Balancing torques with weights hung from a 1 meter rod on a fulcrum

[Quix270]

Homework Statement

A 1 m long rod is initially in equilibrium on a support point located at its midpoint. Imagine that we have 4 identical masses, three of them placed in the following positions: 25cm, 45cm and 95cm. Where should the fourth mass be placed to balance the rod?

Relevant Equations

Balancing torque

I think is 25cm but I'm not sure. Help please

 

[berkeman]

Can you show the math that you used to arrive at that answer please? Always show your work so that we can see what you are doing, and we can check it if it looks like you've made any errors. Thanks.

 

[Quix270]

Here are. Thank you!

 

[berkeman]

Imagine that we have 4 identical masses, three of them placed in the following positions: 25cm, 45cm and 95cm.

I'm a little confused about the coordinate system definition. If there is a diagram in your book, that may be helping you define your coordinates, but I'm not understanding so far.

Are the distances measured from the center of the rod at the fulcrum, with + to the right and - distances to the left of the fulcrum? Initially when I read the problem, I assumed that the distances were from the left end of the rod to the right, so the fulcrum would be at x=0.5m.

 

[berkeman]

Also, it works best if you can type your work into the forum window. That makes it a lot easier to read and for us to be able to quote parts of your work. Thanks.

 

[Quix270]

I'm a little confused about the coordinate system definition. If there is a diagram in your book, that may be helping you define your coordinates, but I'm not understanding so far.

Are the distances measured from the center of the rod at the fulcrum, with + to the right and - distances to the left of the fulcrum? Initially when I read the problem, I assumed that the distances were from the left end of the rod to the right, so the fulcrum would be at x=0.5m.

I assume is like the draw below.
Can you explain me how you get to the answer please?
I think too they give the measurements from left to right of the rod

 

[berkeman]

Can you explain me how you get to the answer please?

The sum of the torques about the fulcrum must equal zero for the rod to be balanced. Take the torques for the masses to the right of the fulcrum to be positive, and the torques from masses to the left to be negative. Be sure to use the distance from the fulcrum to each mass to calculate each torque to add to the sum.

Can you show us your math now?

 

[Quix270]

The sum of the torques about the fulcrum must equal zero for the rod to be balanced. Take the torques for the masses to the right of the fulcrum to be positive, and the torques from masses to the left to be negative. Be sure to use the distance from the fulcrum to each mass to calculate each torque to add to the sum.

Can you show us your math now?

I make that and give me 25cm. So the other mass have to be above another mass. Can you tell me where I get wrong? Thanks again

 

[berkeman]

I make that and give me 25cm. So the other mass have to be above another mass. Can you tell me where I get wrong? Thanks again

That does not look like the distances are from the center of the 1m rod, but it's pretty hard for me to read your picture.

Please type your work into the forum window, and be sure that the distances are measured from the center of the rod. Thanks.

 

[Quix270]

I make it by assuming the distance are taking from the left side of the rod because it's impossible to have a mass 95cm separated from the center of the rod since the rod length is 1 m.

 

[Lnewqban]

If the problem is actually as your diagram shows, you are not resolving it properly.
You are not actually following the advice of post #7.
There is no reason for the negative sign of 0.95mg, as you are referring all the moments to the left end of the bar, rather than to the fulcrum.
Nothing wrong with that convenient approach, only that in that case, you will need to consider the reaction force at the actual fulcrum and to include its moment as well.

 

[Quix270]

If the problem is actually as your diagram shows, you are not resolving it properly.
You are not actually following the advice of post #7.
There is no reason for the negative sign of 0.95mg, as you are referring all the moments to the left end of the bar, rather than to the fulcrum.
Nothing wrong with that convenient approach, only that in that case, you will need to consider the reaction force at the actual fulcrum and to include its moment as well.

Si the answer is at 50cm? At the center of the rod? Can you explain me with numbers how you get there? I will appreciate that. I have test tomorrow

 

[Lnewqban]

The answer is not 50 cm.
Please, relax and try to re-think the problem.
What levers are you using for each force and why?
What is the center of all moments?

 

[Quix270]

Please help me I don't have idea. I'm frustrated right now. Can you give the answer an say me how you get there?

 

[berkeman]

Please help me I don't have idea. I'm frustrated right now. Can you give the answer an say me how you get there?

No, we don't give answers here. We try our best to help you get to the answer.

Draw the 1m rod with the fulcrum in the middle. Draw the 3 weights placed as instructed by the problem statement, and label each of the 3 distances from the center/fulcrum of the rod out to where the 3 weights are hanging. Then write the equation that sums those 3 torques, taking care to get the sign of those torques about the fulcrum correct. Please type that work into the forum window so it is easy for us to read. You can attach a good-quality picture of your sketch of the rod and fulcrum and 3 weights labeled with the distances from the center.

Then think about where you are going to need to hang the 4th weight to "balance" the rod (so that the sum of the torques about the fulcrum adds up to zero. Which side of the fulcrum do you think you will need to hang that 4th weight?

