Data Table help (for a wind tunnel (Lifting force etc.)

[Physics Time]

The original mass is 25.5g.

The mass(G) we got were:
23.4 mass(g) - 7.7 Air speed(km/h)
22 mass(g) - 9.7 Air speed(km/h)
20.3 mass(g) - 12.1 Air speed(km/h)
19.2 mass(g) - 13.5 Air speed(km/h)
18.35 mass(g) - 13.7 Air speed(km/h)

I needed to find the lifting force and the Speed in m s-1

So original mass 25.5 x 9.8 = 249.9

Then the mass' we got:
23.4 x 9.8 = 229.32
22 x 9.8 = 215.6
20.3 x 9.8 = 198.9
19.2 x 9.8 = 188.16
18.35 x 9.8 = 179.83

So take away all the mass' with the original mass (example 249.9 - 229.32 = 20.58)

We get for lifting force (N):
20.58
34.4
51
61.74
70.07

* The teacher told me that this doesn't look like in Newtons.. He told me that I should change the mass(g) to kg.. ? So if I do that will that be the lifting force (N) ?

So 23.4 G is now 0.0234Kg
22 grams = 0.022 kilograms
20.3 grams = 0.0203 kilograms
19.2 grams = 0.0192 kilograms
18.35 grams = 0.01835 kilograms

So do i now 0.022 x 9.8? 0.2156... Then 249.9 - 0.2156 = 249.6844 to find the lifting force??

By the way the air speed in m/s is now:
2.14
2.69
3.36
3.75
3.80

After all this I have to draw two separate graphs one with the lifting force and air speed, one with the lifting force and v^squared (so air speed squared)... on a graph paper and then explain what we see...

 

[Awatarn]

You need to convert the unit of mass to kg to make a consistent.
It may be easily for you to convert every unit into SI unit which mean [mass] = kg, [Force] = N = kg s^2 / m

I'm not sure about your experiment so I can't answer your second question.

 

[Physics Time]

So what do i exactly do now?

Original Mass: 25.55g

Two Columns:
23.4 mass(g) - 7.7 Air speed(km/h)
22 mass(g) - 9.7 Air speed(km/h)
20.3 mass(g) - 12.1 Air speed(km/h)
19.2 mass(g) - 13.5 Air speed(km/h)
18.35 mass(g) - 13.7 Air speed(km/h)

And I need to put these two columns into these two columns:
Lift(N) - Speed (ms-1)..

Could you please show me exactly the way to do this properly and even do just one of them like 23.4 mass(g) - 7.7 Air speed(km/h) ..
Remembering that my teacher told me that G must be converted to Kg..
I really need to find the Lifting force. I need this urgently so I can actually start drawing the required graphs. Thank you.

 

[Physics Time]

My teacher told me that I have to first change the mass (g) to kg then times that by 9.8...

Will that give me the Lift force for everyone of them or do I have to do something else as well?

 

[Physics Time]

Well? Anyone? All I want to do is find out the lifting force?

 

[Physics Time]

 

[rcgldr]

How did you make the measurements? I'm not aware of a common device that measures mass directly. What type of scale did you use and what unit of force does it measure?

 

[Physics Time]

We did it quickly so I wasn't aware of what kind of thing he exactly used. The teacher only wants us to change the things i wrote to the lifting force and to ms-1. I JUST NEED HELP FINDING OUT THE LIFTING FORCE please... THE DATA WASN'T ACCURATE BECAUSE THE TEACHER DID IT QUICKLY...

25.55g is the original mass..

The other data we got was..
23.4 mass(g)
22 mass(g)
20.3 mass(g)
19.2 mass(g)
18.35 mass(g)

Now I want to convert these to the lifting force (N)...

Do we 25.5g - 23.4g = 2.1g... then x9.8 = 20.58g then change that to kg...? which = 0.0206
So that will be the N?

Or do we change 23.4g into kg = 0.0234 Then x it by 9.8 = 0.22932 which will be the N (lifting force?)..?

I just want to know how to change the given data to the LIFTING FORCE, I don't need any explanations or how and why cause that's the data that our teacher has given us.. forget the rest I only want these data and draw the graph and then get over it for now. Thanks.

 

[Physics Time]

Well?!

 

[rcgldr]

What isn't clear is if the measured data is "gram force", or the actual force divided by gravity to get the "gram mass".

As far as the lifting force goes, you have it correct, the lifting force is the difference between the original mass (or original force) and the measured mass (or force) due to the air flow.