Determining Relationship Between Two Variables

[Stormblessed]

Homework Statement

An experiment was conducted to determine the relationship between metre-stick projection from a table (L) on the vertical depression of the free end (y). Using the collected data, graphs had to be plotted until a linear relation was found, and the equation of the line had to be found. To find the relationship, trial and error method had to be used. Note: all graphs are attached.

Here is the method:

1. Clamp a metre stick horizontally so that the first 20 cm project
beyond the lab table. Measure the vertical height of the free end
above the floor under zero load. Record in the table.

2. Attach a mass ( 500 g to 1 kg ) to the free end. Measure the vertical
height of the free end from the floor and record.

3. Repeat the above procedure for projections (L) in steps of 10 cm,
maintaining the constant load. Measure the vertical height of the
free end above the floor at each projection, each time checking the
zero load height.

4. Determine the vertical depression (y) corresponding to each
projection (L) and record.

After finding the relationship, answer this: Do you think the cross-sectional area of a metre-stick (cross sectional area is referring to the area of the rectangular face at one of the ends of the metre stick) has any effect on the vertical depression at a given extension? Discuss. Describe an experiment you might perform to test your prediction.

Homework Equations

Here is the collected data:

L (projection) - 0 cm; y (vertical depression) - 0 cm
L (projection) - 20 cm; y (vertical depression) - 0.1 cm
L (projection) - 30 cm; y (vertical depression) - 0.5 cm
L (projection) - 40 cm ; y (vertical depression) - 1.2 cm
L (projection) - 50 cm ; y (vertical depression) - 3.6 cm
L (projection) - 60 cm ; y (vertical depression) - 5.1 cm
L (projection) - 70 cm ; y (vertical depression) - 8.0 cm
L (projection) - 80 cm ; y (vertical depression) - 11.7 cm

The Attempt at a Solution

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I first plotted a graph using the collected data, which resulted in an exponential relationship. Then, as per the trial and error method, I squared the L values and replotted the data, but the result was still exponential (curve). I then cubed the L values and plotted the points again, which resulted in a somewhat linear relationship. Using this, I found the relation to be: y = 2 x 10^-5(L^3). However, I am not sure this is correct, as when I plug in the L values into the equation, the result is quite off from the y values in my data set. Also, I am not sure if the graph of y vs L^3 is linear (it looks a bit off).

I am also completely stumped for the cross sectional area question (Do you think the cross-sectional area of a metre-stick (cross sectional area is referring to the area of the rectangular face at one of the ends of the metre stick) has any effect on the vertical depression at a given extension? Discuss. Describe an experiment you might perform to test your prediction.) (I have no idea.)

Thanks

All graphs are attached.

 

[Merlin3189]

If you are thinking $y= ax^n$ , you could try plotting the log or ln of x and y.
Because, taking ln of both sides, you get $ln(y) = ln(a) + n ln(x)$ , which should be a straight line of slope n and intercept ln(a)
If you get a reasonable approximation to a straight line (it looks reasonable to me), you can estimate n and a, then put them back into $y= ax^n$

Edit: as for x-section effect:
Why does deflection increase as more of the stick hangs over the edge?
If you made the stick thicker, what effect might that have on the deflection?

On the other side of the coin, if you kept the same stick, but rotated it 90^o how would that affect the deflection?

 

[CWatters]

Also, I am not sure if the graph of y vs L^3 is linear (it looks a bit off).

I bashed the number into Excel and its not too bad. If you aren't happy perhaps try a number a bit less than ^3

I am also completely stumped for the cross sectional area question (Do you think the cross-sectional area of a metre-stick (cross sectional area is referring to the area of the rectangular face at one of the ends of the metre stick) has any effect on the vertical depression at a given extension? Discuss. Describe an experiment you might perform to test your prediction.) (I have no idea.)

Is that, no idea if the cross section has an effect, or no idea what experiment to do?

 

[Stormblessed]

I bashed the number into Excel and its not too bad. If you aren't happy perhaps try a number a bit less than ^3
Is that, no idea if the cross section has an effect, or no idea what experiment to do?

