Inclined Plane Pulley System: Finding Tensions

[sodium40mg]

Homework Statement

Three blocks are connected by two different strings over two pulleys. Assume no friction between the pulleys and the rope segments.

• Green mass = 13 kg
• Red mass = 18 kg
• Blue mass = 23 kg

The coefficient of friction between the masses and the surface is 0.20.

Find the tensions between the masses.

Homework Equations

• fg = mg
• fnet = ma
• fnet = total forces

The Attempt at a Solution

So what I'm thinking of doing is like sectioning each mass.

Green mass

m = 13 kg
Fnet = T1 - Fg1 - Ff
(13)(a) = (13)(9.8)(sinθ1) - (0.20)(13)(9.8)(cosθ1) - T1

So T1 and a is unknown for the Green mass

Red mass

m = 18 kg
Fnet = T2 - T1
(18)(a) = T2 - T1

Three unknowns here.

Blue mass

m = 23 kg
Fnet = T2 - Fg3 - Ff
(23)(a) = T2 - (23)(9.8)(sinθ2) - (0.20)(23)(9.8)(cosθ2)

Two unknowns here.The easiest thing to do now would be to rearrange so that each equals a tension and then substitute that into the other equation to find acceleration. Once I got acceleration, I would just be able to substitute and find the tensions.

But what I'm unsure about is whether the equations are right. Are the signs right? Did I subtract something I should've added or vice versa? Any help would be appreciated.

 

[I like Serena]

Green mass

m = 13 kg
Fnet = T1 - Fg1 - Ff
(13)(a) = (13)(9.8)(sinθ1) - (0.20)(13)(9.8)(cosθ1) - T1

So T1 and a is unknown for the Green mass

Red mass

m = 18 kg
Fnet = T2 - T1
(18)(a) = T2 - T1

Three unknowns here.

Blue mass

m = 23 kg
Fnet = T2 - Fg3 - Ff
(23)(a) = T2 - (23)(9.8)(sinθ2) - (0.20)(23)(9.8)(cosθ2)

Two unknowns here.

For the red mass you have left out friction, which should be added.

For the green mass, you did not substitute properly.
As a result the minus signs are incorrect.
Proper substitution is:
(13)(a) = T1 - (13)(9.8)(sinθ1) - (0.20)(13)(9.8)(cosθ1)
or
(13)(-a) = T1 - (13)(9.8)(sinθ1) - (0.20)(13)(9.8)(cosθ1)
depending on the direction you choose for a.

The direction you choose for a has to be applied in the same direction for the red and blue masses, which you did not do yet.

 

[sodium40mg]

For the red mass you have left out friction, which should be added.

For the green mass, you did not substitute properly.
As a result the minus signs are incorrect.
Proper substitution is:
(13)(a) = T1 - (13)(9.8)(sinθ1) - (0.20)(13)(9.8)(cosθ1)
or
(13)(-a) = T1 - (13)(9.8)(sinθ1) - (0.20)(13)(9.8)(cosθ1)
depending on the direction you choose for a.

The direction you choose for a has to be applied in the same direction for the red and blue masses, which you did not do yet.

Could you explain why the T1 is positive instead of being -T1?

 

[I like Serena]

Could you explain why the T1 is positive instead of being -T1?

Basically T1 is positive, because it says so in your original formula: Fnet = T1 - Fg1 - Ff

Thinking a bit further, you need to choose the direction for all of your forces.
T1 is the tension-force in the string, which pulls the green block upward.
Gravity works in an opposing direction, which is shown in your original formula with the minus sign.
Furthermore you have chosen the friction to work in an opposing direction as well, which is a good choice, because that way it will have a positive value.

T1 also works on the red block, but here it's effectively the reactive force, which pulls the red block to the left. Hence you've put a minus sign in your red block formula (Fnet = T2 - T1), which is correct.

