# Where to place a weight on a bar so that the net torque is 0

• fishturtle1
In summary, the system will balance when the center of mass of the assembly is located at 0.3576 cm from the fulcrum.
fishturtle1

## Homework Statement

A 0.100-kg, 59.9-cm-long uniform bar has a small 0.055-kg mass glued to its left end and a small 0.100-kgmass glued to the other end. You want to balance this system horizontally on a fulcrum placed just under its center of gravity.

no diagram given

Στ = 0 N*m
τ=F×d

## The Attempt at a Solution

Massbar=.100 kg
L=0.599m
m1=0.055kg
m2=0.100kg

I am confused about how to set up my diagram. I have made a bar. The only forces acting on the bar are the two weights from the two masses.

I didn't include the weight of the bar itself because since its uniform I wanted to make the center the pivot axis.

SO my torque equation looks like this:
Στ=0=-m1g(d`) + m2g(d2)

the first torque is negative because it is going in the clockwise direction.

But from this equation i can get infinite many distances depending on what d2 is.

What are the rules/restrictions for this problem? Are either weights' location predetermined?

fishturtle1
fishturtle1 said:
I didn't include the weight of the bar itself because since its uniform I wanted to make the center the pivot axis.
There's your mistake. It's not the CoM of the bar where you will pivot, it's to be the CoM of the assembly.

fishturtle1
TJGilb said:
What are the rules/restrictions for this problem? Are either weights' location predetermined?
the 0.05 kg mass is said to be on the left end, the 0.10 kg mass is said to be on the right end. That is all that is specified. And I assume they do not mean at the very end otherwise the bar would be in motion.

fishturtle1 said:
the 0.05 kg mass is said to be on the left end, the 0.10 kg mass is said to be on the right end. That is all that is specified. And I assume they do not mean at the very end otherwise the bar would be in motion.
I think they mean that the weights are indeed affixed to the very ends of the bar. Your job is to locate the fulcrum so that the system balances horizontally.

fishturtle1
NascentOxygen said:
There's your mistake. It's not the CoM of the bar where you will pivot, it's to be the CoM of the assembly.

Ok so then I have another force that is the weight of the bar, which is at d=0.295m.
gneill said:
I think they mean that the weights are indeed affixed to the very ends of the bar. Your job is to locate the fulcrum so that the system balances horizontally.
Ok using this information and the previous post this is what I did.

Στ=0=0.05gd+0.10g(d-0.295)-0.10g(0.599-d)

rearranging I got to

0.15d - 0.0295 = -0.10d + 0.0599
0.25d = 0.0894
d = 0.3576 =35.8 cm

which my online homework took as the correct answer even though it said exact answer was 35.2 cm.

Thank you for your guidance on clearing up my confusion.

## 1. Where should I place the weight on the bar to achieve a net torque of 0?

The weight should be placed at the center of mass of the bar in order to achieve a net torque of 0. This is the point where the weight distribution is evenly balanced and there is no rotation.

## 2. Can the weight be placed anywhere on the bar to achieve a net torque of 0?

No, the weight cannot be placed anywhere on the bar to achieve a net torque of 0. It must be placed at the center of mass of the bar in order to achieve this balance.

## 3. How do I calculate the center of mass of the bar?

The center of mass can be calculated by finding the average position of all the mass on the bar. This can be done by dividing the total mass of the bar by the distance from one end of the bar to the center of mass.

## 4. Will the center of mass of the bar change if the weight is moved?

Yes, the center of mass will change if the weight is moved. The new center of mass will be determined by the average position of the new weight distribution.

## 5. Why is it important to place the weight at the center of mass for a net torque of 0?

Placing the weight at the center of mass ensures that the bar will be in a state of equilibrium, with no net torque causing rotation. This is important for stability and safety, especially when dealing with heavy weights.

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