Find the load saved by a pulley system

[Ithilrandir]

Homework Statement

None of the identical gondolas can support both the loads of P and F. Giuseppe rigs them up from the mass as shown in the diagram, using massless ropes and massless, frictionless pulleys. He ferries them across before they hit either the mast or the deck. Assuming P's mass is 90kg and F's mass is 60kg, how much load W does Giuseppe save?

Hint: Remember that the tension in a massless cord that passes over a massless, frictionless pulley is the same on both sides of the pulley.

Relevant Equations

...

There are 3 different strings in this system. The one pulling P, the one pulling F and the one pulling a pulley. Since the questions says they're ferried across before hitting mast or deck, I'm assuming that they are not stationary.

g is gravitational acceleration.

P will move twice as fast as F, so a_p = 2 a_f

60a_f = T- 60g

Tension on the string pulling the pulley and the string pulling P are both 2T

90a_p = 90g - 2T

T=60(a_f+g)
T=45(g - a_p)

150a_f = -15g

a_f = - 0.1g

T= 54g

Load = 3T = 162g

From this it doesn't look like any load was saved at all. The answer indicates a load saved of 2.7g.

 

[haruspex]

P will move twice as fast as F

How do you arrive at that?
The reliable way to work these things out, as in the earlier thread, is to look at how movements affect the lengths of string sections and note that the whole string has constant length.

150af = -15g

60+45=?

 

[Ithilrandir]

How do you arrive at that?
The reliable way to work these things out, as in the earlier thread, is to look at how movements affect the lengths of string sections and note that the whole string has constant length.

I arrived at that from looking at how it moves. F go up by 1 unit, it will run down the pulley and go up again resulting in P lowered by 2 units.

As for 60+45 I had substituted the a.

 

[haruspex]

F go up by 1 unit, it will run down the pulley and go up again resulting in P lowered by 2 units.

As I wrote, the reliable way is to add up the lengths. Try it.

As for 60+45 I had substituted the a.

That doesn't answer my question. You seem to have got 60+45=150.

 

[Ithilrandir]

As I wrote, the reliable way is to add up the lengths. Try it.

That doesn't answer my question. You seem to have got 60+45=150.

a_p = 2 a_f
T=60(a_f+g)
T=45(g - a_p)

60(a_f+g) = 45(g - 2a_f), from

a_p = 2 a_f


As for adding lengths I don't find it as straight forward. The right vertical decrease by Δx, the left vertical increase by Δx, the missing Δx I can't see it.

 

[haruspex]

ap = 2 af
T=60(af+g)
T=45(g - ap)

60(af+g) = 45(g - 2af), from

ap = 2 af

I'm not sure why you have copied that out again.
In post #2 I challenged the 150 in "150af = -15g". Seems to me you added 60 and 45 and got 150.

As for adding lengths I don't find it as straight forward. The right vertical decrease by Δx, the left vertical increase by Δx, the missing Δx I can't see it.

You are not finding your current method straightforward, since it is giving the wrong answer.
There are three straight sections, all vertical.
Working from the left, call them L₁, L₂ and L₃.
If the total length is L, what two equations can you write that relate them?

 

[Ithilrandir]

I'm not sure why you have copied that out again.
In post #2 I challenged the 150 in "150af = -15g". Seems to me you added 60 and 45 and got 150.

Well I just replaced the a_p with 2 a_f
45 x 2 a_f is where the 90 comes from.

You are not finding your current method straightforward, since it is giving the wrong answer.
There are three straight sections, all vertical.
Working from the left, call them L₁, L₂ and L₃.
If the total length is L, what two equations can you write that relate them?

ΔL₁ + ΔL₂ + ΔL₃ = 0?

 

[haruspex]

Well I just replaced the ap with 2 af
45 x 2 af is where the 90 comes from.

You seem to be deliberately misunderstanding my point!
I'll try one more time. In post #1 you wrote:

T=60(a_f+g)
T=45(g - a_p)
150a_f = -15g

Where did that 150 come from? 60+45=105, not 150

ΔL1 + ΔL2 + ΔL3 = 0?

Yes, the sum of the lengths is constant, but there is an even simpler equation relating the first two. Just look at the diagram.

 

[Ithilrandir]

You seem to be deliberately misunderstanding my point!

I'll type all the steps I wrote.

a_p = 2 a_f -------1
T=60(a_f+g) -------- 2
T=45(g - a_p) ------- 3

60(a_f + g) = 45(g - a_p) - - - 4

Subbing 1 into 4, replacing a_p with 2a_f,

60a_f + 60g = 45g - 45(2)a_f

60a_f + 60g = 45g - 90a_f

150 a_f = - 15g

Yes, the sum of the lengths is constant, but there is an even simpler equation relating the first two. Just look at the diagram.

I don't think I can see it in terms of length. L₃ contracts by Δx, L₁ contracts by Δx, L₂ extends by 2Δx?

 

[haruspex]

I'll type all the steps I wrote.

My apologies - didn't notice one was a and one was a_r.

I don't think I can see it in terms of length. L3 contracts by Δx, L1 contracts by Δx, L2 extends by 2Δx?

Ignore L₃ for the moment. If the left hand pulley rises by Δx, what happens to L₁ and L₂?

 

[Ithilrandir]

Ignore L₃ for the moment. If the left hand pulley rises by Δx, what happens to L₁ and L₂?

If the left pulley rises by Δx, both L₁ and L₂ will decrease by Δx.

 

[Steve4Physics]

You have had the issue pointed out several times by @haruspex! Here are 3 direct questions for you:

Q1. If P moves down 1cm, what does F do? Ask yourself how much longer or shorter each section of string is, given that the total length is fixed!

Q2. Which has the greater acceleration, P or F?

Q3. Is "ap = 2 af" correct?

 

[haruspex]

If the left pulley rises by Δx, both L₁ and L₂ will decrease by Δx.

So what equation does the give you relating ΔL₁ and ΔL₂?

 

[Ithilrandir]

So what equation does the give you relating ΔL₁ and ΔL₂?

ΔL₁ = ΔL₂, so in that case ΔL₃ = 2 Δx?

 

[haruspex]

ΔL₁ = ΔL₂, so in that case ΔL₃ = 2 Δx?

Yes, assuming you are taking the change in the other two as -Δx.

 

[Ithilrandir]

Yes, assuming you are taking the change in the other two as -Δx.

I was able to solve it with this change. That said I still feel a little iffy about this method. I'll have to read up more on this.