launch-pad relative coordinates means that you are adding or subtracting $g$ from everything. It cancels out in the end.

ok well, the real problem is coming after doing all this.

First lets consider two blocks m1 and m2 and there is Force F on the block m1 such that they accelerate with acceleration "a" { they are attached by massless strings} . Now force F is being applied on it , also the block m1 at acceleration "a" can take force (m1a) { This is what I meant by capability of force} , the whole system is accelerating so of course F > m1a. so it 'takes' force F but it only has a capability of taking force m1a. So the rest of force gets transmitted in the string in form of tension. This can also be expressed in terms of equations like- F-m1a = Tension in the strings .

The same should work in my system right ?

 

[jbriggs444]

ok well, the real problem is coming after doing all this.

First lets consider two blocks m1 and m2

Which are those again? We started with five blocks (capsule plus 4 stages) and lost one stage off the bottom.

and there is Force F on the block m1 such that they accelerate with acceleration "a" { they are attached by massless strings} . Now force F is being applied on it , also the block m1 at acceleration "a" can take force (m1a) { This is what I meant by capability of force}

"can take" is the wrong wording. "Must be subject to" is correct.

It is not a "capability". It is an "actuality". Newton's second law is not optional.

 

[Mohammad Ishmas]

Which are those again? We started with five blocks (capsule plus 4 stages) and lost one stage off the bottom.


"can take" is the wrong wording. "Must be subject to" is correct.

It is not a "capability". It is an "actuality". Newton's second law is not optional

 

[Mohammad Ishmas]

Which are those again? We started with five blocks (capsule plus 4 stages) and lost one stage off the bottom.


"can take" is the wrong wording. "Must be subject to" is correct.

It is not a "capability". It is an "actuality". Newton's second law is not optional.

I've made a situation of the two blocks to ask whether this will work in my system or not ?

 

[kuruman]

Well the blocks are of 200,100,50 and 25 kg respectively and the top of the rocket that pointy thing would have the rest of the mass

Would have? Please supply a coherent statement of the problem. How do you expect us to make sense of this problem if you don't give us the needed information from the start?

 

[Mohammad Ishmas]

Would have? Please supply a coherent statement of the problem. How do you expect us to make sense of this problem if you don't give us the needed information from the start?

Sorry, for that I'll be more careful

 

[Mohammad Ishmas]

Would have? Please supply a coherent statement of the problem. How do you expect us to make sense of this problem if you don't give us the needed information from the start?

Please also consider what I've said in post #24

 

[kuruman]

Sorry, for that I'll be more careful

Good. Now give us the statement of the problem as if we were your students to make sure that all is in order before proceeding with how to solve it and how to teach how to solve it.

 

[Mohammad Ishmas]

Good. Now give us the statement of the problem as if we were your students to make sure that all is in order before proceeding with how to solve it and how to teach how to solve it.

Ok I am giving the statement please tell if I can better it

There is a Rocket capsule of 625 kg attached to blocks of 200kg,100kg,50kg and 25 kg respectively by ideal strings. It is launched from earth such that its velocity at which it is going remains constant which is given as 5000 m/s, if the rocket from the bottom most block is losing gas/mass at the rate of 12.5 kg/s then Find tension between the 50 kg and 100kg block and tension between the 100 kg and 200 kg block, take g= 10 m/s^2


This would be the question and I've found the tensions as shown above you can also confirm by doing it yourself. but a problem is coming after finding these tensions

 

[jbriggs444]

Ok I am giving the statement please tell if I can better it

There is a Rocket capsule of 625 kg attached to blocks of 200kg,100kg,50kg and 25 kg respectively by ideal strings. It is launched from earth such that its velocity at which it is going remains constant which is given as 5000 m/s, if the rocket from the bottom most block is losing gas/mass at the rate of 12.5 kg/s then Find tension between the 50 kg and 100kg block and tension between the 100 kg and 200 kg block, take g= 10 m/s^2


This would be the question and I've found the tensions as shown above you can also confirm by doing it yourself. but a problem is coming after finding these tensions

So the velocity remains constant. This means that the acceleration is zero. This flies in the face of the fact that the exhaust velocity and mass flow rate are given and require that the acceleration be both non-zero and non-constant.

Try again.

 

[kuruman]

Ok I am giving the statement please tell if I can better it

There is a Rocket capsule of 625 kg attached to blocks of 200kg,100kg,50kg and 25 kg respectively by ideal strings. It is launched from earth such that its velocity at which it is going remains constant which is given as 5000 m/s, if the rocket from the bottom most block is losing gas/mass at the rate of 12.5 kg/s then Find tension between the 50 kg and 100kg block and tension between the 100 kg and 200 kg block, take g= 10 m/s^2


This would be the question and I've found the tensions as shown above you can also confirm by doing it yourself. but a problem is coming after finding these tensions

You are confused about your own problem and about how a rocket works.

