How Much Volume Does an Iron Anchor Displace in Water?

  • Thread starter Thread starter A_lilah
  • Start date Start date
  • Tags Tags
    Buoyancy Pressure
AI Thread Summary
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.
A_lilah
Messages
52
Reaction score
0

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
 
Physics news on Phys.org
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.)
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Buoyancy: Cube of Iron - solved this correctly?
Replies
1
Views
3K
Does Throwing an Object Overboard Affect Pond Water Level?
Replies
2
Views
2K
What is the thickness of the wall of a hollow iron sphere submerged in water?
Replies
3
Views
2K
Will a Sealed Steel Container Float in the Ocean?
Replies
12
Views
1K
Finding the density of a block in two layers of liquids
Replies
6
Views
2K
Buoyancy force when a ramp is under water
Replies
15
Views
2K
Volume of object that weighs 200N more in air than in water
Replies
2
Views
2K
Calculate the volume of 1.00 mol of liquid water
Replies
11
Views
3K
Volume Change by Pressure: How Does Pressure Affect the Volume of an Object?
Replies
15
Views
2K
How much sea water is needed to keep a submarine still at 30 meters depth?
Replies
7
Views
3K

Hot Threads

Recent Insights

Back
Top