Net Force of 3 Forces (Equilibrium)

[helpneeded02]

Homework Statement

I need some help with a Physics lab. My teacher wasn't there the only chance I could ask him.

I don't understand how to prove that the net force is zero when you have the three forces in different directions. For example, the first force is 9.7 N [down], the second is 5.3 N [right 46 up] and the final is 7.4 N [left 60 up].

Also, how do you calculate the inaccuracy rate? (eg. [tex]\pm 0.02[/tex])

Homework Equations

When you come up with your answer, how do you know if it is a successful experiment? I know that it should be approximately zero, but I saw somebody doing the same experiment had a result of 102 and he said it should have been smaller than 86. That's a bit far from zero, which confused me.

3. Attempt at the Solution

Well, I know that the sum of all the forces in equilibrium is zero. So, this means that =9.7N + 5.3N + 7.4N is supposed to equal to zero.

I tried to use the perdenicular components approach using 5.3N and 7.4N.

5.3 N:
cos46=x/5.3
x=3.7

sin46=y/5.3
y=3.8

7.4 N:
cos60=-x/7.4
x=-3.7

sin60=y/7.4
y=6.4Sum of x= 3.7 + (-3.7) =0
Sum of y = 3.8+6.4 =10.2

Therefore, my 9.7N should be 10.2N?
Then how do I prove the net force to be 0?
So does that mean that my experiment was inaccurate?

 

[LowlyPion]

Homework Statement

I need some help with a Physics lab. My teacher wasn't there the only chance I could ask him.

I don't understand how to prove that the net force is zero when you have the three forces in different directions. For example, the first force is 9.7 N [down], the second is 5.3 N [right 46 up] and the final is 7.4 N [left 60 up].

Also, how do you calculate the inaccuracy rate? (eg. [tex]\pm 0.02[/tex])

Homework Equations

When you come up with your answer, how do you know if it is a successful experiment? I know that it should be approximately zero, but I saw somebody doing the same experiment had a result of 102 and he said it should have been smaller than 86. That's a bit far from zero, which confused me.

3. Attempt at the Solution

Well, I know that the sum of all the forces in equilibrium is zero. So, this means that =9.7N + 5.3N + 7.4N is supposed to equal to zero.

I tried to use the perdenicular components approach using 5.3N and 7.4N.

5.3 N:
cos46=x/5.3
x=3.7

sin46=y/5.3
y=3.8

7.4 N:
cos60=-x/7.4
x=-3.7

sin60=y/7.4
y=6.4Sum of x= 3.7 + (-3.7) =0
Sum of y = 3.8+6.4 =10.2

Therefore, my 9.7N should be 10.2N?
Then how do I prove the net force to be 0?
So does that mean that my experiment was inaccurate?

Is the object in motion? Is it accelerating? If the object is not in motion. then the net sum of the forces is 0. F = m*a. No a, no F.

Now if you have 3 forces attached to an object, then that object is defined in a plane defined by the points of contact of those 3 forces. (If the object is also subject to gravity, then there is a 4th Force m*g.)

What are the sources of error?

Looks to me like the measurement of the force and the measurement of the angle represent two experimental opportunities for inaccuracy.

Typically the smallest gradation of the measuring instrument, if it's a protractor the degree marks, or if the force gauge it's smallest reading.

Then for addition or subtraction you add the absolute uncertainties and for multiplication or division you add the percentage uncertainties. Your observed values should fit within the range of what these values would allow.

 

[cepheid]

I don't understand how to prove that the net force is zero when you have the three forces in different directions.

You have to add the forces vectorially. Your method above of doing it componentwise is a perfectly good method.

Also, how do you calculate the inaccuracy rate? (eg. [tex]\pm 0.02[/tex])

You need estimates for the uncertainties in your measurements of the individual forces. Typically this takes into account measurement errors and the precision of the instrument. Then you need to know how to correctly propagate these errors through the experiment, i.e. when you add three quantities together, how do their errors combine? This is known as error analysis. For the addition of three quantities, I believe their errors simply add up.

Well, I know that the sum of all the forces in equilibrium is zero. So, this means that =9.7N + 5.3N + 7.4N is supposed to equal to zero.
Therefore, my 9.7N should be 10.2N?

Assuming the math is right, the net force is clearly NOT zero. This is your *experimental result*. Experimental results sometimes differ from what is theoretically predicted.

Then how do I prove the net force to be 0?

You can't. You're not trying to prove anything. From a theoretical standpoint, Newton's laws are what prove that the net force *should* be zero. But your *measured* net force either is or isn't zero, and you have to live with that. You're doing an experiment and hoping that the result will be that the net force is zero (within the limits of experimental error).

