- #1

brotherbobby

- 699

- 163

- Homework Statement
- An iron anchor with mass ##35.0\;\text{kg}## and density ##7860\;\text{kg/m}^3## lies on the deck of a small barge that has vertical sides and floats in a freshwater river. The area of the bottom of the barge is ##8.00\;\text{m}^2##. The anchor is thrown overboard but is suspended above the bottom of the river by a rope; the mass and volume of the rope are small enough to ignore. After the anchor is overboard and the barge has finally stopped bobbing up and down, has the barge risen or sunk down in the water? By what vertical distance?

##\mathbf{\text{Answer = }}\; \boxed{\pmb{5.57\times 10^{-4}\;\text{m}}}##

- Relevant Equations
- 1. ##\text{The law of flotation}## : When a body of mass ##m_B## floats in a liquid, the mass (and weight) of the body is the mass (or weight) of the liquid displaced : ##m_B=\Delta m_L## (and ##w_B=\Delta w_L##).

2. Volume of a body : ##V_B = \dfrac{m_B}{\rho_B}##, where ##\rho_B## is the density of the body.

(I must confess before I can begin that I found this problem difficult to understand, for reasons I will make clear below. I know it appears simple.)

We can see that the in both cases (i) or (ii), the barge and anchor combination is effectively "floating" in water.

Hence, by the law of flotation (see the first of the Relevant Equations above), the mass of water displaced in both occasions : ##\Delta m_W = m_B+m_A## is the same.

If the same amount of water is displaced, the height of water "going up" as a result of the barge-anchor combination also remains the same : ##\Delta h_W = \dfrac{V_W}{A}=\dfrac{m_B+m_A}{\rho_W A}##, where ##A## is the cross sectional area of the river!

The height by which the water level "went up" is the same on both occasions.

Hence, by my reasoning, the barge

A help or hint will be welcome.

**Let me begin by drawing the problem situation alongside, to the best I understand.
Attempt : **We can see that the in both cases (i) or (ii), the barge and anchor combination is effectively "floating" in water.

Hence, by the law of flotation (see the first of the Relevant Equations above), the mass of water displaced in both occasions : ##\Delta m_W = m_B+m_A## is the same.

If the same amount of water is displaced, the height of water "going up" as a result of the barge-anchor combination also remains the same : ##\Delta h_W = \dfrac{V_W}{A}=\dfrac{m_B+m_A}{\rho_W A}##, where ##A## is the cross sectional area of the river!

The height by which the water level "went up" is the same on both occasions.

Hence, by my reasoning, the barge

**will neither rise nor sink further in water**as a result of throwing the anchor overboard. But clearly my answer is incorrect, as per the book.A help or hint will be welcome.