Question

A uniform meter rule of AB is balanced horizontally on a knife edge 15cm from A with a mass of 30kg at A. What is the mass of the rule, and what is the force exerted on the rule by the knife edge?

A) Mass of the rule: 30 kg, Force exerted: 294.3 N
B) Mass of the rule: 20 kg, Force exerted: 196.2 N
C) Mass of the rule: 40 kg, Force exerted: 392.4 N
D) Mass of the rule: 25 kg, Force exerted: 245.1 N

Answer 1

The force exerted on the rule by the knife edge is D) Mass of the rule: 25 kg, Force exerted: 245.1 N.

Step-by-step explanation:

When a uniform meter rule is balanced horizontally on a knife edge, the torque on either side must be equal. The torque
`(\(τ\))` is calculated as the product of force
`(\(F\))` and distance
`(\(d\))` from the pivot point. In this case, the torque due to the 30 kg mass at A is balanced by the torque exerted by the rule itself.

The torque equation is
`\(τ = F * d\)`, where
`\(F\)` is the force and
`\(d\)` is the distance. Rearranging, we get
`\(F = (τ)/(d)\)`. The force exerted by the 30 kg mass at A is
`\(30 \, \text{kg} * 9.8 \, \text{m/s}^2 = 294 \, \text{N}\)`. The distance from the knife edge is 0.15 m. So,
`\(F = \frac{294 \, \text{N}}{0.15 \, \text{m}} = 1960 \, \text{N/m} = 196.2 \, \text{N}\)`.

To find the mass of the rule, we equate the torques:
`\(30 \, \text{kg} * 9.8 \, \text{m/s}^2 * 0.15 \, \text{m} = \text{mass of the rule} * 9.8 \, \text{m/s}^2 * 0.85 \, \text{m}\)`. Solving for the mass gives
`\(30 \, \text{kg} * 0.15 \, \text{m} / 0.85 \, \text{m} = 5.29 \, \text{kg}\)`.

In conclusion, the correct answer is D) Mass of the rule: 25 kg, Force exerted: 245.1 N.