A meter stick is pivoted at the 0.50-m line. A 3.0-kg object is hung from the 0.15-m line. Where should a 5.0-kg object be hung to achieve equilibrium (the meter stick oriented horizontal and motionless)?

Respuesta :

Answer:

0.71 m

Explanation:

Let the 5 kg is hang at a distance d from the midpoint.

use the concept of moments.

take moments about the mid point.

Anticlockwise moments = clockwise moments

3 x (0.5 - 0.15) = 5 x d

1.05 = 5 d

d = 0.21 m

So, the distance of 5 kg from 0 mark on the stick is 0.5 + 0.21 = 0.71 m

Ver imagen Vespertilio

The point at which the second object should be hung to achieve equilibrium is 0.71 m.

To calculate the distance from which the 5.0 kg object should be hung to achieve equilibrium, we use the formula below.

Formula:

  • Clockwise moment = Anti clockwise moment
  • md = m'd'
  • d' = md/m'............. Equation 1

Where:

  • m = mass of the first object hung on the meter stick
  • d = distance of the first object from the pivot
  • m' = mass of the second object hung on the meter stick
  • d' = distance of the second object from the pivot.

From the question,

Given:

  • m = 3 kg
  • d = (0.5-0.15) = 0.35 m
  • m' = 5 kg

Substitute these values into equation 1'

  • d' = (3×0.35)/5
  • d' = 0.21 m.

The point at which the second object should be hung to achieve equilibrium is

  • x = (0.21+0.5) = 0.71 m.

Hence, The point at which the second object should be hung to achieve equilibrium is 0.71 m.

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