# Object Weight given only difference in weights in and out of water

by missnola2a
Tags: difference, object, water, weight, weights
 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)
 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 (mg-Fw(air))/g = m
 HW Helper P: 4,433 Weight of the displaced water = 39 N = m*g = density of water*volume*g.
 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??
 P: 13 -38.96 (which cant be right) *rather than /38.96
HW Helper
P: 4,433
 Quote by missnola2a -38.96 (which cant be right) *rather than /38.96
Fw(in air ) = density of steel*volume*g.
 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) ?
 HW Helper P: 4,433 then to get weight in water I subgract (39N/9.81) ? No. The weight in water = F(air) - 39 N.

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