Question

Water of density 1000 kg/m3 falls without splashing at a rate of 0.373 L/s from a height of 40.5 m into a 0.64 kg bucket on a scale. If the bucket is originally empty, what does the scale read 2.93 s after the first drop contacts the bucket? The acceleration due to gravity is 9.8 m/s 2 .

Answer 1

F_scale = 20.18 N

Step-by-step explanation:

The scale reading corresponds to two factors, the first the weight of the water in the container and the second the force of the liquid that is falling at the moment of reading.

* Let's find the amount of liquid in the container for a time of t = 2.93 s

Let's use a direct proportion rule. If 0.373 l falls in one second at t = 2.93 s, how many liters are there

V_{water} = 2.93 s (0.373 l / 1s) = 1.09 l

V_{water} = 1.09 10⁻³ m³

the amount of water is

ρ = m / V

m = ρ V

m = 1000 1.09 10⁻³

m = 1.09 kg

so the weight of the liquid in the container for this time is

W = mg

W = 1.09 9.8

W = 10.68 N

* Let's look for the force of the falling jet

Let's use Bernoulli's equation, where the subscript 1 is for the container and the subscript 2 is for the water at a height h

P₁ + 1/2 ρ g v₁² + ρ g y₁ = P₂ + 1/2 ρ g v₂² + ρ g y₂

In this case, the water falls freely, so the external pressure is atmospheric.

P₂ = P_{atm}

since they indicate that the water falls, we assume that its initial velocity is zero v₂ = 0

let's use kinematics to find the speed of a drop when it reaches the container y = 0

v² = v₀² - 2 g (y-y₀)

v =
`√(0 -2 g ( 0-y_o))`

let's calculate

v = √(2 9.8 40.5)

v = 28.17 m / s

this is the speed in the container v₁ = 28.17 m / s

the height from where it falls is y₂ = 40.5 and reaches the container y₁ = 0

we substitute in Bernoulli's equation

P₁ +1/2 ρ g v₁² + 0 = P_{atm} + 0 + ρ g y₂

P₁ + ½ ρ g v₁² = P_{atm} + ρ g y₂

P₁ = P_{atm} + ρ g y₂ - ½ ρ g v₁²

P₁ = 1 10⁵ + 1000 9.8 40.5 - ½ 1000 28.17²

P₁ = 1 10⁵ + 3.97 10⁵ - 3.69 10⁵

P₁ = 1.28 10⁵ Pa

The definition of Pressure is

P = F / A

F = P A

We must suppose a time to carry out the reading suppose an average time of the modern equipment t = 0.1 s, in this time how much is now arriving

m₂ = 0.373 0.2 = 0.0746 l = 0.0746 10⁻³ m³

the volume is V = A l

if the length of l = 1 m

A = 0.0746 10⁻³ m³ = 7.45 10⁻⁵ m²

the force of this jet is

F = P A

F = 1.28 10⁵ 7.46 10⁻⁵

F = 9.5 N

with these data let's use the equilibrium equation

F_ scale -W - F = 0

F_scale = W + F

F_scale = 10.682 + 9.5

F_scale = 20.18 N