What is the tension in part GH?

[Suyash Singh]

Homework Statement

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?

 

[ehild]

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.

 

[Suyash Singh]

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

 

[Chestermiller]

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?)

 

[ehild]

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?

 

[Suyash Singh]

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.

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.

 

[Chestermiller]

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.

 

[Suyash Singh]

Bad idea. But, this is not an exam. So, let's see what you get if you use proper free body diagrams.

 

[Chestermiller]

View attachment 224855

OK. Let's see the force balance equations on each of the masses.

 

[Suyash Singh]

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

 

[Chestermiller]

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.

 

[Suyash Singh]

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 :(

 

[Chestermiller]

$$T-(5)g-(1)g=(1)a$$
$$(2)g-T'=(2)a$$
$$T'+(1)g-T=(1)a$$where a is the upward acceleration on the left and downward acceleration on the right. Do these equations make sense to you?

 

[ehild]

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.

 

[Suyash Singh]

2

$$T-(5)g-(1)g=(1)a$$
$$(2)g-T'=(2)a$$
$$T'+(1)g-T=(1)a$$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?

 

[ehild]

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.

 

[haruspex]

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.

 

[Suyash Singh]

i am more confused now
left side
1 kg is going up
T=1(g+a)+5g

 

[Suyash Singh]

sorry for late reply (parents disturb me too much)
2 kg is going down
T'=2(g-a)

 

[Suyash Singh]

but how to find relation between T' and T?

 

[haruspex]

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'

 

[Suyash Singh]

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

 

[ehild]

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.

 

[ehild]

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

 

[haruspex]

T' was meant the tension in the string connected the 2kg mass to the 1 kg one. See Post #8

Ah - my mistake. Thanks.

 

[Suyash Singh]

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.

 

[ehild]

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 = ?

 

[Suyash Singh]

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

 

[ehild]

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.

 

[Suyash Singh]

wrong
Write the equation for the 1 kg mass on the left in the form ma=ΣF.

1a=T-T'