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# Help with analysing SI units in equations watch

1. Question:

The power required by a helicopter when hovering depends only upon the vertical thrust (a force) F provided by the blades, their length l, and the mass density of air, ρ.

a) Establish an equation that relates the helicopters power requirement to these three quantities.

b) By what factor is the power requirement increased such that the helicopter can maintain a constant altitude when it takes on a load that doubles its weight?

We have been doing questions to show the SI units on the left hand side of an equation match the right hand side. These are more advanced questions we are given to manipulate the variables of an equation to match the SI units of the subject of the equation. Part a talks about power depending on F, l and p. The SI units for power are kgm^2s^-3 whereas F, l and p on their own are kg^2m^-1s^-2.

It seems like a complete mess trying to get F,l and p to the right powers to match SI units on the left for power. Any advice on how to do it? Or an easier way to do it?

Thanks!!!
2. (Original post by big_bang_98)
Question:

The power required by a helicopter when hovering depends only upon the vertical thrust (a force) F provided by the blades, their length l, and the mass density of air, ρ.

a) Establish an equation that relates the helicopters power requirement to these three quantities.

b) By what factor is the power requirement increased such that the helicopter can maintain a constant altitude when it takes on a load that doubles its weight?

We have been doing questions to show the SI units on the left hand side of an equation match the right hand side. These are more advanced questions we are given to manipulate the variables of an equation to match the SI units of the subject of the equation. Part a talks about power depending on F, l and p. The SI units for power are kgm^2s^-3 whereas F, l and p on their own are kg^2m^-1s^-2.

It seems like a complete mess trying to get F,l and p to the right powers to match SI units on the left for power. Any advice on how to do it? Or an easier way to do it?

Thanks!!!
The problem you have is that you don't know the powers to which F, l and p should be raised. If you call these powers a, b and c, respectively, then what you need is for the unit of (F^a)(l^b)(p^c) to the the same as that for power. Substitute the base units for F, l and p, expand and simplify, and you will end up with kg, m and s, all to different powers, with each power being some linear combination of a, b and c. You can then compare these to the powers of kg, m and s in the unit for power, which will give you three simultaneous equations you can solve (they are quite easy to solve if you can get this far).

This strikes me as quite a hard problem for an A level - or is this for something else?
3. (Original post by Pangol)
The problem you have is that you don't know the powers to which F, l and p should be raised. If you call these powers a, b and c, respectively, then what you need is for the unit of (F^a)(l^b)(p^c) to the the same as that for power. Substitute the base units for F, l and p, expand and simplify, and you will end up with kg, m and s, all to different powers, with each power being some linear combination of a, b and c. You can then compare these to the powers of kg, m and s in the unit for power, which will give you three simultaneous equations you can solve (they are quite easy to solve if you can get this far).

This strikes me as quite a hard problem for an A level - or is this for something else?
Thanks. I will have a go first and then reply if I can't manage it. It is a-level but it is part of the extension work at the bottom of the work sheet. Most of the questions are "write voltage in terms of SI units only", "show that the SI units on the right hand side are the same as on the left".
4. (Original post by Pangol)
The problem you have is that you don't know the powers to which F, l and p should be raised. If you call these powers a, b and c, respectively, then what you need is for the unit of (F^a)(l^b)(p^c) to the the same as that for power. Substitute the base units for F, l and p, expand and simplify, and you will end up with kg, m and s, all to different powers, with each power being some linear combination of a, b and c. You can then compare these to the powers of kg, m and s in the unit for power, which will give you three simultaneous equations you can solve (they are quite easy to solve if you can get this far).

This strikes me as quite a hard problem for an A level - or is this for something else?
So I have tried this:

(F)a (L)b (p)c = (kgm/s2)a (m)b (kg/m3)c

Expanding would be:

kgama/s(2a) x mb x kgc/m3c

Bringing all three together would give:

kgama/s(2a)mb kgc/m3c

Is what I have done correct so far? I am now unsure what to do - please could you give me some help?

Thank you!
5. (Original post by big_bang_98)
So I have tried this:

(F)a (L)b (p)c = (kgm/s2)a (m)b (kg/m3)c

Expanding would be:

kgama/s(2a) x mb x kgc/m3c

Bringing all three together would give:

kgama/s(2a)mb kgc/m3c

Is what I have done correct so far? I am now unsure what to do - please could you give me some help?

Thank you!
That looks good so far. You will find the next step easier if, rather than /s^(2a), you write s^(-2a), for example.

Collect together the similar terms - so, for example, you should have kg^(a+c). Since your unit for power is kg^1, this means that a+c=1. And so on.
6. (Original post by Pangol)
That looks good so far. You will find the next step easier if, rather than /s^(2a), you write s^(-2a), for example.

Collect together the similar terms - so, for example, you should have kg^(a+c). Since your unit for power is kg^1, this means that a+c=1. And so on.
Great, thanks!