When you made the substitution for the green block, apparently you meant to do:
(13)(-a) = T1 - (13)(9.8)(sinθ1) - (0.20)(13)(9.8)(cosθ1)
If we multiply left and right by -1, we'll get:
(13)(a) = -T1 + (13)(9.8)(sinθ1) + (0.20)(13)(9.8)(cosθ1)

So T1 has a minus sign now, but if you check, you'll see that this is not the same formula you had, since there is a difference in -/+ signs.

Anyway, if you did this, you implicitly made a choice for the direction of a, which is inconsistent with your choice for the red block, where you have chosen a in the opposite direction.
So the proper formula for the green block is:
(13)(a) = T1 - (13)(9.8)(sinθ1) - (0.20)(13)(9.8)(cosθ1)

 

[sodium40mg]

Basically T1 is positive, because it says so in your original formula: Fnet = T1 - Fg1 - Ff

Thinking a bit further, you need to choose the direction for all of your forces.
T1 is the tension-force in the string, which pulls the green block upward.
Gravity works in an opposing direction, which is shown in your original formula with the minus sign.
Furthermore you have chosen the friction to work in an opposing direction as well, which is a good choice, because that way it will have a positive value.

T1 also works on the red block, but here it's effectively the reactive force, which pulls the red block to the left. Hence you've put a minus sign in your red block formula (Fnet = T2 - T1), which is correct.

When you made the substitution for the green block, apparently you meant to do:
(13)(-a) = T1 - (13)(9.8)(sinθ1) - (0.20)(13)(9.8)(cosθ1)
If we multiply left and right by -1, we'll get:
(13)(a) = -T1 + (13)(9.8)(sinθ1) + (0.20)(13)(9.8)(cosθ1)

So T1 has a minus sign now, but if you check, you'll see that this is not the same formula you had, since there is a difference in -/+ signs.

Anyway, if you did this, you implicitly made a choice for the direction of a, which is inconsistent with your choice for the red block, where you have chosen a in the opposite direction.
So the proper formula for the green block is:
(13)(a) = T1 - (13)(9.8)(sinθ1) - (0.20)(13)(9.8)(cosθ1)

I see, and yeah I want my acceleration to be positive. So with that said, would these be the correct equations?

Green mass:

(13)(a) = T1 - (13)(9.8)(sinθ1) - (0.20)(13)(9.8)(cosθ1)

Red mass:

(18)(a) = T2 - T1

Blue mass:

(23)(a) = T2 - (23)(9.8)(sinθ2) - (0.20)(23)(9.8)(cosθ2)

If they are - in which case, I would simply rearrange to find the unknowns.

 

[I like Serena]

I see, and yeah I want my acceleration to be positive. So with that said, would these be the correct equations?

Green mass:

(13)(a) = T1 - (13)(9.8)(sinθ1) - (0.20)(13)(9.8)(cosθ1)

Red mass:

(18)(a) = T2 - T1

Blue mass:

(23)(a) = T2 - (23)(9.8)(sinθ2) - (0.20)(23)(9.8)(cosθ2)

If they are - in which case, I would simply rearrange to find the unknowns.

I have a few corrections.

The green block is ok now.

The red mass needs friction which works against 'a':
(18)(a) = T2 - T1 - (0.20)(18)(9.8)

The blue mass should have:
Fnet = -T2 + Fg3 - Ff
(23)(a) = -T2 + (23)(9.8)(sinθ2) - (0.20)(23)(9.8)(cosθ2)

That is, we have chosen 'a' to move the green mass upward in the same direction as T1.
We have 'a' to move the red mass to the right, against T1, and in the same direction as T2.
And we have 'a' moving the blue mass down, against T2.
In all cases the friction works opposite to 'a'.

 

[ova5676]

I have a few corrections.

The green block is ok now.

