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
D. Nothing will happen; the seesaw will still be balanced.
The force acting on a system with static equilibrium is 0
[tex] \large {\boxed {\bold {\sum F = 0}} [/tex]
(forces acting as translational motion only, not including rotational forces)
[tex] \displaystyle \sum F_x = 0 \\\\\ sum F_y = 0 [/tex]
For objects undergoing rotation, the equilibrium must be met
[tex] \large {\boxed {\bold {\sum \tau = 0}} [/tex]
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:
[tex]\rm Hb\times \dfrac{r_1}{2}= Lg\times \dfrac{r_2}{2}\\\\Hb\times r_1=Lg\times r_2[/tex]
This equation remains the same as equation 1, so the seesaw will still be balanced.
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