Two children of mass 20.0 kg and 30.0 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If the children are separated by a distance of 3.00 m, at what distance from the pivot point is the small child sitting in order to maintain the balance

Respuesta :

Answer:

The distance from the pivot point that the small child will sit in order to maintain the balance is 1.8 m

Explanation:

Given;

mass of the bigger child, M = 30 kg

mass of the smaller child, m = 20 kg

distance between the two children, d = 3 m

This information can be represented diagrammatically;

                                    3m

         |<------------------------------------------------>|

----------------------------------------------------------------------------

         ↓             x            Δ            3-x           ↓

         20kg                                                 30kg

x is the distance from the pivot point that the small child will sit in order to maintain the balance

Take moment about the pivot;

Clockwise moment = anticlockwise moment

30(3-x) = 20x

90 -30x = 20x

90 = 20x + 30x

90 = 50x

x = 90 / 50

x = 1.8 m

Therefore, the distance from the pivot point that the small child will sit in order to maintain the balance is 1.8 m

Lanuel

The distance from the pivot point which the small child must sit in order to maintain the balance is 1.8 meters.

  • Let the first child be A.
  • Let the second child be B.

Given the following data:

  • Mass of A = 20.0 kg
  • Mass of B = 30.0 kg
  • Distance = 3.00 m

To determine what distance from the pivot point is the small child sitting in order to maintain the balance, we would take moment about a pivot:

  • Let the distance from the pivot be n.

Note: The distance of the child from the pivot is equal to [tex]3-n[/tex]

For moment:

Clockwise moment = anticlockwise moment

[tex]30(3-n) = 20n\\\\90-30n=20n\\\\90=20+30n\\\\90=50n\\\\n=\frac{90}{50}[/tex]

n = 1.8 meters

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