 

[Lnewqban]

Sorry, I have to follow the following rule:

• Full solutions on a homework problem should never be given. Only hints and explanations are allowed.

Now, I am here to help you understand what you have not done correctly, as far as I can.
Let's start with the diagram that you have put together, based on the question.

Please, calculate the resulting moment after placing the fourth mass at the point you have determined.
If the result is zero, your answer is correct.
If the result is different from zero, the bar will not be balanced.

(Cross-posted with berkeman's)

 

[Quix270]

No, we don't give answers here. We try our best to help you get to the answer.

Draw the 1m rod with the fulcrum in the middle. Draw the 3 weights placed as instructed by the problem statement, and label each of the 3 distances from the center/fulcrum of the rod out to where the 3 weights are hanging. Then write the equation that sums those 3 torques, taking care to get the sign of those torques about the fulcrum correct. Please type that work into the forum window so it is easy for us to read. You can attach a good-quality picture of your sketch of the rod and fulcrum and 3 weights labeled with the distances from the center.

Then think about where you are going to need to hang the 4th weight to "balance" the rod (so that the sum of the torques about the fulcrum adds up to zero. Which side of the fulcrum do you think you will need to hang that 4th weight?

It's the graphic correct?

 

[berkeman]

It's a little hard to read because it's sideways, but otherwise it's a good start. Now add lines from the center fulcrum out to each of the 3 weight positions, and label those lines with the distance to each from the center. Then write the equation that shows the sum of those 3 torques about the center fulcrum. Thanks.

 

[Quix270]

It's a little hard to read because it's sideways, but otherwise it's a good start. Now add lines from the center fulcrum out to each of the 3 weight positions, and label those lines with the distance to each from the center. Then write the equation that shows the sum of those 3 torques about the center fulcrum. Thanks.

There is where I have trouble. The problem say one weight is at 95cm but not say from what and It couldn't be from the center because the rod is 1m long. Cold be at 45cm if I substract 95 from the center that is 50. That would be right?

 

[Quix270]

There is where I have trouble. The problem say one weight is at 95cm but not say from what and It couldn't be from the center because the rod is 1m long. Cold be at 45cm if I substract 95 from the center that is 50. That would be right?

I make it again with other perspective. Does the answer is 15cm?

 

[Lnewqban]

You have a balanced 100 cm bar.
Fulcrum is located at middle point: 50 cm from left end and 50 cm from right end.
Why is that location important?

Mass 1 is located 25 cm from left end and 100-25=75 cm from right end, how far from fulcrum? Let's call that distance X1.

Mass 2 is located 45 cm from left end and 100-45=55 cm from right end, how far from fulcrum? Let's call that distance X2.

Mass 3 is located 95 cm from left end and 100-95=5 cm from right end, how far from fulcrum? Let's call that distance X3.

Mass 4 is located A cm from left end and 100-A=B cm from right end, how far from fulcrum? Let's call that distance X4.

 

[Quix270]

You have a balanced 100 cm bar.
Fulcrum is located at middle point: 50 cm from left end and 50 cm from right end.
Why is that location important?

Mass 1 is located 25 cm from left end and 100-25=75 cm from right end, how far from fulcrum? Let's call that distance X1.

Mass 2 is located 45 cm from left end and 100-45=55 cm from right end, how far from fulcrum? Let's call that distance X2.

Mass 3 is located 95 cm from left end and 100-95=5 cm from right end, how far from fulcrum? Let's call that distance X3.

Mass 4 is located A cm from left end and 100-A=B cm from right end, how far from fulcrum? Let's call that distance X4.

Ok I understand in this point

 

[Lnewqban]

Ok I understand in this point

Any progress?
Less frustration by now?

 

[Quix270]

Any progress?
Less frustration by now?

I think I got it. Is 35cm
Using net torque equation
(0.25)(9.8)+(0.05)(9.8)+(x)(9.8)-(0.45)(9.8)=0
The answer of that is 15cm from the center but the problem give me distance from left to right so 50-15=35cm

 

[berkeman]

I think I got it. Is 35cm
Using net torque equation
(0.25)(9.8)+(0.05)(9.8)+(x)(9.8)-(0.45)(9.8)=0
The answer of that is 15cm from the center but the problem give me distance from left to right so 50-15=35cm

How can you have 3 weights on the left of the fulcrum balancing one weight on the right of the fulcrum? Can you post your final diagram with unambiguous distances labeled please? Thanks.

 

[Lnewqban]

You did it!

The formula: You need to add "m" between "g" and each lever distance.

 

[Quix270]

How can you have 3 weights on the left of the fulcrum balancing one weight on the right of the fulcrum? Can you post your final diagram with unambiguous distances labeled please? Thanks.

I think it could be because the weight on the left side are very close to the center, that means less down force, and the weight on the right side is very far from the center. So I think it's possible to have 3 weight on one side. The final distance are 25, 35, 45 and 95cm. Can you explain me why that isn't the answer?

 

[berkeman]

Please post your final drawing with distanced labeled from the center, and your final sum of torques (typed into the forum). Thanks.