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A 40 kg beam, 50 m long, is balanced but not attached to a post. A 20 kg boy stands on the right end of the beam. How far to the left of the beam should a 10 kg weight be placed?

Answer :

The 10 kg weight should be placed 50 meters to the left of the boy (or 50 meters from the right end of the beam).

To solve this problem

Any location on the beam must have the total of the clockwise moments equal to the sum of the counterclockwise moments in order to remain balanced.

Let x represent the distance between the 10 kg weight's location and the right end of the beam, where the youngster is standing.

Moments involving the youngster (20 kg) include:

Clockwise moment: 20 kg (boy's weight) * x (distance to the 10 kg weight) * g (acceleration due to gravity)

The moments about the center of mass (midpoint of the beam) will be:

Counterclockwise moment: 40 kg (beam weight) * 25 m (distance from the center of mass to the right end of the beam) * g

Since the beam is balanced, these moments are equal:

20 kg * x * g = 40 kg * 25 m * g

Now, we can solve for x:

x = (40 kg * 25 m) / 20 kg

x = 50 m

Therefore, the 10 kg weight should be placed 50 meters to the left of the boy or 50 meters from the right end of the beam.

Learn more about Clockwise moment here : brainly.com/question/26117248

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