Lab 4 Buoyant Force and Archimedes’ Principle PHYS2091 PDF

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Lab 4 Buoyant Force and Archimedes’ Principle Lab 4 Buoyant Force and Archimedes’ Principle Lab 4 Buoyant Force and Archimedes’ Principle...

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Lab Report IV Buoyant Force and Archimedes’ Principle

Wenzhou Kean University PHYS 2091 W01 Junfeng Wang 1129618 September 29, 2021

Purpose The purpose of this experiment is by applying Archimedes’ principle to: (1) determine the density of the hanging object; (2) determine the density of the given salt solution; Theory When an object is statically hanging on a string, the tension force of the string (T) applied on the object is equal to the objects’ gravitational force (G): T = G = mg (1) When an object is statically hanging on a string while fully immersed in a fluid at the same time, its vertical force equation is: T+F=G (2) The density of a solid object or a liquid solution (ρ) is defined by its mass (m) divide by volume (v): m ρ= V (3) Archimedes’ Principle: When an object is fully immersed in a fluid, the buoyant force (F) is equal to the weight of the fluid it displaces (Gfluid). F = Gfluid = ρ v g (4) Sketch

Hang the object on a tension gauge. Measure its tension respectively when it’s in the air and completely submerged in the fluid. In the first trial, the fluid is water (density is known to be 1000 kg/m3); while in the second trial, change the fluid to salt solution (unknown density) and repeat the procedures.

Data Object ’s Weigh t

Weig ht in Water

Weigh t in Salt Soluti on

Density of Water

1.15N

0.75N

0.7N

1000KG/m^ 3

Object’s Volume (= Displaced Water’s Volume) 0.000040816 m^3

Density of Salt Solution Calculat True ed Valu Result e 1160 1125 3

kg /m

kg /m3

Analysis To get the results in the table above, firstly, the buoyant force F is calculated by formula (2): F1 = G – T1 = 1.15N – 0.75N = 0.4N (In water) F2 = G – T2 = 1.15N – 0.70N = 0.45N (In salt solution) Then, apply Archimedes’ Principle (formula 4) to calculate the object’s volume, which is equal to the volume of water being displaced: Vobject = Vwater =

F1 0.4 N = ρg 9.8 N 1000 kg ×( ) Kg m3

(

)

= 0.000040816 m3

Therefore, the density of the salt solution can be calculated by Archimedes’ Principle because the volume of salt solution displaced is the same as the volume of water displaced in the first trial: salt solution

F2 = = V water g

1.15 N−0.70 N =1125 kg/m 3 −5 3 4.08 ×1 0 m × 9.8 N /kg

Finally, the density of the medal object can be calculated by formula (3): 1.15 N m =2875 kg /m 3 object = = −5 3 V 4.08 ×1 0 m × 9.8 N /kg

In order to find values in table one, the group first converted each of the grams into

% Error

3.01%

kilograms, which is the standard mass measurement unit. Then, we performed the calculations necessary to find the values of the other tables. The calculations are as follow on the next page. To find values in table one, the group first

converted each of the grams into kilograms, which is the standard mass measurement unit. Then, we performed the calculations necessary to find the values of the other tables. The calculations are as follow on the next page.

In order to find values in table one, the group first converted each of the grams into kilograms, which is the standard mass measurement unit. Then, we performed the calculations necessary to find the values of the other tables. The calculations are as

follow on the next page. Results Density of Salt Solution Calculated Result True Value 3 1160 kg /m3 1125 kg /m

|true value−exp value|

%Error=

true value

|1160 kg/m 3−1125 kg/m 3| 3

1160 kg /m

×100 %

% Error 3.01% =

× 100 % ≈ 3.02 %

Possible Source of Error 1. When we moved on from trial 1 to 2, the liquid container and object is not completely dry. That may cause the salt solution sample mixed with a few drops of water, which may lead to a decrease in density. 2. The tension gauge’s dash-board was not fixed, which required a manual calibration. Errors may occur in this process....