After all that I have already said about how the strings can be under tension, give me one good reason why you say that

the rocket from the bottom most block is losing gas/mass at the rate of 12.5 kg/s

and then explain to me what part of the rocket provides the required thrust.

 

[kuruman]

So the velocity remains constant.

OP is confusing the velocity of the rocket with the constant relative velocity of the expelled gas.

 

[Mohammad Ishmas]

OP is confusing the velocity of the rocket with the constant relative velocity of the expelled gas.

sorry but what does "OP" mean ? , And yeah thanks for pointing that error out I'll change that

 

[jbriggs444]

sorry but what does "OP" mean ? , And yeah thanks for pointing that error out I'll change that

OP is internet slang for "original poster". That is you in this case.

It is also used for "original post". Context usually makes it clear which is meant.

 

[Mohammad Ishmas]

OP is internet slang for "original poster". That is you in this case.

It is also used for "original post". Context usually makes it clear which is meant.

ok thanks

 

[kuruman]

sorry but what does "OP" mean ? , And yeah thanks for pointing that error out I'll change that

Good. Let's see your revised problem description.

 

[Mohammad Ishmas]

OP is confusing the velocity of the rocket with the constant relative velocity of the expelled gas.

Let us just reconsider the problem as a whole by removing the rocket. There is a capsule of 625 kg attached to blocks of 200kg,100kg,50kg and 25 kg respectively it is losing its mass from at the rate of 12.5 kg/s from the 25 kg block and then to 50 kg then 100 kg ...

If the acceleration after 5 seconds is 56.666.... m/s^2
then find tensions between 50 kg block and 100 kg block , 100 kg block and 200 kg block and also find contact force between 200 kg block and 625 kg capsule

 

[Mohammad Ishmas]

Let us just reconsider the problem as a whole by removing the rocket. There is a capsule of 625 kg attached to blocks of 200kg,100kg,50kg and 25 kg respectively it is losing its mass from at the rate of 12.5 kg/s from the 25 kg block and then to 50 kg then 100 kg ...

If the acceleration after 5 seconds is 56.666.... m/s^2
then find tensions between 50 kg block and 100 kg block , 100 kg block and 200 kg block and also find contact force between 200 kg block and 625 kg capsule

Also in the reconsidered problem everything is being on the x axis so the mg doesn't mean anything

 

[jbriggs444]

Let us just reconsider the problem as a whole by removing the rocket. There is a capsule of 625 kg attached to blocks of 200kg,100kg,50kg and 25 kg respectively it is losing its mass from at the rate of 12.5 kg/s from the 25 kg block and then to 50 kg then 100 kg ...

If the acceleration after 5 seconds is 56.666.... m/s^2
then find tensions between 50 kg block and 100 kg block , 100 kg block and 200 kg block and also find contact force between 200 kg block and 625 kg capsule

Can you please state the complete problem as you would pose it to your students?

As it stands, you have not stated that an external force applied to the 200 kg block is responsible for the observed 56.666... m/s² acceleration.

Stepping back a bit...

A variable mass problem is sometimes posed in terms of a railroad car (a hopper car) which is pouring out its contents into a container below as it rolls past. Or one can invert it and pour material onto a conveyer belt.

A multiple mass problem is normally posed in terms of stacked blocks being pushed sideways. Perhaps on a frictionless surface.

A realistic rocket problem would have the large stage at the bottom pushing on the smaller stages and payload above. The mass flow rate would not be constant from one stage to the next. Instead, the mass flow rate for the initial stage would be large compared to the mass flow rate for the final stage.

Although Robert Goddard's initial design had the rocket nozzle near the top of the assembly with the rocket frame dangling behind, this is not at all standard. I am not entirely sure how much of the classic picture below is rocket and how much is launch stand. Goddard's later designs moved the nozzle to the bottom.

 

[kuruman]

Let us just reconsider the problem as a whole by removing the rocket. There is a capsule of 625 kg attached to blocks of 200kg,100kg,50kg and 25 kg respectively it is losing its mass from at the rate of 12.5 kg/s from the 25 kg block and then to 50 kg then 100 kg ...

If the acceleration after 5 seconds is 56.666.... m/s^2
then find tensions between 50 kg block and 100 kg block , 100 kg block and 200 kg block and also find contact force between 200 kg block and 625 kg capsule

Your reconsidered problem ignores the issue with your design that I pointed out right from the start.
You cannot push with a string.

I will refrain from spending more time to post here unless I see that you take our advice and suggestions under consideration.

