Answer:
Approximately [tex]50.5\; {\rm g}[/tex] (rounded to three significant figures.)
Explanation:
Let [tex]m_\text{p}[/tex] and [tex]m_\text{a}[/tex] denote the mass of the pineapple and the apple, respectively.
Let [tex]r_{\text{p}}[/tex] denote the distance between the pineapple and the pivot. Let [tex]r_{\text{a}}[/tex] denote the distance between the apple and the pivot.
Because of gravity, both the pineapple and the apple would exert a normal force on the seesaw. The magnitude of that force is equal to the weight of the fruit. Let [tex]g[/tex] denote the gravitational field strength.
- Normal force from the pineapple: [tex]F_\text{p} = m_{\text{p}}\, g[/tex].
- Normal force from the apple: [tex]F_\text{a} = m_{\text{a}}\, g[/tex].
Since these two forces are perpendicular to the seesaw, the magnitude of the torque exerted by the pineapple and the apple would be:
- From the pineapple: [tex]\tau_{\text{p}} = F_\text{p}\, r_\text{p} = m_{\text{p}}\, g\, r_{\text{p}}[/tex].
- From the apple: [tex]\tau_{\text{a}} = F_\text{a}\, r_\text{a} = m_{\text{a}}\, g\, r_{\text{a}}[/tex].
For the seesaw to maintain a rotational equilibrium, these two torques need to balance each other. Thus:
[tex]m_{\text{p}} \, g\, r_{\text{p}} = m_{\text{a}} \, g\, r_{\text{a}}[/tex].
Rewrite and simplify this equation to find an expression for the unknown mass of this apple, [tex]m_{\text{a}}[/tex]:
[tex]\begin{aligned}m_{\text{a}} &= \frac{m_{\text{p}}\, g\, r_\text{p}}{g\, r_\text{a}} \\ &= \frac{m_{\text{p}}\, r_{\text{p}}}{r_\text{a}}\end{aligned}[/tex].
Substitute in the values [tex]m_\text{p} = 121\; {\rm g}[/tex], [tex]r_\text{p} = 1.23\; {\rm m}[/tex], and [tex]r_{\text{a}} =2.95\; {\rm m}[/tex] and evaluate:
[tex]\begin{aligned}m_{\text{a}} &= \frac{m_{\text{p}}\, r_{\text{p}}}{r_\text{a}} \\ &= \frac{121\; {\rm g} \times 1.23\; {\rm m}}{2.95\; {\rm m}} \\ &\approx 50.5\; {\rm g}\end{aligned}[/tex].
Thus, the mass of the apple should be approximately [tex]50.5\; {\rm g}[/tex].
