A block of wood floats in a beaker of water. The block experiences an upward buoyant force. If the beaker (with water and block) is weighed, would the measured weight be less than the sum of the weights of the individual components?


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  • $\begingroup$ Where would the mass go? $\endgroup$ – pela Feb 9 '16 at 19:49
  • $\begingroup$ I'm sorry, I didn't mean to be rude. I just wanted a clarification of what you thought would make it weigh less, since it would make it easier to formulate an answer. I realize that my comment seems condencending; that wasn't the purpose. Anyway, tmwilson provided an excellent answer, which I encourage you to accept/upvote :) $\endgroup$ – pela Feb 10 '16 at 7:21

The measured weight would be the same as the sum of the two individual components as pointed out by "pela" in the comments section due to conservation of mass.

A simple reasoning to think about in this case would Newton's third law.

For every action, there is an equal and opposite reaction

When you measure the weight of something, you are measuring the force that it exerts on a scale (this force being due to gravity in this case). Now, the water exerts a buoyant force on the wood to hold it afloat, but at the same time, the wood block exerts an equal an opposite force on the water.

The same thing can be said for your scale. It is exerting a force to hold up your container, water and the block to keep it from falling under the influence of gravity. All of these forces must balance in order to maintain the equilibrium of the system. The force that the scale must apply is equal to the sum of the weights of all of the objects its holding up, otherwise this equilibrium cannot be maintained.


A similar question was asked in last couple of days in terms of a falling mass inside a liquid filled container.

The answer is no, the weight will not be less than the sum of the weights of the individual components.


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