if a 250 g mass is placed at the 20 cm mark (30 cm from the fulcrum), at what mark should a 500 g mass be placed so that the system balances?
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To solve this problem, we can use the principle of moments, which states that for an object to balance, the sum of the clockwise moments must be equal to the sum of the anticlockwise moments.
Let's assign the following variables:
m1 = mass of the 250 g mass (250 g)
d1 = distance of the 250 g mass from the fulcrum (30 cm)
m2 = mass of the unknown mass (500 g)
d2 = distance of the unknown mass from the fulcrum (unknown)
Since the system is in equilibrium, the moments on both sides of the fulcrum will be equal:
m1 * d1 = m2 * d2
Plugging in the values we know:
(250 g) * (30 cm) = (500 g) * d2
Rearranging the equation:
d2 = (250 g * 30 cm) / 500 g
Calculating the result:
d2 = 15 cm
Therefore, the 500 g mass should be placed at the 15 cm mark from the fulcrum in order to balance the system.
To find the mark at which the 500 g mass should be placed so that the system balances, we need to consider the principle of moments. The principle of moments states that for a system to be in equilibrium, the sum of the clockwise moments must be equal to the sum of the anticlockwise moments.
In this case, we have a fulcrum and two masses. Let's say the distance of the 250 g mass from the fulcrum is x cm. Since the 250 g mass is placed at the 20 cm mark, we have:
Clockwise moment: 250 g * (30 cm - x cm)
Anticlockwise moment: 500 g * (x cm)
To achieve balance, the clockwise moment must be equal to the anticlockwise moment. Therefore, we can set up the equation:
250 g * (30 cm - x cm) = 500 g * (x cm)
Now, let's solve this equation to find the value of x, which represents the mark at which the 500 g mass should be placed:
250 g * (30 cm - x cm) = 500 g * (x cm)
Simplifying the equation:
(30 cm - x cm) = 2 * x cm
30 cm = 3 * x cm
x cm = 30 cm / 3
x cm = 10 cm
Therefore, the 500 g mass should be placed at the 10 cm mark so that the system balances.
I assume the fulcrum is at the center of a 100-cm bar. If so, then
250*30 = 500*15
so, the heavy weight is 15cm on the other side of the fulcrum, at the 65-cm mark.
If the fulcrum is not at the center of the bar, then you have to adjust for the difference in mass of the bar on each side.