# physics balancing problem?

Here is the problem: A uniform meter stick supported at the 25-cm mark is in equilibrium when a 1=kg rock is suspended at the 0-cm end. Is the mass of the meter stick greater than equal to, or less than the mass of the rock? Explain your reasoning.

My thinking: I figured that because the moment arm of the rock was less than the moment arm of the meter stick, that the rock would weigh more but apparently it is equal to. I'm just having a hard time comprehending why it is equal to and not the rock greater than the meter stick. Any help is much appreciated, thank you.

### 5 Answers

- Anonymous10 years agoBest Answer
The meter stick is uniform, so there is a direct relationship between mass and length.

1 meter = 100 cm, so 25% of the mass of the stick can be considered centered at the 25/2 cm mark, and the remaining 75% of the mass of the stick can be considered centered at the (25+75/2) cm mark.

Static equilibrium: sum of moments about the 25 cm mark is zero.

1 kg * 25 cm + 0.25 m_stick * 25/2 cm = 0.75 m_stick * (75/2) cm

Solve for m_stick, then compare m_stick with 1-kg (the mass of the rock.)

- 10 years ago
lets call the

mass of the stick, M

mass of the rock, m

Being a meter stick, it is 100 cm long. If it is balanced at the 25 cm mark, then 1/4 of the stick is hanging to one side and 3/4 is hanging on the other. The force balance equation becomes...

> (25 cm)(m) + (25/2 cm)(M/4) - (75/2 cm)( (3/4)*M) = 0

Solve for M

- 10 years ago
so the stick is a total of 50 cm long sing the balancing point is at 25 cm? and at the 0 cm end there is a rock being balanced? seems like if there is an equilibrium then the mass of the stick would far outweigh the mass of the rock. but what the **** do i know.

- Anonymous10 years ago
________________________________________ << meter stick * note meter means 100 cm

^

0cm 25cm 100cm

Force = mass

moment = Force X Dist

moment of the rock is 25 X 1000 =25000

for the rule to stay in equilibrium

the 75 cm part on the right must have the same force acting on it , other wise it would topple .

hence it is equal to

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- 10 years ago
You can understand it this way:

Mass of rock is 'm' and of stick is 'M' .

Now by momentum balance;

mx1/4 + (M/4)x1/8 = (3M/4)x3/8

By taking the center of mass (COM) of either sides of the stick at mid of them. Hence the COM of 3/4th part will lie at distance of 3/8th length of the stick measured from the center of fulcrum.Vice-versa for the other side.

Source(s): Yeah so it'll turn out that m=M.