So does that mean that my experiment was inaccurate?

Possibly. If, as you say above, the net force was *supposed* to be zero, and it wasn't, then experimental error could be one explanation for the discrepancy. Or it could be something else. Without more details about your lab setup, I can't comment further.

 

[HallsofIvy]

Yes, the calculation is correct. You mention the "inaccuracy" but yuo don't say anything about the accuracy of the individual measurements. You give each force to one decimal place so I would guess that you measured them with a scale have marks at each 0.1 Newton. If that is the case, then you were rounding to the nearest 0.1 Newton and so have an error of [itex]\pm 0.05[/itex]N. There is also a possible error in measuring the angles and that is harder to judge. You are probably measuring angles to the nearest half degree so, for example, "sin(60)" may be as large as sin(60.5)= 0.87 or as low as sin(59.5)= 0.86. That means that "7.4 sin(60) could be as large as 7.45 sin(60.5)= 6.48 ior as low as 7.36 sin(59.5)= 6.34. That is, the calculation of each component of force may be off as much as 0.1 N so the net force may be off as much as 0.3 N.

Since your calculations are off by 0.5 N, there has clearly been some problem with the experiment!

 

[helpneeded02]

Thank you everyone for helping me. I have a much better understand of what is expected of me in a lab. I was really confused about what I needed to do with the information.

I was using two spring scales to measure the force acting on the string when I put a 1kg weight on the string. From what I understood in LowlyPion's answer, to get the sources of error, I have to take the smallest mark (from the whole scale? and from the protractor? -eg. 1 degree) and add those. Then add that to the percentage of all of my uncertainties? I'm not sure what you mean when he/she are saying "for addition and subtraction" and "for multiplication or division"? Would somebody be able to provide me with an example so that I can understand better?

 

[LowlyPion]

Thank you everyone for helping me. I have a much better understand of what is expected of me in a lab. I was really confused about what I needed to do with the information.

I was using two spring scales to measure the force acting on the string when I put a 1kg weight on the string. From what I understood in LowlyPion's answer, to get the sources of error, I have to take the smallest mark (from the whole scale? and from the protractor? -eg. 1 degree) and add those. Then add that to the percentage of all of my uncertainties? I'm not sure what you mean when he/she are saying "for addition and subtraction" and "for multiplication or division"? Would somebody be able to provide me with an example so that I can understand better?

What you are concerned with is the propagation of uncertainties in measurement through calculations.

In your experiment you have error associated with angle and force.

The proper way to propagate the uncertainty is for your math operations that involve addition or subtraction of measured quantities then just add the absolute uncertainties. If you are adding weights that have ±10grams and you add 2 such quantities then your uncertainty is ±20grams. Multiplication and division should be treated differently. For these operations you need to calculate the percentage uncertainty or sometimes referred to as the relative uncertainty (as opposed to the absolute uncertainty mentioned above).

Hence for a division operation if you have 8 ±1 divided 2±.2 then your relative uncertainties are 12.5% and 10%. Your resulting uncertainty would be represented as 22.5% of the nominal result.

When measurements are taken by independent means, there are other treatments that involve taking the Root Sum of the Squares (RSS) of the absolute error uncertainties for addition and subtraction and the RSS of the relative uncertainties for multiplication and division operations. There are likewise more complicated statistical treatments to identify systematic and observational populations of readings, but let's just not get more complicated than what may be called for in your lab.

With regard to your scale or protractor there may be several contributing causes to error as well. Both the type of the scale being marked say in 1 gram increments, the calibration of the scale being incorrect, and if using multiple scales simultaneously there may be unwanted variation and finally there is observer error associated with how you read the scale.

 

[helpneeded02]

With regard to your scale or protractor there may be several contributing causes to error as well. Both the type of the scale being marked say in 1 gram increments, the calibration of the scale being incorrect, and if using multiple scales simultaneously there may be unwanted variation and finally there is observer error associated with how you read the scale.

So, by unwanted variation, do you mean magnetic declination?

 

[LowlyPion]

So, by unwanted variation, do you mean magnetic declination?

Such things as viewing angle for reading a needle pointer on a scale, or the scales being at an angle to gravity, but magnetic declination? I have no idea about that unless you are using something depending on magnetics.

Unwanted variation might come as regards to using multiple instruments to simultaneously measure the forces, in which case if they are not all perfectly linear and mutually calibrated you are introducing a systematic error that would affect all similar measurements with that set up.