What is the tension in part GH?

AI Thread Summary
The discussion centers on calculating the tension in part GH of a system involving multiple masses and a spring. Participants analyze the forces acting on each mass, emphasizing the importance of free body diagrams for accurate calculations. The tension in the spring is highlighted as a critical factor, particularly when the system experiences changes, such as when a mass is cut. The conversation reveals confusion regarding the relationships between the tensions in different parts of the string and the effects of gravity on the system's acceleration. Ultimately, the tension in part GH is derived from a series of equations based on the forces acting on the masses involved.
Suyash Singh
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Homework Statement


upload_2018-4-28_9-21-50.png

upload_2018-4-28_9-27-47.png


Homework Equations


g is the acceleration due to gravity
a is the acceleration of system

The Attempt at a Solution


g is the acceleration due to gravity
Force=3(g+a)
Force=6(g-a)
solving both equations,
a=g/3
tension =force=4g

How do i calculate tension in the particular part GH?
and
what does the spring do here?
 

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Suyash Singh said:

Homework Statement


View attachment 224796
View attachment 224797

Homework Equations


g is the acceleration due to gravity
a is the acceleration of system

The Attempt at a Solution


g is the acceleration due to gravity
Force=3(g+a)
Force=6(g-a)
solving both equations,
a=g/3
tension =force=4g

How do i calculate tension in the particular part GH?
and
what does the spring do here?
GH and FE are pieces of the same string. The mass of the pulley is negligible. What about the tensions in the pieces of the string?
There is tension in the spring, acting at both masses attached at its ends. Does that tension change at the instant when AB is cut? Watch this video



You determined the acceleration when all masses move together. But it does not happen at once when AB is cut, because of the spring.
 
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Force=1(g-a)
Force=3(g+a)
0=g-a-3g-3a
0=-2g-4a
a=g/2
Tension=1(g/2)
Its wrong
 
I don't see any free body diagrams or any force balances on the masses. If T1 is the tension in CD and T2 is the tension in EFGH, what is the force balance equation for each of the 4 masses? (Or, do you think you have advanced to the point where you no longer need to use free body diagrams?)
 
Suyash Singh said:
Force=1(g-a)
Force=3(g+a)
0=g-a-3g-3a
0=-2g-4a
a=g/2
Tension=1(g/2)
Its wrong
What is a? Do you think the spring does not matter at all?
 
ehild said:
What is a? Do you think the spring does not matter at all?
a is the acceleration of the system.
Spring cancels out the 5kg mass according to the video you shared.
Chestermiller said:
I don't see any free body diagrams or any force balances on the masses. If T1 is the tension in CD and T2 is the tension in EFGH, what is the force balance equation for each of the 4 masses? (Or, do you think you have advanced to the point where you no longer need to use free body diagrams?)
We only get like 30 seconds per question in the exam so i naturally try to avoid free body diagrams.
 
Suyash Singh said:
a is the acceleration of the system.
Spring cancels out the 5kg mass according to the video you shared.

We only get like 30 seconds per question in the exam so i naturally try to avoid free body diagrams.
Bad idea. But, this is not an exam. So, let's see what you get if you use proper free body diagrams.
 
Chestermiller said:
Bad idea. But, this is not an exam. So, let's see what you get if you use proper free body diagrams.
upload_2018-4-29_8-46-59.png
 

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  • #10
Chestermiller said:
OK. Let's see the force balance equations on each of the masses.
For right hand side,
Force=3(g-a) its falling
for left side,
Force=1(g+a) its rising
 
  • #11
Suyash Singh said:
For right hand side,
Force=3(g-a) its falling
for left side,
Force=1(g+a) its rising
I asked for it on each individual mass. And, please don't use the word force. I want to see some ma's.
 
  • #12
Chestermiller said:
I asked for it on each individual mass. And, please don't use the word force. I want to see some ma's.
i don't know what to do :(
 
  • #13
$$T-(5)g-(1)g=(1)a$$
$$(2)g-T'=(2)a$$
$$T'+(1)g-T=(1)a$$where a is the upward acceleration on the left and downward acceleration on the right. Do these equations make sense to you?
 
