A uniform meter strick is found to balance at 50cm mark when placed on fulcrum. When 50g mass is attached at the 10cm mark, the fulcrum must be moved to the 40cm mark for balance. what is the mass of the meter strick?
moments about fulcrum
50(40 - 10 ) = m (50 - 40)
10 m = 1500
m = 150 grams
To solve this problem, we can use the principle of moments, which states that the sum of the clockwise moments is equal to the sum of the anticlockwise moments.
Let's assume the mass of the meter stick is M.
When the meter stick is balanced at the 50cm mark when placed on the fulcrum, we can say that the moment on the left side of the fulcrum is equal to the moment on the right side of the fulcrum.
Moment on the left side (before the 50g mass is attached):
= (Length of the meter stick on the left side of the fulcrum) * (Mass of the meter stick)
= (50cm - 0cm) * M
Moment on the right side (before the 50g mass is attached):
= (Length of the meter stick on the right side of the fulcrum) * (Mass of the meter stick)
= (100cm - 50cm) * M
After attaching the 50g mass at the 10cm mark, the fulcrum needs to be moved to the 40cm mark for balance, which means the moments on both sides should be equal. Let's calculate the moments after attaching the 50g mass.
Moment on the left side (after the 50g mass is attached):
= (Length of the meter stick on the left side of the fulcrum) * (Mass of the meter stick) + (Length of the meter stick on the left side of the fulcrum) * (Mass of the 50g mass)
= (50cm - 0cm) * M + (40cm - 10cm) * 50g
Moment on the right side (after the 50g mass is attached):
= (Length of the meter stick on the right side of the fulcrum) * (Mass of the meter stick)
= (100cm - 40cm) * M
Now, we can set up an equation by equating the moments on both sides:
(50cm - 0cm) * M + (40cm - 10cm) * 50g = (100cm - 40cm) * M
Simplifying the equation:
50cm * M + 30cm * 50g = 60cm * M
Dividing both sides by 10cm:
5M + 150g = 6M
Subtracting 5M from both sides:
150g = M
Therefore, the mass of the meter stick is 150 grams.
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 balanced, the sum of the clockwise moments must be equal to the sum of the anticlockwise moments.
Let's assume the mass of the meter stick is "M" in grams.
When the meter stick is balanced at the 50cm mark on the fulcrum:
The clockwise moments = 0 (since the meter stick is balanced)
The anticlockwise moments = (M grams) × (50 cm) = 50M
When a 50g mass is attached at the 10cm mark and the fulcrum is moved to the 40cm mark for balance:
The clockwise moments = (50g) × (30cm) = 1500
The anticlockwise moments = (M grams) × (10 cm) + (50g) × (40 cm) = 10M + 2000
According to the principle of moments, the clockwise moments must be equal to the anticlockwise moments:
1500 = 10M + 2000
-500 = 10M
M = -50g
The mass of the meter stick is -50 grams. However, it is important to note that mass cannot be negative. Hence, it is likely that there was an error in the measurements or the information provided. Please double-check the values and reevaluate the problem.