 

[Mohammad Ishmas]

Your reconsidered problem ignores the issue with your design that I pointed out right from the start.
You cannot push with a string.

I will refrain from spending more time to post here unless I see that you take our advice and suggestions under consideration.

well let us say that this system is being pulled from 625 kg block with the data given below

 

[jbriggs444]

well let us say that this system is being pulled from 625 kg block with the data given below

I notice that you have not given any data below. So I will harvest the data from up-thread.

We begin with blocks of 625, 200, 100, 50 and 25 kg arranged in a row from left to right on a frictionless surface, all connected with ideal massless cords.

We apply a leftward force of 62,500 N on the 625 kg block. We are invited to imagine that this is from a rocket
nozzle on that block. The nozzle is supplied with 12.5 kg/s of liquid propellant which is expelled with a relative velocity of 5000 m/s rightward.

The propellant is supplied in turn from the smaller blocks, starting at the right and working left. The blocks are pure propellant. Any tank is negligibly massive. Once the propellant in a block is exhausted, the block itself is discarded.

Consider the situation after 5 seconds has elapsed.

a) What is the acceleration of the assembly at this point?
b) What is the tension in the cord connecting the 625 kg block to the 200 kg block behind it?


I would analyze the problem by first calculating the system state at 5 seconds into the scenario. The trailing 25 kg tank would have been exhausted in the first two seconds. It will have been discarded and is now irrelevant. The 50 kg tank will have lost 37.5 kg of its initial mass. 12.5 kg remains.

So we have 625 + 200 + 100 + 12.5 kg = 937.5 kg total assembly mass at this time.

This assembly is subject to a net external force of 62,500 N.

Assuming that the assembly moves rigidly (as it obviously should) this means that it is accelerating at a rate of $a = \frac{f}{m_\text{tot}} = \frac{62500}{937.5} \approx 66.7$ m/sec²

[Since gravity is not a factor, we will not be adjusting the acceleration to 56.7 m/sec²]

Ignoring the 625 kg block in front, the mass of the remainder of the assembly is currently 200 + 100 + 50 + 12.5 = 312.5 kg. This assembly is accelerating leftward at approximately 66.7 m/sec². So it must be subject to a leftward net force of $f_\text{net} = m_\text{tail} a \approx 312.5 \times 66.7 = 20833$ N.

Naively, this is the tension in the cord between the 625 kg and 200 kg blocks: about 20,833 N.

If I've not muffed the calculation. [Edit: One muff corrected. I counted the 50 kg block when I should not have].


However, there is a momentum flow that we have not accounted for.

We are piping liquid propellant forward from the 12.5 kg (remaining) tank to the 625 kg block where the rocket nozzle is attached.

If the pipe is narrow, the velocity of propellant in the pipe times the 12.5 kg/sec flow rate may amount to a significant force. This would be a retarding force on the trailing block and a propulsive force on the leading block. The tension in the cord[s] would need to increase to reflect this.

If the pipe is wide and long, the mass of propellant in the pipe may be significant. This messes up our accounting for the mass of the various assembly bits at any point in time -- where do we count the mass of the propellant in the pipe?

So let us assume that the whole assembly is short and the pipe is wide so that the mass and momentum transfer from the unburned propellant in the pipeline may both be safely neglected.

 

[Mohammad Ishmas]

Guys I sorted out how to find tension in the initial problem consisting of rocket, but there is still one issue that If I take net acceleration that is subtracting value of g from acceleration in upward direction, and then do not account for weight force for FBD of each block as I have already taken net acceleration then the values of tensions are uniform.

But let us say that we take acceleration only in upwards direction and do not ignore weight force then the value of tension force is not force

NOTE that I've also attached a string between 625 kg and 200 kg block which was not in initial problem

 

[jbriggs444]

Guys I sorted out how to find tension in the initial problem consisting of rocket, but there is still one issue that If I take net acceleration that is subtracting value of g from acceleration in upward direction, and then do not account for weight force for FBD of each block as I have already taken net acceleration then the values of tensions are uniform.

Please show your work.

Are you saying that the calculated tensions are identical regardless of the acceleration of gravity? Yes, I agree. So what?

 

[Mohammad Ishmas]

Please show your work.

Are you saying that the calculated tensions are identical regardless of the acceleration of gravity? Yes, I agree. So what?

I'll post the work ,
the calculated tensions are coming identical if we take net acceleration and then as we don't have to subtract mg. in my pdf , I have taken net acceleration but have also subtracted mg which is wrong. BUT STILL what I am saying is that if we take acceleration in upward direction only and then subtract mg and calculate tensions the tensions coming are not uniform means they are not matching with the previous method of taking net acceleration