Object Weight given only difference in weights in and out of waterby missnola2a Tags: difference, object, water, weight, weights 

#1
Jul2709, 12:39 AM

P: 13

1. The problem statement, all variables and given/known data
A chunk of carbon steel with density, ρ = 7.84 g/cm3 , is completely submerged in fresh water. The chuck of steel weighs 39 N more in air than in water. Please answer the following: (a) Find the buoyant force acting on the chuck of steel. N (b) Find the the volume of the chuck of steel. m3 (c) What is the mass of the chuck of steel? 2. Relevant equations Fw(air)=Fw(water) + 39 N d=m/v Fb=gp(f)V BUT I dont know how to find the volume. density of air is 1.16 kg/m3 density of water is 1.0x10^3 3. The attempt at a solution once I get the VOLUME I can find the relative weights by using rho*g*v and Fw(water)+ 39N = Fw(air) 



#2
Jul2709, 12:45 AM

P: 295

The buoyant force (upward direction) in the air is less than in the water, therefore
Fw(air)=Fw(water) 39 N And you know that (mgFw(air))/g = m 



#3
Jul2709, 12:46 AM

HW Helper
P: 4,439

Weight of the displaced water = 39 N = m*g = density of water*volume*g.




#4
Jul2709, 01:04 AM

P: 13

Object Weight given only difference in weights in and out of water
ok, a little hazy. Fw(water) is 39N=dens water*v*g? than V = .00398
if so, Fw(water)39N=Fw (air) than Fw(air) is /38.96?? 



#5
Jul2709, 01:05 AM

P: 13

38.96 (which cant be right) *rather than /38.96




#6
Jul2709, 01:11 AM

HW Helper
P: 4,439





#7
Jul2709, 01:20 AM

P: 13

OHH,...
so what you are saying is that I use previously acquired volume of .00398 in this eq. I am sorry if I am frustrating. I mean well! OH no... do I convert 7.84 g/cm^3 to kg/m^3 ?? if so 7.84 g/cm^3 (100cm*100cm*100cm) * 1kg = 7840 kg/m^3 (1m^3) 1000 G Fw(air) = 7.84 g/cm3 OR Fw (air) = 7840 kg/m^3 * .00398 m^3 * 9.81 m/s is that right? then to get weight in water I subgract (39N/9.81) ? 



#8
Jul2709, 01:33 AM

HW Helper
P: 4,439

then to get weight in water I subgract (39N/9.81) ?
No. The weight in water = F(air)  39 N. 


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