How Much Volume Does an Iron Anchor Displace in Water?

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An iron anchor with a density of 7870 kg/m³ is reported to be 290 N lighter in water than in air, prompting a calculation of its volume. The buoyant force equation, F_buoyant = density of water * g * volume, is central to solving this problem. By equating the buoyant force to the weight difference, the volume can be derived directly from the equation. The correct approach simplifies the calculation by focusing solely on the buoyant force and the weight difference, neglecting the density of air. The final volume of the anchor is determined to be approximately 0.00431 m³, although the initial calculation was incorrect.
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Homework Statement


An iron anchor of density 7870 kg/m3 appears 290 N lighter in water than in air.
What is the volume of the anchor?



Homework Equations


Fnet = 0 (we are doing static fluids) = forces up - forces down

Fbuoyant = density of liquid*g*volume of object

density = mass/ volume, so mass = density*volume

Weight in the air = mg, so weight in the water = (mg - 290N)

density of water = 1000 kg/m^3

The Attempt at a Solution



Fnet = Fbuoyant - (mg - 290N)
0 = 1000 * g * volume - (7870* 9.8*volume - 290N)
-290N = 9800 * volume - 77126 * volume
factor out volume, and

-290/ -67326 = volume = .00431 m^3

(the units work out right, I just left them out here to try to make things clearer)
unfortunately this answer is wrong...
any help would be great!
thanks
 
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It's simpler than it looks.

All you need is the Fbouyant equation, with 290 N on the left hand side and the values for g and the density of water on the right. Then solve for V. (This assumes that you can neglect the density of air (=1.29 kg/m**3) in comparison to the density of water, and, since the bouyant force is only given to 3 figures, this seems to be the case.)
 
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