Question

On a see-saw, two children of masses 50 kg are sitting on one side of it at distances 2 m and 2.5 m respectively from its middle. Where should a man of mass 74 kg sir to balance it?


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Answer 1

Answer:

2.5 m

Explanation:

Here we have been provided with the values of masses of two children and distances at which they are sitting.

Those are as follows:-

• Mass of first children = 30 kg
• Mass of second children = 50 kg
• Distances = 2m and 2.5 m

As the both children are sitting on left arm. Therefore anticlockwise movement has been produced.

Let's calculate the anticlockwise moment produced by first children.

→ 30kgf × 2m

→ 60 kgf × m

Now let us calculate the anticlockwise moment produced by second children.

→ 50 kgf × 2.5 m

→ 125 kgf × m

Thus total anticlockwise moment produced is as follows:-

→ Anticlockwise moment produced by first children + anticlockwise moment produced by second children.

Substituting values,

→ 60kgf × m + 125 kgf × m

→ 185 kgf × m

Here we are calculating the clockwise moment produced which are as follows,

But before that let us assume distance from middle by y m.

Thus,

→ Clockwise moment = 74 kgf × y m

→ Clockwise moment = 74 y kgf × m

At last we are calculating the distance where should a man of mass 74 kg sir to balance it that I y m.

Using principal of moments, in equilibrium:-

Which states that,

→ Anticlockwise moment = Clockwise moment

Substituting required values,

→ 185 = 74 y

→ y = 185 / 74

→ y = 2.5 m

Therefore, he should sit at 2.5 m to balance it