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  • #14
Suyash Singh said:
a is the acceleration of the system.
Spring cancels out the 5kg mass according to the video you shared.
No, the spring does not "cancel out the 5kg mass". The video shows that the tension in the spring is present even after releasing it, and decreases gradually as the length changes. At the instant when AB is cut the 5 kg mass does not accelerate. so the tension in the spring is Ts=5g, and that tension acts also at the 1 kg mass above it.
 
  • #15
2
Chestermiller said:
$$T-(5)g-(1)g=(1)a$$
$$(2)g-T'=(2)a$$
$$T'+(1)g-T=(1)a$$where a is the upward acceleration on the left and downward acceleration on the right. Do these equations make sense to you?
not exactly but i will try

Mass 1 kg(left) travels upwards, 3kg(right) travels downwards, 5 kg mass is stationary but is attracted by gravity
T=1(g+a)+3(g-a)+5(g)

2kg is falling
T'=2(g-a)

how come the system is falling on the right side?Is it because the 5 kg mass is being held by the spring at that instant?
 
  • #16
Suyash Singh said:
2

not exactly but i will try

Mass 1 kg(left) travels upwards, 3kg(right) travels downwards, 5 kg mass is stationary but is attracted by gravity
T=1(g+a)+3(g-a)+5(g) Wrong !

2kg is falling
T'=2(g-a)

how come the system is falling on the right side?Is it because the 5 kg mass is being held by the spring at that instant?
If you solve @Chestermiller 's equations in Post #15 correctly, you will get negative acceleration. The system will move in the opposite direction as it was assumed.
 
  • #17
Suyash Singh said:
T=1(g+a)+3(g-a)+5(g)
This is a common blunder. The tension in a string is not the sum of the forces applied at the ends.
Assuming the string is either light or stationary, the forces at the ends are each equal to the tension.
 
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  • #18
i am more confused now :cry:
left side
1 kg is going up
T=1(g+a)+5g
 
  • #19
sorry for late reply (parents disturb me too much)
2 kg is going down
T'=2(g-a)
 
  • #20
but how to find relation between T' and T?
 
  • #21
Suyash Singh said:
but how to find relation between T' and T?
You are told the pulley is light (and assume a frictionless axle). If the tensions are T and T', what is the net torque on the pulley? If its moment of inertia is effectively zero, what is its angular acceleration?
Edit: my confusion re definition of T'
 
Last edited:
  • #22
haruspex said:
You are told the pulley is light (and assume a frictionless axle). If the tensions are T and T', what is the net torque on the pulley? If its moment of inertia is effectively zero, what is its angular acceleration?
torque=I alpha
but I=0
torque=0
alpha=a/r
but radius is not given
 
  • #23
Suyash Singh said:
but how to find relation between T' and T?
write the equation for the 1 kg mass on the left. You can cancel T'. See @Chestermiller's equations.
 
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  • #24
haruspex said:
You are told the pulley is light (and assume a frictionless axle). If the tensions are T and T', what is the net torque on the pulley? If its moment of inertia is effectively zero, what is its angular acceleration?
T' was meant the tension in the string connected the 2kg mass to the 1 kg one. See Post #8
 
  • #25
ehild said:
T' was meant the tension in the string connected the 2kg mass to the 1 kg one. See Post #8
Ah - my mistake. Thanks.
 
  • #26
ehild said:
write the equation for the 1 kg mass on the left. You can cancel T'. See @Chestermiller's equations.
1 kg is going up
"something"=1(g+a)

i don't know what "something" is but it some kind of force or tension.
 
  • #27
Suyash Singh said:
1 kg is going up
"something"=1(g+a)

i don't know what "something" is but it some kind of force or tension.
Write the equations according to the free-body diagrams in the form ma= ∑F. See your own picture Post #8. The forces acting on the 1 kg mass are the tension T upward , the tension T' downward and the weight 1*g. So ma = ?
 
