Question

A heavy seesaw (i.e., not massless) is out of balance. A light girl sits on the end that is tilted downward, and a heavy body sits on the other side so that the seesaw now balances. If they both move forward so that they are one-half their original distance from the pivot point (the fulcrum) what will happen to the seesaw?

• A. The side the body is sitting on will tilt downward
• B. The side the girl is sitting on will once again tilt downward
• C. Nothing; the seesaw will still be balanced.
• D. It is impossible to say without knowing the masses and the distances.

Answer 1

The correct option is B The side the girl is sitting on will once again tilt downward

Let the mass distribution on the left of seesaw be x kg and that on the right be y kg such that $x>y$. And let the mass of the girl be $m_g$ and that of heavy object be $m_h$ such that $m_h>m_g$
Now during the equilibrium position we have the moments about the fulcrum as
$m_h\times \frac{l}{2}+y\times \frac{l}{4}=m_g\times \frac{l}{2}+x\times \frac{l}{4}$
or
$m_h-m_g=\frac{x-y}{2}$
Now when both the girl and the heavy object move half the distance we see that the moment about the pivot by the girl and the heavy object are $m_g\times \frac{l}{4}+x\times \frac{l}{4}$ and $m_h\times \frac{l}{4}+y\times \frac{l}{4}$
Subtracting moment of the girl from the moment of the heavy object
$(m_h+y)\frac{l}{4}-(m_g+x)\frac{l}{4}$
or
$\frac{l}{4}(m_h-m_g+y-x)$
or
$\frac{l}{4}(\frac{x}{2}-\frac{y}{2}+y-x)$
or
$\frac{l}{4}(\frac{y-x}{2})$
We get this value as negative because $x>y$
Thus the moment applied by the girl is more than that of the heavy object implying that the side of the girl tilts downwards