A uniform half metre rule is freely pivoted at the 15 cm mark and it balances horizontally when a body of mass 40g is hung from 2 cm mark
A. Draw a clear first diagram of the arrangement
B. Calculate the mass of the object placed 10 cm to the end of the meter rule
A uniform half metre rule is freely pivoted at the 15cm mark and it is balanced horizontally when a body of mass 40g is hung from the 2cm mark draw a clear first diagram of the arrangement and calculate the mass of the rule
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To draw a clear first diagram of the arrangement, you'll need to depict the uniform half-meter rule with the body of mass 40g hung from the 2 cm mark and the pivot point at the 15 cm mark. Here's a simple representation:
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In this diagram, the horizontal line represents the half-meter rule, with markings for different positions. The circular dot at the 15 cm mark represents the pivot point, while the vertical line at the 2 cm mark represents the body of mass 40g hanging from it.
Now, to calculate the mass of the object placed 10 cm from the end of the meter rule, following these steps:
1. Given that the system is balanced horizontally, the moments on each side of the pivot must be equal. The moment of a force is calculated by multiplying the force by its distance from the pivot point.
2. Let's assume that the mass at 10 cm from the end of the meter rule has a mass of "m" grams. The moment on the left side of the pivot is given by the product of the mass (40g) and the distance (13 cm = 15 cm - 2 cm).
3. The moment on the right side of the pivot is given by the product of the unknown mass (m g) and its distance (10 cm).
4. Set up an equation to represent the balance of moments:
40g × 13 cm = m g × 10 cm
5. Rearrange the equation to solve for the unknown mass "m":
m g = (40g × 13 cm) / 10 cm
6. Cancel out the units and solve for "m":
m = (40 × 13) / 10
= 520 / 10
= 52 g
Therefore, the mass of the object placed 10 cm from the end of the meter rule is 52 grams.