So I had as far as:

kgama/s(2a)mb kgc/m3c

kgamas(-2a)mb kgcm-3c

Shuffling to get like terms together:

kgakgcmambm-3cs(-2a)

Putting like terms together:

kg(a+c) m(a+b-3c) s(-2a)

The SI units for power are:
kg1m2s-3

So now I would make some equations from the powers:

The raised power of kg = 1 therefore a+c = 1
The raised power of m = 2 therefore a+b-3c = 2
The raised power of s = -3 therefore -2a = -3

Is this correct so far? Now do I use simultaneous equations? I haven't dome those for a while will look them up and have a go if the above is correct?

Thanks again for the help!
7. (Original post by Pangol)
That looks good so far. You will find the next step easier if, rather than /s^(2a), you write s^(-2a), for example.

Collect together the similar terms - so, for example, you should have kg^(a+c). Since your unit for power is kg^1, this means that a+c=1. And so on.

Ok, so simultaneous equations are not as bad as I remember at GCSE. Would this be correct:

I had got as far as:

1) a+c = 1
2) a+b-3c = 2
3) -2a = -3

So equation 3 shows us that a = 2/3
Putting this into equation 1 shows us that c = 3/2
I am not sure if my substitution and solving for b is correct as looks like a really strange number. So I did

a+b-3c = 2

b = 2 + 3c - a
b = 2 + 3(3/2) - (2/3)
b = 2 + 9/2 - 2/3
b = 2 + (27/6 - 4/6)
b = 2 + (23/6)
b = 12/6 + 23/6
b = 35/6

Is that correct? What a stupid number b comes out as!
8. (Original post by big_bang_98)
3) -2a = -3

So equation 3 shows us that a = 2/3
All the right ideas, but look again at what you have done here.
9. (Original post by Pangol)
All the right ideas, but look again at what you have done here.
Ahhhhh how silly of me!

a = 3/2

so a+c = 1
c = 1-a = 1 - 3/2 = -1/2

Using equation 2:

b = 2 + 3c - a
b = 2 + 3(-1/2) - 3/2
b = 2 - 3/2 - 3/2
b = -1

So going back to original equation:

Power = FaLbpc we can now say that in order for it to work:

Power = F3/2L-1p-1/2

Is that correct? I was just thinking that if this is correct it confuses me a bit. If length (L) and density (p) have negative powers does this not mean that as these quantities go up, the helicopter power goes down? Because L-1 is the same as 1/L isn't it? I would have thought that all three variables would have a positive relationship with power?

Thank you again, you have been super super helpful!
10. (Original post by big_bang_98)
Power = F3/2L-1p-1/2

Is that correct? I was just thinking that if this is correct it confuses me a bit. If length (L) and density (p) have negative powers does this not mean that as these quantities go up, the helicopter power goes down? Because L-1 is the same as 1/L isn't it? I would have thought that all three variables would have a positive relationship with power?
That's what I get, so unless we're both wrong, that is it. Note that you do not know for sure that power = this, just that it is proportional to it. There might be a constant in there as well, but as constants have no dimensions, we won't be able to find them using this method. We are also assuming that the three quantities involved are the only things that the power depends on (although we are told that we can assume this in the question).

Shockily, I didn't stop to reflect on whether this makes sense, which should always be done in physics questions, so thanks for raising this. Length first - if the length of the blades is increased, they would be able to rotate more slowly and still sweep out the same volume of air per second, and so the required power would decrease. Similarly, if the density of the air was greater, then less air would be required to be swept downwards per second for the same change in momentum, and so again less power would be required. So this does make sense. I think!
11. (Original post by Pangol)
That's what I get, so unless we're both wrong, that is it. Note that you do not know for sure that power = this, just that it is proportional to it. There might be a constant in there as well, but as constants have no dimensions, we won't be able to find them using this method. We are also assuming that the three quantities involved are the only things that the power depends on (although we are told that we can assume this in the question).

Shockily, I didn't stop to reflect on whether this makes sense, which should always be done in physics questions, so thanks for raising this. Length first - if the length of the blades is increased, they would be able to rotate more slowly and still sweep out the same volume of air per second, and so the required power would decrease. Similarly, if the density of the air was greater, then less air would be required to be swept downwards per second for the same change in momentum, and so again less power would be required. So this does make sense. I think!
Brilliant, thanks for all your help - you are a legend! Hopefully my teacher will be impressed I managed to do his extension question
12. (Original post by big_bang_98)
Brilliant, thanks for all your help - you are a legend! Hopefully my teacher will be impressed I managed to do his extension question
You did most of it yourself, but there's nothing wrong with being honest and admitting that you asked for a few pointers...
13. (Original post by Pangol)
You did most of it yourself, but there's nothing wrong with being honest and admitting that you asked for a few pointers...
I don't think he would believe me if I said I did it on my own

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