Question 15 Marks
A $40 \ kg$ boy sits $1.2\ m$ from the fulcrum of a see saw. Where should another boy of $30 \ kg$ sit on the other side in order to balance it ?
Answer
View full question & answer→The see$-$saw will be balanced when the moment of force of both the boys is equal.
For $40 \ kg$ boy,
Weight$, F = 40 \ kg \times 10\ m/s^2 = 400\ N$
Moment arm$, d = 1.2\ m$
Moment of force $= F \times d = 400\ N \times 1.2\ m = 480\ N m$
For $30 \ kg$ boy,
Weight$,F = 30\ kg \times 10\ m/s^2= 300\ N$
Moment of force $= 480\ N$ Moment arm, $d=\frac{\text { Momesy of force }}{F}=\frac{480\ Nm }{300\ N } 1.6\ m$
The $30\ kg$ boy should sit at a distance of $1.6\ m$ from the fulcrum of the see$-$saw.
For $40 \ kg$ boy,
Weight$, F = 40 \ kg \times 10\ m/s^2 = 400\ N$
Moment arm$, d = 1.2\ m$
Moment of force $= F \times d = 400\ N \times 1.2\ m = 480\ N m$
For $30 \ kg$ boy,
Weight$,F = 30\ kg \times 10\ m/s^2= 300\ N$
Moment of force $= 480\ N$ Moment arm, $d=\frac{\text { Momesy of force }}{F}=\frac{480\ Nm }{300\ N } 1.6\ m$
The $30\ kg$ boy should sit at a distance of $1.6\ m$ from the fulcrum of the see$-$saw.