# Homework Help: Torque Equilibrium

1. Nov 13, 2008

### Kandycat

1. The problem statement, all variables and given/known data
A meter stick is pivoted at its 50 cm mark but does not balance because of the nonuniformities in its material that cause its center of gravity to be displaced from its geometrical center. However, when weights of 150 g and 300 g are place at the 10 cm and 75 cm marks, respectively, balance is obtained. The weights are then interchanged and balance is again obtained by shifting the pivot point to the 43 cm mark. Find the mass o the meter stick and the location of its center of gravity.

2. Relevant equations
t1 = t2
m1gr1 = m2gr2

3. The attempt at a solution

I tried to figure out the mass of the meter stick first by using the equation below:

m1gr1 = m2gr2

So... I plugged in the numbers (g cancels out)

(300 kg)(25 cm) = (150+x)(25 cm)

I thought it was 25 cm because from 0 to 50 cm, the diameter is 50 cm and the radius must be 25 cm. I don't know... my teacher lost me. X is the mass of the meter stick.

But I got lost and confused since there is another piece of information that I think I need to use. I don't really understand torque very well.

2. Nov 13, 2008

### physics girl phd

Part of the problem is that you have THREE weights: One from each mass, and one from the meter-stick (at some unknown location down the stick, say at position L). You need to include all torques in your equation -- the weights and the upward force from the pivot... and you really have two systems (saying that you'll probably have to solve a system of equations). Draw the free body diagrams for each case.

I also predict that you'll have to use not only counter-clockwise torque equals clockwise torqu (or net torque equals 0), but also upward force equals downward force (or net force equals 0).

3. Nov 13, 2008

### Kandycat

So this is what I figure out... somewhat:

T1 = T2 + T3

When the fulcrum is at 50 cm:
(.300 kg)(25 cm) = XL + (.150 kg)(40 cm)

When the fulcrum is at 43 cm:
(.150 kg)(32 cm) + XL = (.300 kg)(33 cm)

X= the mass of the meter stick
L= the position where the center of gravity from the fulcrum

I sort of understand, but I'm still a bit lost on how to solve for my unknowns.

4. Nov 13, 2008

### LowlyPion

Maybe take a different approach?

The given is that you have a stick that has uncertain mass distribution. But whatever its distribution it does have a weight and a Center of Mass.

Now you also have these 2 masses 150 and 300 g. And they are placed 65 cm apart. And they also have a center of mass. What you are doing then is stacking 2 objects, one with a known weight and CoM and the other not known and you are given the resulting CoM.

Since they do that twice and move the known CoM and give you another resulting CoM you should be able to solve for the 2 unknowns - the weight and CoM of the unknown meterstick.

5. Nov 13, 2008

### Kandycat

Just to make sure I'm right... I figured this:

At the 50 cm fulcrum:
0 = (.300 kg)(25 cm) - (.150 kg)(40 cm) - x(50 cm-L)
0 = 1.5 - x(50-L)
1.5 = x(50-L)

At the 43 cm fulcrum:
0 = (.150 kg)(32 cm) + x(L - 43 cm) - (.300 kg)(33 cm)
0 = -5.1 + x(L - 43 cm)
5.1 = x(L - 43 cm)

So using substitution... I found the mass of the meter stick to be .94 kg and the position of CoG to be 48.4 cm. Did I do it right?

Also, why isn't the force of the stand that balances the meter stick not included?

Last edited: Nov 13, 2008