The red mass needs friction which works against 'a':
(18)(a) = T2 - T1 - (0.20)(18)(9.8)

The blue mass should have:
Fnet = -T2 + Fg3 - Ff
(23)(a) = -T2 + (23)(9.8)(sinθ2) - (0.20)(23)(9.8)(cosθ2)

That is, we have chosen 'a' to move the green mass upward in the same direction as T1.
We have 'a' to move the red mass to the right, against T1, and in the same direction as T2.
And we have 'a' moving the blue mass down, against T2.
In all cases the friction works opposite to 'a'.

If you wouldn't mind, can you tell me about the active/reactive thing from another thread:

That's like three blocks hooked up by three pulleys. Would the active/reactive thing apply here in that (if the middle one is the heaviest)

The first mass: Fnet = T - Fg1 --> since it looks like the block is accelerating upwards (or if it's at rest does that change anything), the force gravity is acting against it so it's negative.The middle mass: Fnet = Fg2 - T --> since it looks like the block is accelerating downwards, the tension is working against it.

The third mass: The first mass: Fnet = T - Fg3 --> same as the first mass, right?

 

[I like Serena]

If you wouldn't mind, can you tell me about the active/reactive thing from another thread:

That's like three blocks hooked up by three pulleys. Would the active/reactive thing apply here in that (if the middle one is the heaviest)

The first mass: Fnet = T - Fg1 --> since it looks like the block is accelerating upwards (or if it's at rest does that change anything), the force gravity is acting against it so it's negative.


The middle mass: Fnet = Fg2 - T --> since it looks like the block is accelerating downwards, the tension is working against it.

The third mass: The first mass: Fnet = T - Fg3 --> same as the first mass, right?

Sound reasoning.
You've missed only 1 thing: there are 2 strings and not 1.
The tensile force in the string that carries the middle weight will be different.

 

[ova5676]

Sound reasoning.
You've missed only 1 thing: there are 2 strings and not 1.
The tensile force in the string that carries the middle weight will be different.

I was wondering about the two strings thing, but I was told that if pulleys were held at a ceiling instead of an inclined plane like this thread, that it's just one string? How positive are you that it's two strings? If it's two strings, two tensions so:

First mass = T1 - Fg1 - same thing basically

Second mass = Fg2 - T1 + T2 --> I'm not sure about this one but I would think since both tensions acting upwards you would add them but is it that or is it

Fg2 - (T1 + T2) = Fg2 - T1 - T2?

Third mass = T2 - Fg3 - same thing basically again.

 

[I like Serena]

I was wondering about the two strings thing, but I was told that if pulleys were held at a ceiling instead of an inclined plane like this thread, that it's just one string? How positive are you that it's two strings? If it's two strings, two tensions so:

First mass = T1 - Fg1 - same thing basically

Second mass = Fg2 - T1 + T2 --> I'm not sure about this one but I would think since both tensions acting upwards you would add them but is it that or is it

Fg2 - (T1 + T2) = Fg2 - T1 - T2?

Third mass = T2 - Fg3 - same thing basically again.

They are physically 2 separate strings.
Assuming the middle pulley to be weightless, the tensile force in the middle string will be exactly twice the tensile force in the string connecting the left weight to the right weight.
(Otherwise there would be a resulting force on the pulley, making it speed out of this universe ;).
So T2 = 2 x T1.

When looking at forces we only look at the forces that act directly on a given object.
That is:
First mass = T1 - Fg1
Second mass = Fg2 - T2
Third mass = T1 - Fg3

 

[ova5676]

They are physically 2 separate strings.
Assuming the middle pulley to be weightless, the tensile force in the middle string will be exactly twice the tensile force in the string connecting the left weight to the right weight.
(Otherwise there would be a resulting force on the pulley, making it speed out of this universe ;).
So T2 = 2 x T1.

When looking at forces we only look at the forces that act directly on a given object.
That is:
First mass = T1 - Fg1
Second mass = Fg2 - T2
Third mass = T1 - Fg3

So in other words, that would be:

First mass = T1 - Fg1

Second mass = Fg2 - 2(T1)

Third mass = T1 - Fg3

?