No idea if the cross section has an effect. I was thinking that it does not affect the depression at all, because if you placed two metre-sticks side by side (essentially doubling the x-sectional area), the vertical depression would be the same. But I'm not sure.

 

[Stormblessed]

If you are thinking $y= ax^n$ , you could try plotting the log or ln of x and y.
Because, taking ln of both sides, you get $ln(y) = ln(a) + n ln(x)$ , which should be a straight line of slope n and intercept ln(a)
If you get a reasonable approximation to a straight line (it looks reasonable to me), you can estimate n and a, then put them back into $y= ax^n$

Edit: as for x-section effect:
Why does deflection increase as more of the stick hangs over the edge?
If you made the stick thicker, what effect might that have on the deflection?

On the other side of the coin, if you kept the same stick, but rotated it 90^o how would that affect the deflection?

I am actually not sure why deflection increases as more of the stick hangs over the edge. But if the stick were to be thicker, would deflection stay the same?

 

[CWatters]

I am actually not sure why deflection increases as more of the stick hangs over the edge.

Have you studied levers?

But if the stick were to be thicker, would deflection stay the same?

Try putting the ruler on edge instead of flat. Does it still bend downwards the same?

 

[Merlin3189]

You might think, why does the stick bend at all? Why is it not straight however much is not supported by the table?
If you place it so that more than half is off the table, does it stay there or do you have to hold it? Why?

 

[Stormblessed]

If you are thinking $y= ax^n$ , you could try plotting the log or ln of x and y.
Because, taking ln of both sides, you get $ln(y) = ln(a) + n ln(x)$ , which should be a straight line of slope n and intercept ln(a)
If you get a reasonable approximation to a straight line (it looks reasonable to me), you can estimate n and a, then put them back into $y= ax^n$

Edit: as for x-section effect:
Why does deflection increase as more of the stick hangs over the edge?
If you made the stick thicker, what effect might that have on the deflection?

On the other side of the coin, if you kept the same stick, but rotated it 90^o how would that affect the deflection?

I came up with this: The cross sectional area of a beam does have an effect on the vertical depression at a given extension, as if you increase the dimensions of the beam, and therefore its thickness and cross-sectional area increase as well, then the amount of depression will vary based on the orientation of the beam. Horizontal: as the height of the beam is less than the breadth, there will be more depression. Vertical: as the height is greater than the breadth, and there is more thickness, there will be less depression. An experiment to test this would be to use beams of various thicknesses and dimensions and to measure vertical depression at a given extension. If there is a change, the hypothesis is proven.

 

[Stormblessed]

Have you studied levers?
Try putting the ruler on edge instead of flat. Does it still bend downwards the same?

I came up with this: The cross sectional area of a beam does have an effect on the vertical depression at a given extension, as if you increase the dimensions of the beam, and therefore its thickness and cross-sectional area increase as well, then the amount of depression will vary based on the orientation of the beam. Horizontal: as the height of the beam is less than the breadth, there will be more depression. Vertical: as the height is greater than the breadth, and there is more thickness, there will be less depression. An experiment to test this would be to use beams of various thicknesses and dimensions and to measure vertical depression at a given extension. If there is a change, the hypothesis is proven.

 

[Merlin3189]

I think you're right about a stick with a non-square rectangular cross-section: it will bend more when the long side (of the x-section) is horizontal and bend less when the long side is vertical.

What I think is still missing from your account, is some insight into why it bends in the first place and what factors affect the amount of bend.
If I take a straight stick - a rectangular x-section metre stick say - and place it part on a flat horizontal table, part overhanging and hold the part on the table firmly to the table so that it is horizontal, then why is the overhanging part not dead straight and horizontal? Why does it bend at all?
I hope that sounds like a stupid question and that is why you have not commented on it, but until you acknowledge what is making it bend, how can you decide what determines how much it will bend?

BTW I see you posted a previous thread about log-log graphs, so you should know the method I suggested for finding the constants of its graph.