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Question

# In lab you find that a 1-kg rock suspended above water weighs $10 \mathrm{~N}$. When the rock is suspended beneath the surface of the water, the scale reads $8 \mathrm{~N}$. a. What is the buoyant force on the rock? b. If the container of water weighs $10 \mathrm{~N}$ on the weighing scale, what is the scale reading when the rock is suspended beneath the surface of the water? c. What is the scale reading when the rock is released and rests at the bottom of the container?

Solution

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$\textbf{Given}$

$m = 1 \text{ kg}$

$F_{g_1} = 10 \text{ N}$

$F_{g_2} = 8 \text{ N}$

$\textbf{Solution}$

$\textbf{a)}$ The buoyant force can be determined by subtracting the weight of a rock before and after it is submerged. Thus, the force is equal to :

$10 \text{ N} - 8 \text{ N} = \boxed{2 \text{ N}}$

$\textbf{b)}$ When a body is completely submerged, the buoyant force acts on it in the upward direction. According to the Newtons third law, there must be an opposing force of equal magnitude. This force was calculated in the previous part of the task and equals 2 N, which combined with the weight a container equals to:

$2 \text{ N} + 10 \text{ N} = \boxed{12 \text{ N}}$

$\textbf{c)}$ When the rock is resting on the bottom of a container, there are four forces that need to be taken into account. Two of these are the buoyant force and its reaction which cancel each other out. This leaves two forces which are the weight of the container and the rock. So the net force is equal to:

$10 \text{ N} + 10 \text{ N} = \boxed{20 \text{ N}}$

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