Calculating distance to balance see-saw

[guyvsdcsniper]

Homework Statement

Calculate the position of the mass m3 by using the torque balance equation.

Relevant Equations

ΣT=0

I am trying to find out where I need to place mass 3 to balance this see-saw. I am measuring all distances with respect to the fulcrum, so positive is to the right of the fulcrum and negative is to the left of it. Experimentally I have found the distance that mass 3 needs to be to balance this system is -2m from the fulcrum. I have set up a net torque equation and am coming up with -4m. I can see that when I subtract (.750m x 784N) by (1m x 196N) and divide by 196N I get -2m. It seems that putting (-1m) into my equation is stopping me from getting the right answer. I can't see why I would leave that value positive though if it is on the negative x-axis of the fulcrum.

Could someone help me understand where I am going wrong?

 

[haruspex]

Homework Statement:: Calculate the position of the mass m3 by using the torque balance equation.
Relevant Equations:: ΣT=0

I am trying to find out where I need to place mass 3 to balance this see-saw. I am measuring all distances with respect to the fulcrum, so positive is to the right of the fulcrum and negative is to the left of it. Experimentally I have found the distance that mass 3 needs to be to balance this system is -2m from the fulcrum. I have set up a net torque equation and am coming up with -4m. I can see that when I subtract (.750m x 784N) by (1m x 196N) and divide by 196N I get -2m. It seems that putting (-1m) into my equation is stopping me from getting the right answer. I can't see why I would leave that value positive though if it is on the negative x-axis of the fulcrum.

Could someone help me understand where I am going wrong?

View attachment 281138
View attachment 281139

It's a sign problem.
You can approach it in either of two ways:
1. Keep everything positive and write the equation that says the sum of the clockwise torques equals the sum of the anticlockwise torques.
2. Use signs on the displacements, and write that the sum of all torques is zero.
You have mixed the two.

 

[BvU]

Experimentally

You actually collected three 20 kg weights and a 80 kg one ?

I'm somewhat missing the problem statement, but I suppose you are at 2. and do your best to make life difficult by using a numerical value for $g$, which should simply divide out of the torque balance
$$\sum_{left} mgr = \sum_{right} mgr\Leftrightarrow $$
$$ 20\cdot 2 + 20 \cdot 1 = 80 \cdot 0.75$$

[edit]Ah, Hello Haru!
$\$

 

[guyvsdcsniper]

It's a sign problem.
You can approach it in either of two ways:
1. Keep everything positive and write the equation that says the sum of the clockwise torques equals the sum of the anticlockwise torques.
2. Use signs on the displacements, and write that the sum of all torques is zero.
You have mixed the two.

Gotcha, I see where I went wrong now. Thank you!

 

[guyvsdcsniper]

It's a sign problem.
You can approach it in either of two ways:
1. Keep everything positive and write the equation that says the sum of the clockwise torques equals the sum of the anticlockwise torques.
2. Use signs on the displacements, and write that the sum of all torques is zero.
You have mixed the two.

Haru,
I tried you approach and I did get 2 but the answer is positive. It should be negative. Am I missing something?

 

[guyvsdcsniper]

You actually collected three 20 kg weights and a 80 kg one ?

I'm somewhat missing the problem statement, but I suppose you are at 2. and do your best to make life difficult by using a numerical value for $g$, which should simply divide out of the torque balance
$$\sum_{left} mgr = \sum_{right} mgr\Leftrightarrow $$
$$ 20\cdot 2 + 20 \cdot 1 = 80 \cdot 0.75$$

[edit]Ah, Hello Haru!
$\$

Ha, I am using a simulation https://phet.colorado.edu/sims/html/balancing-act/latest/balancing-act_en.html .

 

[haruspex]

Haru,
I tried you approach and I did get 2 but the answer is positive. It should be negative. Am I missing something?

Again, it is a matter of picking one of the two approaches and sticking to it.
In method 1, you place the third mass at a certain distance to the left of the fulcrum. This will give a positive number.
In method 2, you place it a certain displacement to the right. This will give a negative number.
If you are still getting the wrong sign, please post your detailed calculation and say which method you are using.

