An ideal meter stick is balanced on a pencil which is at the 60 cm mark. A 6 gram mass sits at the 90 cm mark. What is the mass of the meter stick?
Again, I have the answer and it is 18g. I have no clue where to even begin with this questions. Thank you in advance.
To find the mass of the meter stick, we can use the principle of moments. The principle of moments states that for an object to be in equilibrium, the sum of the clockwise moments must be equal to the sum of the anticlockwise moments.
In this scenario, the meter stick is balanced on the pencil at the 60 cm mark. We can assume that the mass of the meter stick is concentrated at its center, which would be at the 50 cm mark (since a meter stick is 100 cm long).
Let's calculate the clockwise and anticlockwise moments using the formula:
Moment = Force x Distance
Clockwise Moments:
The 6 gram mass at the 90 cm mark creates a clockwise moment. The distance of the mass from the fulcrum (60 cm mark) is (90 - 60) = 30 cm.
Clockwise Moment = (6 g) x (30 cm) = 180 g*cm
Anticlockwise Moments:
The mass of the meter stick concentrated at its center (50 cm mark) will create an anticlockwise moment. The distance of the meter stick's mass from the fulcrum is (60 - 50) = 10 cm.
Anticlockwise Moment = (Mass of meter stick) x (Distance) = (Mass of meter stick) x (10 cm)
According to the principle of moments, the clockwise moments equal the anticlockwise moments. So we can set up the equation:
180 g*cm = (Mass of meter stick) x (10 cm)
Now we can solve for the mass of the meter stick:
(Mass of meter stick) = (180 g*cm) / (10 cm)
(Mass of meter stick) = 18 g
Hence, the mass of the meter stick is 18 grams.