Solve PVg Volume-Weight Problem: 28 N Required

  • Thread starter Thread starter Izekid
  • Start date Start date
  • Tags Tags
    Volume Weight
AI Thread Summary
The discussion revolves around calculating the power needed to submerge a bathball with a volume of 3.0 dm³, which is currently floating with 6.0% of its volume submerged. The mass of the bathball is correctly calculated as 0.18 kg. The main issue is determining the force required to push the entire bathball underwater, which is initially miscalculated as 2.946 N instead of the expected 28 N. Participants suggest that the forces acting on the bathball—gravity, the applied force, and buoyancy—must be balanced to find the correct solution. A formula is proposed to relate these forces, emphasizing that the net force should equal zero for equilibrium.
Izekid
Messages
40
Reaction score
0
I have a problem that I can't solve,

A bathball with the volume 3,0dm^3 is floating on water and has 6,0% of it's volume down in the water.

a) Count the bathball's mass , Easy : 0,06*3,0= 1,8hg = 0,18 kg

Now comes the real problem

b) how much power do you need to use to force the whole bathball down under water?

Well for this I though i'd use p*v*g /p (where p is water density)
Water density 1,0 * 10^3 = 1000*0,3m^3*9,82 = 2946 / 1000 = 2,946 N

Which totaly wrong it should be 28 N how do I solve this?
 
Physics news on Phys.org
Draw a diagram. You should have three forces.

"p*v*g /p"
How did you get this?
 
Eh?

Don't know what you're suggesting please understand I'm no science freak :cry: :cry: :bugeye: :smile: :zzz:

Yeah I have 3 forces gravity is one my power pressing the bathball down is one and the will to go up is the third. But I do not not the force pressing the item down!?
And how do I get to know that ?
Give me a formula or something?
 
There isn't much help I can give without solving the problem. Here goes, though:

Well, the forces should equal zero (for the minimal force).
So:
mg + F = \rho _{water} Vg
 
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...
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...

Similar threads

Back
Top