 

[guyvsdcsniper]

Again, it is a matter of picking one of the two approaches and sticking to it.
In method 1, you place the third mass at a certain distance to the left of the fulcrum. This will give a positive number.
In method 2, you place it a certain displacement to the right. This will give a negative number.
If you are still getting the wrong sign, please post your detailed calculation and say which method you are using.

I used method 2.

 

[Shreya]

In the step 3, when you divided 20 from numerator & denominator, you forgot to carry the negative sign to the next step.
Sorry if I am wrong.

 

[guyvsdcsniper]

In the step 3, when you divided 20 from numerator & denominator, you forgot to carry the negative sign to the next step.
Sorry if I am wrong.

I just divided it at the end, which still gives me a positive number. I probably could have been a little more clear on that step.

 

[Shreya]

I don't think you must start by taking x as negative. Take it as x and then solve.
Sorry if I am wrong

 

[guyvsdcsniper]

-40 m kg / 20 kg = -2 m

Yes but the unknown distance is in the negative direction, so when I divide that by -2 m it becomes 2 m.

This is the diagram I am working with and I am considering everything to the left of the fulcrum a distance on the negative x axis. I am not suppose to know the distance of the boy at the very end and that is what I am solving for. So the way I was interpreting it was that since he is on the x-axis and i don't know his distance, I would use negative sign for his displacement.

 

[Shreya]

As you said, You are not supposed to know where the boy must be placed, (he could even be on the +x axis, though not in this case). So you assume that the boy is at x displacement and the math tells you the sign.
When you wrote -x, you have already assumed that the value you will obtain for x is the magnitude of x.

The boy is at a -x distance, where |x|is 2m. Your answer is right. You just assumed that he is at a -x displacement and solved for the magnitude of x.
Sorry if I am wrong

 

[guyvsdcsniper]

As you said, You are not supposed to know where the boy must be placed, (he could even be on the +x axis, though not in this case). So you assume that the boy is at x displacement and the math tells you the sign.
When you wrote -x, you have already assumed that the value you will obtain for x is the magnitude of x.

The boy is at a -x distance, where |x|is 2m. Your answer is right. You just assumed that he is at a -x displacement and solved for the magnitude of x.
Sorry if I am wrong

...
I feel so dumb. I believe you are absolutely right. I have to do better at catching these small mistake.

 

[Shreya]

I am no better. All the best!

 

[guyvsdcsniper]

It's a sign problem.
You can approach it in either of two ways:
1. Keep everything positive and write the equation that says the sum of the clockwise torques equals the sum of the anticlockwise torques.
2. Use signs on the displacements, and write that the sum of all torques is zero.
You have mixed the two.

Haru,

I am a bit confused on why #2 works.

You state to take the sum of all torques. I did that and with the help of Shreya point out what i was doing wrong I found the right answer.

But why would I have to add the the torque of the object on the right. If I think of it as the only force it would rotate clockwise making it negative which would mean I need to subtract the clockwise torque from the counter clockwise torque giving me a completely different answer.

 

[Shreya]

Haru,

I am a bit confused on why #2 works.

You state to take the sum of all torques. I did that and with the help of Shreya point out what i was doing wrong I found the right answer.

But why would I have to add the the torque of the object on the right. If I think of it as the only force it would rotate clockwise making it negative which would mean I need to subtract the clockwise torque from the counter clockwise torque giving me a completely different answer.

By using signs of displacement in method #2, you are just pointing out the direction of the torque (clockwise or anticlockwise).
If you rearrange the equation you got in #2 you get the same equation as #1.
Both the methods are the same,in the first one you consciously separate the clockwise and anti clockwise and in #2 you do it mathematically.
Sorry if I am wrong

 

[Shreya]

 

[guyvsdcsniper]

By using signs of displacement in method #2, you are just pointing out the direction of the torque (clockwise or anticlockwise).
If you rearrange the equation you got in #2 you get the same equation as #1.
Both the methods are the same,in the first one you consciously separate the clockwise and anti clockwise and in #2 you do it mathematically.
Sorry if I am wrong

So I am dealing with a different question but it is very similar. I have to calculate the distance of the object on the left from the fulcrum in order to achieve equilibrium.

