Buoyancy Lab
4/23/2013
Lab Partner: Norma
The mass obtained is considered to be the “apparent weight” when it is in a fluid (as opposed to its actual
weight when it is suspended in the air), so the force on the block is:
ΣF=ma
Fscale+FB-W=0
FB=W-Fscale
FB= Wair-Wfl
we solve for the buoyant force by using the last equation. In the last line we change W to Wair, which is the object’s actual weight (NOT the weight of the air).
weight when it is suspended in the air), so the force on the block is:
ΣF=ma
Fscale+FB-W=0
FB=W-Fscale
FB= Wair-Wfl
we solve for the buoyant force by using the last equation. In the last line we change W to Wair, which is the object’s actual weight (NOT the weight of the air).
Mass of object with the string: 69.1 gr
Mass of object when submerged in water: 59.9 gr
Δ mass= 69.1-59.9= 9.2 gr
Volume= 9 ml= 9 cm^3
B= Δmg= ρVg
B= Δm= ρV
ρ=Δm/V
9.2 gr/9 cm^3=1.022 gr/cm^3= 1022 kg / m3
We got the density of water to be 1022 kg/m^3, which is close to the real value (1000 kg/m^3).
Mass of object when submerged in water: 59.9 gr
Δ mass= 69.1-59.9= 9.2 gr
Volume= 9 ml= 9 cm^3
B= Δmg= ρVg
B= Δm= ρV
ρ=Δm/V
9.2 gr/9 cm^3=1.022 gr/cm^3= 1022 kg / m3
We got the density of water to be 1022 kg/m^3, which is close to the real value (1000 kg/m^3).
Reflection: This lab was helpful to me, and helped me understand the Buoyancy force. Our calculated value for the density of the water was close to the true value, but it was not exact because we did not take the tension between the string and the object into account.