A meter stick weighs 200 grams and is supported at the center. Three masses of 100, 200, and 150 grams are placed respectively, at the 10-cm, 20-cm and 60 cm points of the meter stick. To balance the meter stick, where must a 300 grams weight be placed on the meter stick?
55 cm
To find the position where the 300 grams weight should be placed to balance the meter stick, we need to consider the principle of moments.
The principle of moments states that for an object to be in rotational equilibrium, the sum of the clockwise moments must be equal to the sum of the counterclockwise moments.
In this case, we have four masses on the meter stick: the 200 grams of the meter stick itself, the 100 grams at the 10-cm point, the 200 grams at the 20-cm point, and the 150 grams at the 60-cm point.
Let's label the position where we will place the 300 grams weight as "x" cm.
The clockwise moments are:
200 grams * distance from the center (0 cm) = 0
100 grams * distance from the center (10 cm) = 1000 cm * grams
200 grams * distance from the center (20 cm) = 4000 cm * grams
150 grams * distance from the center (60 cm) = 9000 cm * grams
The counterclockwise moments are:
300 grams * distance from the center (x cm) = 300x cm * grams
According to the principle of moments, the sum of the clockwise moments is equal to the sum of the counterclockwise moments.
1000 cm * grams + 4000 cm * grams + 9000 cm * grams = 300x cm * grams
Simplifying this equation, we get:
14000 cm * grams = 300x cm * grams
Dividing both sides by 300 grams, we find:
14000 cm = x cm
Therefore, the 300 grams weight should be placed at the 140-cm mark on the meter stick to balance it.