A uniform half metre rule is freely pivoted at 15 cm mark and it balances horizontally when a body of mass 40g is hung from the 2cm mark.
(a) Draw a clear force-diagram of the arrangement
(b) Calculate the mass of the rule
(a)
Here is a force diagram of the arrangement:
<-- 15 cm --> <-- 25 cm -->
O---------------|---------------O
| |
| |
2 cm 40 g
The point O represents the pivot point, or the fulcrum. The left side of the half meter rule is represented by the line to the left of O, with the lengths labeled in centimeters. The right side of the half meter rule is represented by the line to the right of O, with the lengths labeled in centimeters as well. The mass of 40 g is represented by the downward arrow on the right side of the ruler at the 2 cm mark.
(b)
To calculate the mass of the rule, we can use the principle of moments. The principle of moments states that the clockwise moments about a point equal the anti-clockwise moments about the same point.
In this case, the weight of the body of mass 40 g creates a clockwise moment about the pivot, and the weight of the ruler creates an anti-clockwise moment about the pivot. Since the ruler is balanced horizontally, these moments must cancel each other out.
Weight × Distance = Weight × Distance
(40 g) × (2 cm) = (m_rule) × (15 cm)
(40 g) × (0.02 m) = (m_rule) × (0.15 m)
0.8 g m = (m_rule) × 0.15 m
m_rule = (0.8 g m) / (0.15 m)
m_rule = 5.33 g
Therefore, the mass of the rule is approximately 5.33 g.