Physics Practical Investigation using a mass, a fulcrum and a plank

In summary: On the table or on the fulcrum?2. Which end of the ruler?3. Which side of the table?4. Which end of the fulcrum?I don't think so. d=(mr x D) 1/mq, where d= distance mr=mass ruler (22g) D is the distance that's on the other side of the fulcrum and MQ, is mass of qoauter coin (5.0g). A diagram or picture would probably help...Maybe you could post a picture of your setup here so that others can see it?
  • #1
Jeff97
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5
Homework Statement
Carry out an investigation in to the relationship between the mass of an object (m) and the distance from the fulcrum (d) at which it will cause the plank to tip.
Relevant Equations
You will use a ruler to model the plank and a pile of quarters (6) to model the mass of the cyclist
Homework Statement: Carry out an investigation into the relationship between the mass of an object (m) and the distance from the fulcrum (d) at which it will cause the plank to tip.
Homework Equations: You will use a ruler to model the plank and a pile of quarters (6) to model the mass of the cyclist

I kind of know what to do but not really. My first step is data gathering, so it says to place your ruler on a flat surface (i.e a table) then place a coin on the tipping point then add more coins and move them to find the new tipping point. My question is in my data table I need to make what are the values I need to collect? distance (d) and mass(m) the mass in the independent variable correct? and distance is the dependant? Also, how should I set up my data gathering table?
 
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  • #2
Cyclist on a teeter-totter? A diagram or picture would probably help...
 
  • #3
I'm not given a diagram...sorry.
 
  • #4
Jeff97 said:
I'm not given a diagram...sorry.
Here you go. Happy to help...

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  • #5
Ok, thanks. Still, I need some guidance, please.
 
  • #6
Jeff97 said:
@berkeman are they any sites that can help me than?
We are happy to help you here, but you really need to do a better job of defining the problem and showing your work.

Draw a diagram of what you are asking. Then scan it to a PDF file, or take a good quality picture of it with your cell phone to make a JPG file. Then "attach" the file to a reply here where you describe in a lot more detail what you are trying to do in this experiment.

Thank you.
 
  • #7
What coins? What tipping point? If the ruler is flat on a table, there is no fulcrum.

Are you trying to set up a teeter-totter with coins to add weight to one side and something else on the other side?

And was the word "cycle" a typo in your first post to start this thread?
 
  • #8
Jeff97 said:
1. My first step is to set up a data gathering table, it says to place your ruler on a flat surface (i.e a table) then place a coin on the tipping point then add more coins and move them to find the new tipping point.
As the problem suggests, your first step is to gather a bunch of quarters and a ruler. Although the problem does not suggest it, you will also need a fulcrum. A small rigid metal or plastic tube, a thickish piece of wire, the back of a wooden spoon (it's stable on a flat surface), a round pencil would all be good. Then you need to decide on your independent variable. For that you need to write down the appropriate equation at the tipping point and decide what you are going to vary at will (independent variable) and what you are going to measure at the tipping point (dependent variable). Choices for these could be
1. Number of coins to the left of the fulcrum.
2. Number of coins to the right of the fulcrum.
3. Distance of fulcrum to the left side coins.
4. Distance of fulcrum to the right side coins.
5. The ratio of distances to the fulcrum.
6. The ratio of number of coins.
At this point you need to consider the theory. What does the theory predict about the expression relating all of the above quantities? You need to design an experiment such that if you vary only one of the quantities, your dependent variable will be uniquely determined. This means that you may have to keep some of the quantities fixed as you vary your independent variable. The equation relating these quantities should guide your thinking in that regard.
 
  • #9
@berkeman Here. ... I don't know how to compare my experimental results with theory. I was thinking along the lines of of substituting a weight value and using 10cm (i.e 20cm on the table which is what D is) into d=(Mr x D)1/Mc d=(22x20)1/5.0= 8.8

So if this is right do I say in theory my balancing point for one coin should be 8.8cm off the fulcrum but I found it to be 10cm?

Note they give me d=(mrxD) 1/mq, where d= distance mr=mass ruler (22g) D is the distance that's on the other side of the fulcrum and MQ, is mass of qoauter coin (5.0g)
 
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  • #10
@berkman do you need me to supply you with my tables? graphs? or gradient ect ect?
 
  • #11
kuruman said:
Although the problem does not suggest it, you will also need a fulcrum.
Perhaps the ruler is supposed to project from the edge of the table. That would make a better defined fulcrum than would a pencil.
It leaves open which end the coins are supposed to be on.
 
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  • #12
@haruspex Yes this is correct.

Jeff97 said:
@berkeman Here. ... I don't know how to compare my experimental results with theory. I was thinking along the lines of substituting a weight value and using 10cm (i.e 20cm on the table which is what D is) into d=(Mr x D)1/Mc d=(22x20)1/5.0= 8.8

So if this is right do I say in theory my balancing point for one coin should be 8.8cm off the fulcrum but I found it to be 10cm?

Note they give me d=(mrxD) 1/mq, where d= distance mr=mass ruler (22g) D is the distance that's on the other side of the fulcrum and MQ, is mass of quarter coin (5.0g)
do you think you could assist me with this?
 
