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

[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

Homework Equations

Στ = 0 N*m
τ=F×d

The Attempt at a Solution

Mass_bar=.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.

 

[TJGilb]

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

 

[NascentOxygen]

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]

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.

 

[gneill]

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]

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.

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.