Question

Тело движется вверх по наклонной плоскости (угол альфа =30 градусов) найти ускорение и силу трения, если сила тяги 10кН коэф трения 0.2, масса 1 т,

Answer 1

Ответ:

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Answer 2

Problem Statement

We are given the following information: - The body is moving upwards on an inclined plane with an angle of α = 30 degrees. - The force of traction is 10 kN. - The coefficient of friction is 0.2. - The mass of the body is 1 ton.

We need to find the acceleration and the force of friction.

Solution

To solve this problem, we can break down the forces acting on the body into components parallel and perpendicular to the inclined plane.

Let's start by finding the acceleration of the body.

Step 1: Finding the gravitational force The gravitational force acting on the body can be calculated using the formula: F_gravity = m * g, where m is the mass of the body and g is the acceleration due to gravity.

Given that the mass of the body is 1 ton (1000 kg) and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the gravitational force: F_gravity = 1000 kg * 9.8 m/s^2.

Step 2: Finding the component of the gravitational force parallel to the inclined plane The component of the gravitational force parallel to the inclined plane can be calculated using the formula: F_parallel = F_gravity * sin(α), where α is the angle of the inclined plane.

Given that α = 30 degrees, we can calculate the component of the gravitational force parallel to the inclined plane: F_parallel = F_gravity * sin(30 degrees).

Step 3: Finding the force of traction The force of traction is given as 10 kN (10,000 N).

Step 4: Finding the net force parallel to the inclined plane The net force parallel to the inclined plane can be calculated by subtracting the force of traction from the component of the gravitational force parallel to the inclined plane: F_net_parallel = F_parallel - F_traction.

Step 5: Finding the acceleration The acceleration can be calculated using Newton's second law: F_net_parallel = m * a, where a is the acceleration.

Given that the mass of the body is 1000 kg and the net force parallel to the inclined plane is F_net_parallel, we can calculate the acceleration: a = F_net_parallel / m.

Step 6: Finding the force of friction The force of friction can be calculated using the formula: F_friction = μ * F_normal, where μ is the coefficient of friction and F_normal is the normal force.

The normal force can be calculated using the formula: F_normal = F_gravity * cos(α).

Given that the coefficient of friction is 0.2, we can calculate the force of friction: F_friction = 0.2 * F_normal.

Calculation

Let's plug in the values and calculate the acceleration and the force of friction.

Step 1: Finding the gravitational force F_gravity = 1000 kg * 9.8 m/s^2 = 9800 N.

Step 2: Finding the component of the gravitational force parallel to the inclined plane F_parallel = 9800 N * sin(30 degrees).

Step 3: Finding the force of traction F_traction = 10,000 N.

Step 4: Finding the net force parallel to the inclined plane F_net_parallel = F_parallel - F_traction.

Step 5: Finding the acceleration a = F_net_parallel / m.

Step 6: Finding the force of friction F_normal = F_gravity * cos(30 degrees). F_friction = 0.2 * F_normal.

Answer

The acceleration of the body moving upwards on the inclined plane is a m/s^2, and the force of friction is F_friction N.

Please note that the exact values of the acceleration and the force of friction can be calculated by substituting the values into the respective formulas.

Answer 3

Для решения этой задачи мы можем использовать физические законы, связанные с движением тела по наклонной плоскости.

1. Нарисуем силовую диаграмму тела, чтобы проанализировать силы, действующие на него:

- Сила тяжести (F_г): направлена вертикально вниз и равна \(m \cdot g\), где \(m\) - масса тела, \(g\) - ускорение свободного падения (примерно \(9.81 \, м/с^2\)). - Нормальная реакция (N): перпендикулярна наклонной плоскости и равна вертикальной компоненте силы тяжести. - Сила трения (F_тр): действует вдоль наклонной плоскости и противоположно направлению движения тела. - Сила тяги (F_тяги): приложена вдоль наклонной плоскости в направлении движения тела.

Учитывая, что угол наклона плоскости \( \alpha = 30^\circ \), мы можем найти составляющие силы:

- \( F_г = m \cdot g \cdot \cos \alpha \) - \( N = m \cdot g \cdot \sin \alpha \)

Для начала нужно выразить массу тела в килограммах, так как дана в тоннах. \( 1 \, т = 1000 \, кг \).

Масса тела \( m = 1 \, т = 1000 \, кг \)

Ускорение свободного падения \( g \approx 9.81 \, м/с^2 \)

Угол наклона \( \alpha = 30^\circ \)

Сила тяги \( F_тяги = 10 \, кН = 10,000 \, Н \)

Коэффициент трения \( \mu = 0.2 \)

Теперь, найдем \( F_г \) и \( N \):

\( F_г = m \cdot g \cdot \cos \alpha = 1000 \, кг \cdot 9.81 \, м/с^2 \cdot \cos 30^\circ \) \( F_г ≈ 1000 \, кг \cdot 9.81 \, м/с^2 \cdot 0.866 ≈ 8,525 \, Н \)

\( N = m \cdot g \cdot \sin \alpha = 1000 \, кг \cdot 9.81 \, м/с^2 \cdot \sin 30^\circ \) \( N ≈ 1000 \, кг \cdot 9.81 \, м/с^2 \cdot 0.5 ≈ 4,905 \, Н \)

Теперь мы можем рассчитать силу трения \( F_тр \) и ускорение тела:

\( F_тр = \mu \cdot N = 0.2 \cdot 4,905 \, Н = 981 \, Н \)

\( F_{\text{рез}} = F_тяги - F_тр = 10,000 \, Н - 981 \, Н = 9,019 \, Н \)

Далее, ускорение \( a \) можно найти через второй закон Ньютона: \( F_{\text{рез}} = m \cdot a \)

\( a = \frac{F_{\text{рез}}}{m} = \frac{9,019 \, Н}{1000 \, кг} = 9.019 \, м/с^2 \)

Таким образом, ускорение тела составляет примерно \( 9.019 \, м/с^2 \), а сила трения составляет примерно \( 981 \, Н \).