  • #13
i am in a sumular situation. and i am having trouble with the relationship with x and y and i have the independent variable as Distance and the dependent variable as the weight of the coin. is this corect
 
  • #14
mehaha said:
i am in a sumular situation. and i am having trouble with the relationship with x and y and i have the independent variable as Distance and the dependent variable as the weight of the coin. is this corect
Since you can vary the distance continuously but only vary the weight in fixed steps, it will be more workable to make the weight the independent variable.
 
  • #15
mehaha said:
i am in a sumular situation. and i am having trouble with the relationship with x and y and i have the independent variable as Distance and the dependent variable as the weight of the coin. is this corect
You did not explain in detail how you carried out your measurements, i.e. what you kept fixed (if anything) and what you varied, because that will determine the choice of dependent and independent variable. Let's assume that you have mass ##m_L## (representing the cyclist in the original problem) on the left side of the fulcrum at distance ##x_L## from it and mass ##m_R## on the right side of the fulcrum at distance ##x_R## from it. At the tipping point, the relevant equation is $$m_L*x_L=m_R*x_R~.$$Here is where I differ philosophically with @haruspex. In a given trial, the independent variable is the quantity that is set before performing the measurement and the dependent variable is the value that that is obtained as a result of the measurement. Cause precedes effect.

Keeping that in mind, I mention here three possibilities that might describe how you conducted your experiment. In all cases, the measurements can be summarized in a plot where you have a straight line passing through the origin in the form ##y=ax## where ##x## is the independent variable, ##y## is the dependent variable and ##a## is the constant slope.

You fix ##m_L## representing the mass of the cyclist, you put a certain number of coins on the other side at a fixed position and you find the value of ##x_L## at which tipping occurs. Then, ##x_L## is the dependent variable that goes on the left-hand side and the rest of the stuff goes on the right-hand side. $$x_L=\frac{m_R*x_R}{m_L}.$$ That's one trial. An important question is "what do you do next for the rest of trials?" Here are three possibilities when mass ##m_L## does not change from one trial to the next:

(a) Keep ##x_R## fixed and change ##m_R## to new values. Then the independent variable is ##m_R## whilst the constant slope is ##a=\dfrac{x_R}{m_L}##.

(b) Keep ##m_R## fixed and change ##x_R## to new values. Then the independent variable is ##x_R## whilst the constant slope is ##a=\dfrac{m_R}{m_L}##.

(c) Change both ##m_R## and ##x_R##. Then the independent variable is the product ##\cancel{m_R*x_R}## whilst the constant slope is ##\cancel{a=\dfrac{1}{m_L}}##.

There are other combinations of variables, which I am not going to list. However, the fact remains that, in all possibilities above, ##x_R## is determined last after the other three have been set. Being determined last in all trails is what makes ##x_R## the dependent variable. The independent variable is the variable (or combination of variables) that changes from one trial to the next and the slope is what remains the same in all trials. You can see how that idea is applied in the three possibilities above.

I am including what follows for completion. What if you varied all 4 variables from one trial to the next? Suppose that for each trial you start with a random choice of two masses and two positions. If you do that, the probability will be very high that the system will not be at the tipping point. Now suppose that from one trial to the next you randomly select which three to set and which to determine last in order to get to the tipping point. How do you include the results from all your trials in a single plot when there is no clearly identifiable variable that was determined last?

Answer: Your dependent variable is the ratio ##y=\dfrac{m_R*x_R}{m_L*x_L}## and the independent variable is the number of the trial ##n=1,2,3,~\dots~## A plot of ##y## vs. ##n## will give you a straight line with zero slope at value ##y=1##.

As you see from the above rather lengthy exposition only you can determine which of the four variables is your dependent variable because only you know what you did, how you did it and which of the 4 variables you consistently determined last. If none of them was consistently determined last, then I have already explained how to deal with it.
 
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  • #16
kuruman said:
Here is where I differ philosophically with @haruspex. In a given trial, the independent variable is the quantity that is set before performing the measurement and the dependent variable is the value that that is obtained as a result of the measurement. Cause precedes effect.
We have no disagreement on that. My point is that it will be more convenient to set the mass and measure the distance. Doing it the other way around, you get a much wider uncertainty in the measurement.

I don't understand your option (c). What is being measured? In the problem statement, ##m_L## and ##x_L## are fixed. If we arbitrarily set ##m_R## and ##x_R## then all that can be measured is the binary result: tips/doesn't tip.
 
  • #17
haruspex said:
We have no disagreement on that. My point is that it will be more convenient to set the mass and measure the distance. Doing it the other way around, you get a much wider uncertainty in the measurement.
And we have no disagreement on that either. I was not suggesting a lab procedure. I was just listing possibilities to illustrate how the labeling of the dependent and independent variables depends on the lab procedure that one follows.
haruspex said:
I don't understand your option (c). What is being measured? In the problem statement, ##m_L## and ##x_L## are fixed. If we arbitrarily set ##m_R## and ##x_R## then all that can be measured is the binary result: tips/doesn't tip.
You are correct. I was thinking ahead to the situation of the cyclist with variable mass.
 

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