Question

When checking the weight of the solid used in the laboratory experiment by suspending the solid from the spring scale in air per the specifications in Part 1 of the Procedure section of the laboratory assignment, the tension the spring scale registers for the solid is not exactly equal to the magnitude of the weight of the solid. This is due to the buoyant force of the air acting on the solid. The density of air is 1.204kg/m3 at 20°C at 1.000 atmospheres. Note: All processed data must obey the rules of significant figures; consequently, processed data will contain between 2 and 5, inclusive, significant figures A) (3%) Using appropriate values from this problem and from Table 1 of the Data section of the laboratory assignment in addition to an appropriate constant, calculate the buoyant force of the air acting on the solid EquationObject Using the answer from Part section of the laboratory assignment, calculate the tension the spring scale registers Hint: Use the following partial solution to assist in the complete solutiorn. B) (5%) and the appropriate value from Table 1 of the Data when measuring the weight of the solid. aM Note: Enumerate all force terms on the summation side of Newton's Second Law. Equationobject Using the answer from Part B, fully explain the reason the buoyant force due to air acting on the solid can be ignored for the weight calculation (using the sum of the forces from Part B) in this laboratory experiment. EquationObject C) (3%)

Answer 1

Buoyant force of the air acting on the solid FB = (weight of object) + (density of object) times volume of object FB = -9.81 + (1.201 - 6666.67) times 0.00015 FB = -10.81N T + FB = W T = W - FB -9.81 - (-10.81 = 1N Tension force when in air is 9.81N (from table). Total tension produced will be (9.81 + 1) = 10.81N which completes the buoyant force. Hence it can be ignored