On the left I have followed Haru's convention and I did receive a negative displacement as my answer. On the right was my though process to this problem. The torque on the right is negative because it would rotate clockwise if it was the only force but it ultimately leaves me with a positive displacement.

How the method on the right wrong? I've been using that for multiple problems and have been getting them correct but now when I have a fulcrum and torques on both sides of it, it is not working..

 

[Shreya]

So I am dealing with a different question but it is very similar. I have to calculate the distance of the object on the left from the fulcrum in order to achieve equilibrium.
View attachment 281167

On the left I have followed Haru's convention and I did receive a negative displacement as my answer. On the right was my though process to this problem. The torque on the right is negative because it would rotate clockwise if it was the only force but it ultimately leaves me with a positive displacement.

View attachment 281166

How the method on the right wrong? I've been using that for multiple problems and have been getting them correct but now when I have a fulcrum and torques on both sides of it, it is not working..

Your method is right.
The method that haru suggested is correct, but in your solution, you have not applied different signs for clockwise and counterclockwise torques.
That must be the reason for the wrong sign

 

[guyvsdcsniper]

Your method is right.
The method that haru suggested is correct, but in your solution, you have not applied different signs for clockwise and counterclockwise torques.
That must be the reason for the wrong sign

I think I am missing something could you help me understand a little better? I feel like in my method I am applying the the different signs for the clockwise and counterclockwise torque. The (x)(10kg*g) is the CCW torque which makes it positive. The (1m)(5kg*g) Torque is the CW torque making it negative. Hence the CCW Torque is being subtracted from the CW torque.

 

[Shreya]

I think I am missing something could you help me understand a little better? I feel like in my method I am applying the the different signs for the clockwise and counterclockwise torque. The (x)(10kg*g) is the CCW torque which makes it positive. The (1m)(5kg*g) Torque is the CW torque making it negative. Hence the CCW Torque is being subtracted from the CW torque.

You should use + 5kg *g, because you are dealing with clockwise torque.
In usual sign convention clockwise is + and counter clockwise is -
Sorry if I am wrong

 

[guyvsdcsniper]

You should use + 5kg *g, because you are dealing with clockwise torque.
In usual sign convention clockwise is + and counter clockwise is -
Sorry if I am wrong

Yea I am not too sure about that. I believe when you are dealing with torque counter clock wise is positive and clockwise is negative.

 

[Shreya]

Yea I am not too sure about that. I believe when you are dealing with torque counter clock wise is positive and clockwise is negative.

Sorry about that... I didn't know about that convention. Thanks for letting me know.
So you should add a negative sign to 10kg *g because the force in this case is downward. Therefore, for 10kg the force is - and position is unknown and for 5kg the force is - and position is +. (I think this is the reason why clockwise is said to be negative )

 

[haruspex]

I am a bit confused on why #2 works.

Suppose the weights w_i are at displacements x_i, these being signed values, positive on the right of the fulcrum.
With the convention that CCW is positive, the torques are -w_ix_i.
Using method 2, the torque balance equation is Σ-w_ix_i=0.
But we can drop the minus sign and write Σw_ix_i=0.
To switch to method 1, we can collect together the terms in which x is negative, i.e. for which x=-|x|. To distinguish these I'll write y instead of x, so the method 2 form is
Σw_ix_i+Σw_iy_i=0.
Σw_ix_i+Σw_i(-|y_i|)=0.
Σw_ix_i-Σw_i|y_i|=0.
Σw_ix_i=Σw_i|y_i|.
Which is the method 1 form.

 

[Lnewqban]

Yea I am not too sure about that. I believe when you are dealing with torque counter clock wise is positive and clockwise is negative.

You can create your own meaning of the signs, as two oposite conventions exist regarding positive and negative moments or torques.
What is important is that both directions of potential rotation about a common fulcrum are opposite.

To understand that concept is important because your future problems may not involve a straight horizontal beam with forces acting perpendicularly to it.

 

[guyvsdcsniper]

You can create your own meaning of the signs, as two oposite conventions exist regarding positive and negative moments or torques.
What is important is that both directions of potential rotation about a common fulcrum are opposite.

To understand that concept is important because your future problems may not involve a straight horizontal beam with forces acting perpendicularly to it.

View attachment 281244

You're right I didnt consider that. I will keep that in mind.