Basic torque fulcrum problem assitance needed

[PsychonautQQ]

Homework Statement

A uniform rod with a length of 10 meters and a mass of 20 kilograms has a 20 kilogram mass on one end and a 40 kilogram mass on the other end. Where should the fulcrum be located for this beam to balance?
________________
| V |
|20kg| |40Kg|

Homework Equations

Torque = Force * Radius

The Attempt at a Solution

Usually when i did problems like this I'm pretty sure the board was considered mass-less, and I'm not sure how to do it now that it has mass. Is calculus needed?

Anyway, let's consider:
20 kg mass = m1 @ r1
40 Kg mass = m2 @ r2

and then to figure in the boards mass can we just consider it to have a center of mass on each side (that obviously depends on where the fulcrum is)?
if so...
Left side of board = m3 @ r3
right side of board = m4 @ r4

(m1g*r1) + (m3g*r3) - (m2g*r2) - (m4g*r4)=0
m1g + m2g + m3g + m4g = Normal force of fulcrum

Anyway I'm a little stuck obviously ..

 

[dreamLord]

You can assume the rod to be a point particle with mass 20kg located at it's center of mass, since that's where gravity can be assumed to be acting.

In your method, the torque due to 'm3' and 'm4' (about the center of the rod) will be the same in magnitude, but opposite in direction, and will cancel out.

 

[PsychonautQQ]

You can assume the rod to be a point particle with mass 20kg located at it's center of mass, since that's where gravity can be assumed to be acting.

In your method, the torque due to 'm3' and 'm4' (about the center of the rod) will be the same in magnitude, but opposite in direction, and will cancel out.

Oh okay cool. So i now have two equations m1gr1 + m2gr2 -m3gr3 = 0

and

m1g+m2g+m3g = Normal force of fulcrum (not sure if this equation is useful)

I'm just trying to figure out a systematic way to solve this problem, i could plug in numbers and play the "hotter and colder" game until i get something that fits but i want to find a method more mechanical.

 

[SteamKing]

You are just writing equations without giving thought to finding the desired location of the fulcrum. The equations you have, while strictly correct, are not leading you towards finding the location of the fulcrum which balances the rod.

Assume:
x - the location of the fulcrum from the left end of the rod (this choice is arbitrary)
Let's also say that the 20 kg mass is on the left end of the rod, and the 40 kg mass is on the opposite end.

We are using masses here, but the requirement for the rod to be in equilibrium means that the forces acting on the rod must balance, just as the total torque acting on the rod at the fulcrum must equal zero.

Write the torque equation using the fulcrum as a reference point. When you plug in the known quantities into the torque equation, you should have one equation with one unknown, namely x, the location of the fulcrum. Solve for x, and Bob's you uncle.

 

[PsychonautQQ]

You are just writing equations without giving thought to finding the desired location of the fulcrum. The equations you have, while strictly correct, are not leading you towards finding the location of the fulcrum which balances the rod.

Assume:
x - the location of the fulcrum from the left end of the rod (this choice is arbitrary)
Let's also say that the 20 kg mass is on the left end of the rod, and the 40 kg mass is on the opposite end.

We are using masses here, but the requirement for the rod to be in equilibrium means that the forces acting on the rod must balance, just as the total torque acting on the rod at the fulcrum must equal zero.

Write the torque equation using the fulcrum as a reference point. When you plug in the known quantities into the torque equation, you should have one equation with one unknown, namely x, the location of the fulcrum. Solve for x, and Bob's you uncle.

Thanks a ton! does this look correct to you?



m1g*x - m2g*(10-x) + m3g * (x-5)

where m3 is the mass of the board. Btw bob is my uncles name and i didn't know that was a saying I was like wtf haha

 

[dreamLord]

Looks correct to me.