  • #28
1a=T-T'-1g
a=T-T'-g
a=1(g+a)+5g-1(g+a)-5g-g
a=g+a+5g-g-a-6g
a=-g
 
  • #29
Suyash Singh said:
1a=T-T'-1g
a=T-T'-g
a=1(g+a)+5g-1(g+a)-5g-g
a=g+a+5g-g-a-6g
a=-g
wrong
Write the equation for the 1 kg mass on the left in the form ma=ΣF.
 
  • #30
ehild said:
wrong
Write the equation for the 1 kg mass on the left in the form ma=ΣF.
1a=T-T'
 
  • #31
Suyash Singh said:
1a=T-T'
You left out the gravitational force on the 1 kg mass.

(How could you possibly have thought you could solve this problem without using free body diagrams?)
 
  • #32
Chestermiller said:
You left out the gravitational force on the 1 kg mass.

(How could you possibly have thought you could solve this problem without using free body diagrams?)
a=T-T'-g
Chestermiller said:
(How could you possibly have thought you could solve this problem without using free body diagrams?)
I only get around 30 seconds for one question :(
 
  • #33
Suyash Singh said:
a=T-T'-g

I only get around 30 seconds for one question :(
Your objective is to learn the subject, not to complete problems in a certain amount of time. In my judgment, this problem can’t reasonably be analyzed in 30 seconds. How much time have you spent on it so far?
 
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  • #34
Suyash Singh said:
a=T-T'-g
(
Very nice. You've made progress. Now, let's the force balances on the other 3 masses. After that, I'll give you some ideas on how to solve the equations to get the acceleration and tensions.
 
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  • #35
Chestermiller said:
Very nice. You've made progress. Now, let's the force balances on the other 3 masses. After that, I'll give you some ideas on how to solve the equations to get the acceleration and tensions.
on 1kg left side,
a=T-T'-g
on 1 kg right side,
a=-T+g
on 2kg right side,
2a=-T-T'+2g
 
  • #36
Suyash Singh said:
on 1kg left side,
a=T-T'-g
on 1 kg right side,
a=-T+g
on 2kg right side,
2a=-T-T'+2g
This is not what I get. Using the symbols and diagram in your post #8, I corresponding equations get are:

$$a=T-5g-g$$
$$a=T'-T+g$$
$$2a=-T'+2g$$

Do you see where T is not acting on the 2 kg mass, only T'?
Do you see where T' is acting on the 1 kg mass on the right?
Do you see where T' is not acting on the 1 kg mass on the left, but 5g is (the force exerted by the spring)?
 
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  • #37
Chestermiller said:
This is not what I get. Using the symbols and diagram in your post #8, I corresponding equations get are:

$$a=T-5g-g$$
$$a=T'-T+g$$
$$2a=-T'+2g$$

Do you see where T is not acting on the 2 kg mass, only T'?
Do you see where T' is acting on the 1 kg mass on the right?
Do you see where T' is not acting on the 1 kg mass on the left, but 5g is (the force exerted by the spring)?
oh ok i understand now

1 kg mass on left,
T up, 5g down ,g down
a=T-5g-g

1 kg on right,
2g down , 1g down, T up
a=2g+g-T

2 kg right,
T' up,2g down
2a=-T'+2g

second equation is not matching
 
  • #38
Suyash Singh said:
1 kg on right,
2g down , 1g down, T up
a=2g+g-T

2g down , 1g down, T up
No. T' down, 1g down, T up
 
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  • #39
Chestermiller said:
2g down , 1g down, T up
No. T' down, 1g down, T up
a=T'+g-T
 
  • #40
Suyash Singh said:
a=T'+g-T
Good. Now, add the three equations together. What do you get?
 
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  • #41
Chestermiller said:
Good. Now, add the three equations together. What do you get?
4a=T-5g-g+T'+g-T-T'+2g
4a=-6g+3g
a=-3g/4
 
  • #42
Suyash Singh said:
4a=T-5g-g+T'+g-T-T'+2g
4a=-6g+3g
a=-3g/4

from equation 1
a=T-6g
T=-3g/4+6g
T=21g/4
 
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