 

[I like Serena]

So in other words, that would be:

First mass = T1 - Fg1

Second mass = Fg2 - 2(T1)

Third mass = T1 - Fg3

?

Yes, that is correct.
Now you need to factor in that the strings have a constant length.
That constrains there relative coordinates (and there relative accelerations).

 

[ova5676]

Yes, that is correct.
Now you need to factor in that the strings have a constant length.
That constrains there relative coordinates (and there relative accelerations).

So acceleration is the same for each mass, is that what you mean?

 

[I like Serena]

So acceleration is the same for each mass, is that what you mean?

Something like that.
But the acceleration of each mass can only be the same if a displacement of each mass results in the same displacement of any other mass if the third mass is kept in place.
Is that the case?

 

[ova5676]

Something like that.
But the acceleration of each mass can only be the same if a displacement of each mass results in the same displacement of any other mass if the third mass is kept in place.
Is that the case?

Well in my specific case, they are at rest initially (three people are holding them) and then all three let go BUT one of them doesn't move (the one with the heaviest weight, by the way).

So, would I assume the first mass and the last mass have the same acceleration but the middle mass would have an acceleration equal to 0?

 

[I like Serena]

Well in my specific case, they are at rest initially (three people are holding them) and then all three let go BUT one of them doesn't move (the one with the heaviest weight, by the way).

So, would I assume the first mass and the last mass have the same acceleration but the middle mass would have an acceleration equal to 0?

Well, according to my calculations the middle mass would be stationary if its mass equals 4.m1.m3/(m1+m3).
Can that be the case?

Anyway, what I meant is that if you move the left mass down, and keep the right mass in place, the middle mass will go up by only half as much.
Put into other words, if the distance to the roof of each mass is y1, y2, resp. y3.
Then y1 + 2.y2 + y3 will be constant.

 

[ova5676]

Well, according to my calculations the middle mass would be stationary if its mass equals 4.m1.m3/(m1+m3).
Can that be the case?

Anyway, what I meant is that if you move the left mass down, and keep the right mass in place, the middle mass will go up by only half as much.
Put into other words, if the distance to the roof of each mass is y1, y2, resp. y3.
Then y1 + 2.y2 + y3 will be constant.

Ah, I see what you mean.

Hope you don't mind answering one last question please. If you look @ this situation:

Basically an object moves from that one inclined plane with friction across a frictionless surface then up a frictionless plane.

I'm not asking about calculating anything, but I'm wondering if I used a rotated coordinate system (as you can see) such that my Fgx on the left side was + and the Fgy on the left side was -, should I still use the same coordinate system for the same question when I'm "isolating" the second inclined plane?

In other words, would Fgx on the right side be - and the Fgy on the right side be +? Or would I have to use another coordinate system?

 

[I like Serena]

Ah, I see what you mean.

Hope you don't mind answering one last question please. If you look @ this situation:

Basically an object moves from that one inclined plane with friction across a frictionless surface then up a frictionless plane.

I'm not asking about calculating anything, but I'm wondering if I used a rotated coordinate system (as you can see) such that my Fgx on the left side was + and the Fgy on the left side was -, should I still use the same coordinate system for the same question when I'm "isolating" the second inclined plane?

In other words, would Fgx on the right side be - and the Fgy on the right side be +? Or would I have to use another coordinate system?

You're entirely free to use any coordinate system you want, and you can change your coordinate system whenever you feel like it.
It's only that if you have calculated a position and/or speed in one coordinate system and you want to switch coordinate system, you usually have to "transform" the coordinates.

In your case I don't think you need transformations, since everything is frictionless and distances do not seem to matter.
The only important thing is the direction of gravity. You will probably want to keep track how high the mass is as related to gravity. You need that to keep track of its potential energy (given by m.g.h). But you can do that in any coordinate system.