To calculate the weight of the meter rule, we need to use the principle of moments. The principle of moments states that for a body in equilibrium, the sum of the clockwise moments is equal to the sum of the anticlockwise moments.
Let's assume the weight of the meter rule is W (unknown weight). We know that the object of weight 3.15N at the 10cm mark creates a clockwise moment and the meter rule pivoted at the 20cm mark creates an anticlockwise moment.
The clockwise moment is given by the weight of the object multiplied by its distance from the pivot:
Moment clockwise = 3.15N * 10cm = 31.5 Ncm
The anticlockwise moment is given by the weight of the meter rule (W) multiplied by its distance from the pivot:
Moment anticlockwise = W * 20cm = 20W Ncm
Since the meter rule is balanced horizontally, the sum of the clockwise moments should be equal to the sum of the anticlockwise moments:
31.5 Ncm = 20W Ncm
Dividing both sides of the equation by 20cm, we get:
31.5/20 = W
W ≈ 1.575 N
Therefore, the weight of the meter rule is approximately 1.575 Newtons.
A figure shows a uniform metre rule which is pivoted at the 20cm mark and balanced horizontally by an object of weight 3.15N placed at the 10cm mark.
Calculate the weight of the metre rule.
1 answer