Question

9.2 На оба конца легкой тонкой нити помещены шарики из меди и олова. ШapИКИ расположены точки опоры, относительно симметрично горизонтальны и находятся в равновесии с поверхностью воды. Расстояние между между центрами шариков 16 см. В каком направлении и на сколько нужно переместить оловянный шарик, чтобы сохранить равновесие в воздухе? 894 кг/м3, плотность олова - 730 кг/м3, плотность воды (плотность меди 1000 кг/м3) ​

Answer 1

Ответ:

Чтобы понять, как нужно переместить оловянный шарик, чтобы сохранить равновесие в воздухе, давайте воспользуемся принципом Архимеда. Этот принцип гласит, что поддерживающая сила, действующая на погруженное в жидкость тело, равна весу выталкиваемой из этой жидкости тела.

Для начала найдем вес медного шарика (W1) и вес оловянного шарика (W2).

Вес медного шарика:

W1 = объем * плотность * ускорение свободного падения

W1 = (4/3 * π * (0.08 м)³) * 1000 кг/м³ * 9.8 м/с²

W1 ≈ 267.09 Н

Вес оловянного шарика:

W2 = объем * плотность * ускорение свободного падения

W2 = (4/3 * π * (0.08 м)³) * 730 кг/м³ * 9.8 м/с²

W2 ≈ 190.75 Н

Общий вес системы шариков в воздухе:

W_total = W1 + W2

W_total ≈ 457.84 Н

Теперь, чтобы сохранить равновесие в воздухе, нужно, чтобы вес системы шариков был равен поддерживающей силе, действующей на них в воде. Поддерживающая сила в воде равна весу выталкиваемой из воды жидкости и зависит от плотности воды, объема выталкиваемой жидкости и ускорения свободного падения.

W_total = плотность воды * объем выталкиваемой жидкости * ускорение свободного падения

Теперь мы можем решить это уравнение для объема выталкиваемой жидкости (V):

V = W_total / (плотность воды * ускорение свободного падения)

V ≈ 0.0466 м³

Расстояние между центрами шариков равно 16 см, что составляет 0.16 м. Объем выталкиваемой жидкости равен объему, смещенному оловянным шариком. Пусть это смещение будет х метров. Тогда:

V = π * (0.08 м)² * (0.16 м - х)

Теперь мы можем решить это уравнение относительно х:

0.0466 м³ = π * (0.08 м)² * (0.16 м - х)

х ≈ 0.06 м

Таким образом, чтобы сохранить равновесие в воздухе, оловянный шарик должен быть смещен на приблизительно 6 см вверх.

Answer 2

Problem Analysis

We have a system consisting of two balls made of copper and lead, respectively, attached to the ends of a light thread. The balls are symmetrically positioned with respect to the horizontal and are in equilibrium with the surface of the water. We need to determine in which direction and by how much the lead ball needs to be moved to maintain equilibrium in the air.

Solution

To solve this problem, we need to consider the forces acting on the system. The system is in equilibrium when the net force and net torque acting on it are both zero.

Let's analyze the forces acting on the system: 1. Buoyant force on the copper ball: The buoyant force on the copper ball is equal to the weight of the water displaced by the copper ball. Since the copper ball is in equilibrium with the water surface, the buoyant force on it is equal to its weight. 2. Buoyant force on the lead ball: The buoyant force on the lead ball is equal to the weight of the water displaced by the lead ball. Since the lead ball is in equilibrium with the water surface, the buoyant force on it is equal to its weight. 3. Tension in the thread: The tension in the thread is the same throughout its length and acts in the upward direction.

Since the system is in equilibrium, the net force acting on the system is zero. Therefore, the sum of the buoyant forces on the copper and lead balls must be equal to the tension in the thread.

Let's denote the mass of the copper ball as m_copper, the mass of the lead ball as m_lead, and the tension in the thread as T.

From the problem statement, we know that the distance between the centers of the balls is 16 cm. Let's denote this distance as d.

To maintain equilibrium in the air, the lead ball needs to be moved in such a way that the sum of the buoyant forces on the copper and lead balls is equal to the tension in the thread.

The buoyant force on an object can be calculated using the formula:

Buoyant force = (density of fluid) * (volume of fluid displaced) * (acceleration due to gravity)

The volume of fluid displaced by a ball can be calculated using the formula:

Volume of fluid displaced = (4/3) * π * (radius of ball)^3

Let's denote the radius of the copper ball as r_copper and the radius of the lead ball as r_lead.

Using the above formulas, we can write the following equations:

m_copper * g = (density of water) * (4/3) * π * (r_copper)^3 * g

m_lead * g = (density of water) * (4/3) * π * (r_lead)^3 * g

m_copper * g + m_lead * g = T

We can rearrange the equations to solve for the radii of the balls:

r_copper = ((3 * m_copper) / (4 * π * (density of water)))^(1/3)

r_lead = ((3 * m_lead) / (4 * π * (density of water)))^(1/3)

Substituting the values of the densities of copper, lead, and water, we can calculate the radii of the balls.

Once we have the radii of the balls, we can calculate the distance between their centers as d = 2 * r_copper.

To maintain equilibrium in the air, the lead ball needs to be moved in such a way that the distance between the centers of the balls is equal to the calculated value of d.

Calculation

Let's calculate the radii of the balls and the distance between their centers.

Given: Density of water = 1000 kg/m^3 Density of copper = 894 kg/m^3 Density of lead = 730 kg/m^3

We can calculate the radii of the balls using the formulas mentioned above.

r_copper = ((3 * m_copper) / (4 * π * (density of water)))^(1/3) r_lead = ((3 * m_lead) / (4 * π * (density of water)))^(1/3)

Substituting the densities of copper and lead, we can calculate the radii of the balls.

Once we have the radii of the balls, we can calculate the distance between their centers as d = 2 * r_copper.

Finally, we can determine the direction and distance by which the lead ball needs to be moved to maintain equilibrium in the air.

Answer

To determine the direction and distance by which the lead ball needs to be moved to maintain equilibrium in the air, we need to calculate the radii of the balls and the distance between their centers.

Unfortunately, the masses of the copper and lead balls are not provided in the question. Without the masses, we cannot calculate the radii of the balls and the distance between their centers. Please provide the masses of the copper and lead balls so that we can proceed with the calculation.

If you have any further questions, please let me know.