Question

A heavy boy and a lightweight girl are balanced on a massless seesaw. If they both

Answer 1

D. Nothing will happen; the seesaw will still be balanced.

Further explanation

The force acting on a system with static equilibrium is 0

$\large {\boxed {\bold {\sum F = 0}}$

(forces acting as translational motion only, not including rotational forces)

$\displaystyle \sum F_x = 0 \\\\\ sum F_y = 0$

For objects undergoing rotation, the equilibrium must be met

$\large {\boxed {\bold {\sum \tau = 0}}$

A heavy boy (Hb) and a lightweight girl (Lg) are balanced on a mass-less seesaw

Because there is a balance of rotation, the torque equation:

Στ = 0

Hb.r1-Lg.r2 = 0

Hb.r1 = Lg.r2 (equation 1)

If they both move forward so that they are one-half their original distance from the pivot point, then the distance of the two children to the pivot point is reduced to half

Then the torque equation:

$\rm Hb\times \dfrac{r_1}{2}= Lg\times \dfrac{r_2}{2}\\\\Hb\times r_1=Lg\times r_2$

This equation remains the same as equation 1, so the seesaw will still be balanced.

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Answer 2

Answer:

D. Nothing will happen; the seesaw will still be balanced.

Explanation:

D. Nothing will happen; the seesaw will still be balanced. Since both toruqes or momentums respect to the center have changed in the same amount (one-half their original distance) the seesaw will remain balanced, if the children change distance in a different amount then it will be out of balance