Question

1. a salvage ship is attempting to raise an iron anchor off the ocean floor. the anchor has a mass of 560 kg and the density of iron is 7 800 kg/m³. the cable used to lift the anchor can support a weight of 5 000 n before breaking. use 1 025 kg/m³ as the density of seawater. (a) can the cable support the anchor while it is completely submerged? (b) can the cable support the anchor when it is completely out of the water? (c) what percentage of the anchor will be out of the water when the cable breaks?

Explanation:

Part (a)

Step 1: Find the volume of the anchor

The density formula is $
ho = \frac{m}{V}$, so $V=\frac{m}{
ho_{iron}}$. Substituting $m = 560\space kg$ and $
ho_{iron}=7800\space kg/m^3$, we get $V=\frac{560}{7800}\space m^3\approx0.0718\space m^3$.

Step 2: Find the buoyant force

The buoyant force $F_b=
ho_{seawater}gV$, where $g = 9.8\space m/s^2$, $
ho_{seawater}=1025\space kg/m^3$, and $V = 0.0718\space m^3$. So $F_b=1025\times9.8\times0.0718\approx727\space N$.

Step 3: Find the tension in the cable when submerged

The weight of the anchor $W = mg=560\times9.8 = 5488\space N$. The tension $T=W - F_b=5488 - 727 = 4761\space N$.

Step 4: Compare with cable's capacity

Since $4761\space N<5000\space N$, the cable can support it.

Part (b)

Step 1: Find the weight of the anchor

$W = mg=560\times9.8 = 5488\space N$.

Step 2: Compare with cable's capacity

Since $5488\space N>5000\space N$, the cable cannot support it.

Part (c)

Step 1: Let the volume of the anchor out of water be $V_{out}$, so submerged volume is $V - V_{out}$. The tension $T = W-

ho_{seawater}g(V - V_{out})$. We know $T = 5000\space N$, $W = 5488\space N$, $V = 0.0718\space m^3$.
Substituting: $5000=5488 - 1025\times9.8\times(0.0718 - V_{out})$.

Step 2: Solve for $V_{out}$

First, simplify: $1025\times9.8\times(0.0718 - V_{out})=5488 - 5000 = 488$.
Then, $0.0718 - V_{out}=\frac{488}{1025\times9.8}\approx0.048$. So $V_{out}=0.0718 - 0.048 = 0.0238\space m^3$.

Step 3: Find the percentage

Percentage $=\frac{V_{out}}{V}\times100=\frac{0.0238}{0.0718}\times100\approx33.1\%$.

Answer:

s:
(a) Yes, the cable can support the anchor while submerged.
(b) No, the cable cannot support the anchor out of water.
(c) Approximately $\boldsymbol{33.1\%}$ of the anchor will be out of water when the cable breaks.