A sphere of mass of m = 1.69 kg is first placed directly on the plate of an electronic scale. The scale shows 16.58 N as the weight of the object. A large beaker containing a liquid with a density of 783 kg/m^3 is then placed on the scale and the scale is tared (i.e. zeroed to this weight as a reference point, it has a reading of zero with the beaker on the scale). The sphere, hung by a thin string, is lowered into the liquid and submerges below the surface. The scale reads 2.19 N with the sphere not touching the beaker. Calculate the density of the sphere.
My work so far:
F_buoyant = 16.58N - 2.19N = 14.39
F_buoyant = (Rho_liquid)(g)(V_liquid displaced)
14.39N = (783kg/m^3)(9.81m/s^2)(V_liquid displaced)
V_liquid displaced = 0.001873398 m^3 = 0.00187 m^3
So the density of the sphere:
rho_sphere = m_sphere/V_sphere
V_sphere = V_liquid displaced = 0.00187 m^3
rho_sphere = 1.69kg/0.00187m^3
= 902.1040911 kg/m^3
= 902 kg/m^3
... which is what I got for the final answer but apparently it is incorrect. If there is anything, what did I do wrong?
The buoyant force is the force up on the sphere exerted by the fluid and down on the fluid by the sphere Nothing is accelerating. The string is still holding the rest of the sphere weight (most of it). Therefore 2.19 N is the buoyant force = rho fluid *g * volume of sphere
g here = 16.58/1.69 = 9.81 m/s^2 so I guess we are on earth.
In other words use 2.19 not 14.39
Your calculations seem correct based on the information provided. However, there is another factor that needs to be considered when calculating the density of the sphere. Since the sphere is hanging from a string and submerged in the liquid, there is tension in the string due to the buoyant force acting on the sphere.
To account for the tension in the string, you need to subtract it from the apparent weight of the sphere. The apparent weight can be calculated as the weight of the sphere out of the liquid minus the buoyant force:
Apparent weight = Weight out of liquid - Buoyant force
The weight out of the liquid can be calculated by multiplying the mass of the sphere (m_sphere) by the acceleration due to gravity (g):
Weight out of liquid = m_sphere * g
Substituting the values given:
Weight out of liquid = 1.69 kg * 9.81 m/s^2 = 16.5889 N
Now, subtracting the buoyant force from the apparent weight:
Apparent weight = 16.5889 N - 14.39 N = 2.1989 N (approximated to 2.20 N)
Next, you can calculate the tension in the string by subtracting the apparent weight from the weight of the liquid displaced by the sphere:
Tension in string = Weight of liquid displaced - Apparent weight
Weight of liquid displaced can be calculated using the density of the liquid (ρ_liquid) and the volume of the liquid displaced (V_liquid displaced):
Weight of liquid displaced = ρ_liquid * g * V_liquid displaced
Substituting the values given:
Weight of liquid displaced = 783 kg/m^3 * 9.81 m/s^2 * 0.00187 m^3 = 14.3949 N
Finally, calculating the tension in the string:
Tension in string = 14.3949 N - 2.1989 N = 12.196 N (approximated to 12.20 N)
Now, to calculate the density of the sphere, you need to account for the tension in the string. The net force acting on the sphere is the tension in the string minus the buoyant force:
Net force = Tension in string - Buoyant force
Substituting the values:
Net force = 12.196 N - 14.39 N = -2.194 N (approximated to -2.19 N)
Since the net force is negative, this means the buoyant force is greater than the tension in the string. In this case, the buoyant force is partially supporting the weight of the sphere.
To find the density of the sphere, you can use the net force equation:
Net force = (Density of sphere * g * Volume of sphere) - (Density of liquid * g * Volume of liquid displaced)
Since the sphere is completely submerged, the volume of the sphere is equal to the volume of the liquid displaced:
Volume of sphere = Volume of liquid displaced = 0.00187 m^3
Rearranging the equation:
Density of sphere = (Net force / (g * Volume of sphere)) + Density of liquid
Using the given values:
Density of sphere = (-2.194 N / (9.81 m/s^2 * 0.00187 m^3)) + 783 kg/m^3 ≈ 780.62 kg/m^3
Therefore, the correct density of the sphere is approximately 780.62 kg/m^3.