Solve the Balance: Meter Stick Mass Calculation

[ah-huh]

"A meter stick is found to balance at the 49.7cm mark when placed on a fulcrum. When a 50 gram mass is attached at the 10cm mark, the fulcrum must be moved to the 39.2 cm mark for balance. What is the mass of the meter stick?"

thats the question and the last one on my homework and its given me the u-know-what. I've tried workin it - one answer i got was 245g... but some reason i am quite sure its wrong.. so any hints or help would be greeeat

 

[hotvette]

Maybe you can describe your method?

 

[quicknote]

Based on the information provided, we can use the principle of moments to solve for the mass of the meter stick. The principle of moments states that the sum of the moments on one side of a fulcrum must be equal to the sum of the moments on the other side for the system to be in equilibrium.

In this case, we can set up the equation as follows:

(49.7 cm - x) * m = (39.2 cm - 10 cm) * 50 g

Where:
- x is the distance from the fulcrum to the center of mass of the meter stick
- m is the mass of the meter stick

Solving for m, we get:

m = (39.2 cm - 10 cm) * 50 g / (49.7 cm - x)

Now, we know that the meter stick balances at the 49.7 cm mark when placed on the fulcrum, which means that the center of mass must be at the 49.7 cm mark. Therefore, x = 49.7 cm.

Substituting this value into the equation, we get:

m = (39.2 cm - 10 cm) * 50 g / (49.7 cm - 49.7 cm) = 245 g

Therefore, the mass of the meter stick is 245 g.

It is always a good idea to double check your calculations and make sure they make sense. In this case, we can see that the mass of the meter stick is significantly larger than the mass of the 50 g weight added to it, which makes sense since the fulcrum had to be moved closer to the weight to achieve balance.