🪨 Rockin' the boat 🛶

A buoyancy brain teaser

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👶 Meme Of The Week

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🙋‍♂️ Interview Practice Question of the Week

Company: SpaceX

You’re a brown bear, testing your brand new boat in a bog, full to the brim, when you find a rock in your boat. Without thinking twice you angrily toss it into the water below you. What happens to the water line?

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✅ The Answer

Rock the boat, don't rock the boat baby

This question is a classic physics brain teaser testing Archimedes principle. Simply put:

“The buoyant force equals the weight of the fluid displaced.”

Or

\(F_b = \rho Vg \)

At first glance, someone might be inclined to say that the rock has sunk in the water, and caused the bog to overflow. A more thoughtful listener may realize that is not the case, but the real key lies in being able to easily and concisely prove what happens.

Starting at the free body diagram (FBD), prior to the throw we know that the total buoyant force exerted on the body is equal to the combined weight of the bear, boat, and rock (neglecting surface effects).

\(F_g = (m_{\text{bear}} + m_{\text{boat}} + m_{\text{rock}})g \)

\(F_b = \rho g V_{\text{sub}}\)

\(V_{\text{bog}} = V_{\text{water}} + V_{\text{sub}} \)

\( F_g = F_b \Longrightarrow m_{\text{bear}} + m_{\text{boat}} + m_{\text{rock}} = \rho V_{\text{sub}}\)

After throwing the rock, the formulas are as below:

\(F_{g2} = (m_{\text{boat}} + m_{\text{bear}}) g \)

\(F_{b2} = \rho g V_{\text{sub2}} \)

\(V_{\text{bog2}} = V_{\text{sub2}} + V_{\text{water}} + V_{\text{rock}} \)

\(F_{g2} = F_{b2} \quad \rightarrow \quad m_{\text{boat}} + m_{\text{bear}} = \rho V_{\text{sub2}} \)

\(\rho V_{\text{sub}} - m_{\text{rock}} = \rho V_{\text{sub2}} \)

\( \rho (V_{\text{bog}} - V_{\text{water}}) - m_{\text{rock}} = \rho V_{\text{sub2}} \)

\( \rho (V_{\text{bog}} - V_{\text{water}}) - m_{\text{rock}} = \rho (V_{\text{bog2}} - V_{\text{rock}} - V_{\text{water}})\)

\(\rho g V_{\text{bog}} - m_{\text{rock}} g = \rho g V_{\text{bog2}} - \rho g V_{\text{rock}}\)

\(m_{\text{rock}} g = F_{g, \text{rock}} \hspace{1cm} \)

\(\rho g V_{\text{rock}} = F_{b, \text{rock}} \hspace{1cm} \rho g V_{\text{bog}} = \rho g V_{\text{bog2}}\)

\(F_g = F_N + F_B \quad \rightarrow \quad F_g > F_B\)

Eventually, we reach a point where Fg, rock > Fb, rock, a fact that becomes abundantly clear if an FBD of the rock is drawn after it has sunk to the bog floor. Therein lies the crux of solving this problem!

Prior to the throw, the entire weight of the boat, bear, and rock was supported by Archimedes principle. By throwing the rock overboard, the ground of the bog provides some level of normal force on the rock (assuming it has sunk), supplementing its buoyant force. Plugging in the rest of the algebra:

\(F_g = F_N + F_B \quad \rightarrow \quad F_g > F_B \)

\(\rho g V_{\text{bog}} - F_g = \rho g V_{\text{bog2}} - F_B \quad \rightarrow \quad \rho g V_{\text{bog}} > \rho g V_{\text{bog2}}\)

\(V_{\text{bog}} = V_{\text{bog2}} + V_{\text{additional}}\)

\(V_{\text{additional}} > 0\)

A positive volume must be contributed to Vbog2 to equal the initial volume. From this we can conclude that the water level has dropped! If you realized off the bat that no longer having to carry the rock, and allowing the ground’s normal force to carry it instead of displacing additional water volume via boat submersion would cause the boat to rise out of the water more than the added volume of the rock to the bog, then kudos! If not, no worries!

These problems are meant not to be immediately obvious, but should be doable with a few deep breaths and solid first principles!