A meterstick is found to balance at the 49.7 cm mark when placed on a fulcrum. When a 50.0g mass is attached at the 10.0cm mark, the fulcrum must be moved to the 39.2 cm mark for balance. What is the mass of the meterstick?
All of the problems ive been doing i needed to find length of meter stick, so I am not too sure on how to solve this. thanks
You know the length of the meter stick, one meter (100 cm).
First we need to find the center of gravity. We know that is at 49.7 am because that is where it balances all by itself.
now we put the 50 g mass on,
take moments about the 39.2 balance point.
50 (39.2 - 10) = m (49.7-39.2)
solve for m
To solve this problem, we can use the principle of moments, which states that the sum of the moments on one side of a balanced lever is equal to the sum of the moments on the other side.
Let's assume that the mass of the meterstick is M grams and its length is L cm.
First, let's calculate the moment of the meterstick about the fulcrum when it balances at the 49.7 cm mark:
Moment of the meterstick = M × (49.7 cm)
Next, let's calculate the moment of the 50.0 g mass about the fulcrum when it's attached at the 10.0 cm mark:
Moment of the 50.0 g mass = (50.0 g) × (10.0 cm)
Since the moments on both sides of the fulcrum must be balanced, we can set up the following equation:
Moment of the meterstick = Moment of the 50.0 g mass.
M × (49.7 cm) = (50.0 g) × (10.0 cm)
Now, let's solve for the mass of the meterstick (M):
M = (50.0 g) × (10.0 cm) / (49.7 cm)
M = 1000 g cm / 49.7 cm
M ≈ 20.1 g
Therefore, the mass of the meterstick is approximately 20.1 grams.
To solve this problem, we can use the principle of moments which states that the moment of a force about a fulcrum is equal to the product of the force and its perpendicular distance from the fulcrum.
First, let's define some variables:
- Let L be the length of the meterstick in centimeters.
- Let x be the distance from the fulcrum to the 10.0 cm mark.
- Let y be the distance from the fulcrum to the 49.7 cm mark.
- Let m be the unknown mass of the meterstick in grams.
According to the problem, the meterstick balances at the 49.7 cm mark when placed on the fulcrum. This means that the sum of the clockwise moments (when viewed from the left side) is equal to the sum of the counterclockwise moments.
The counterclockwise moment due to the mass at the 49.7 cm mark is:
49.7 cm * m = 49.7m
Now, when the 50.0g mass is attached at the 10.0 cm mark, the fulcrum needs to be moved to the 39.2 cm mark for balance. Again, the sum of the clockwise moments is equal to the sum of the counterclockwise moments.
The clockwise moment due to the 50.0g mass at the 10.0 cm mark is:
(39.2 cm - 10.0 cm) * 50.0g = 29.2 cm * 50.0g = 1460g·cm
Now, since the system is in balance, the sum of the clockwise and counterclockwise moments must be equal:
49.7m = 1460
Now we can solve for m:
m = 1460 / 49.7 ≈ 29.39 g
Therefore, the mass of the meterstick is approximately 29.39 grams.