Application of the Newton’s law of the motion in an elevator – problems and solutions

1. A 50-kg person in an elevator. Ускорение силы тяжести = 10 м / с². Обозначить нормальная сила exerted on the object by the elevator, if :

(a) the elevator is at rest

(b) the elevator is moving downward at a постоянная скорость

(c) elevator accelerated upward at a постоянное ускорение 5 /с²

(d) elevator accelerated downward at a constant 5 m/s²

(e) elevator in a свободное падение

Решение

Известно:

Человек масса (м) = 50 кг

Ускорение свободного падения (g) = 10 м/с²²

Вес (w) = mg = (50)(10) = 500 Ньютона

Разыскивается: Нормальная сила (Н)

решение:

(a) the elevator is at rest

The elevator is at rest so there is no acceleration (a = 0)

We choose the upward direction in the positive direction and the downward direction in the negative direction.

ΣF = ма

Н – в = 0

Н = w

Н = 500 Ньютонов

(b) the elevator is moving downward at a constant velocity

Constant velocity so there is no acceleration (a = 0)

We choose the upward direction in the positive direction and the downward direction in the negative direction.

ΣF = ма

Н – в = 0

Н = w

Н = 500 Ньютонов

(c) elevator accelerated upward at a constant 5 m/s²

The direction of the acceleration is upward, so we choose the positive direction as up.

N – w = m a

N = w + m a

N = 500 + (50)(5)

N = 500 + 250

Н = 750 Ньютонов

The person feels the floor pushing up harder than when the elevator is stationary or moving with a constant velocity.

If the person stands on a scale, the scale reads the magnitude of the downward force exerted by the person on the scale. By Newton’s third law, this equals the magnitude of the upward normal force exerted by the scale on the person.

(d) elevator accelerated downward at a constant 5 m/s²

The direction of the acceleration is downward, so we choose the positive direction as down.

w – N = m a

N = w – m a

N = 500 – (50)(5)

N = 500 – 250

Н = 250 Ньютонов

The person’s weight is 250 N, less than actual weight w = 500 N.

(e) elevator in a free fall

Free fall means the elevator’s acceleration is the same as the acceleration due to gravity. The magnitude of the acceleration due to gravity is 9,8 m/s², it’s direction is downward toward the center of the Earth. The speed increases linearly in time by 9,8 m/s during each second.

The direction of the acceleration is downward, so we choose the positive direction as down.

w – N = m a

N = w – m a

N = 500 – (50)(10)

N = 500 – 500

N = 0

2. Determine tension in an elevator cable. Elevator’s mass = 2000 kg.

(a) elevator is at rest

(B) elevator accelerated downward at a constant 5 m/s²

(c) elevator accelerated upward at a constant 5 m/s²

(d) elevator in a free fall

Ускорение свободного падения (g) = 10 м/с²²

Решение

Известно:

Elevator’s mass (m) = 2000 kg

Ускорение свободного падения (g) = 10 м/с²²

weight (w) = m g = (2000)(10) = 20,000 Newton

В розыске: The tension force (T)

решение:

(a) elevator is at rest

лифт is at rest so there is no acceleration (a = 0)

We choose the upward direction as the positive direction and the downward direction as the negative direction.

ΣF = ма

T – w = 0

T = w

T = 20,000 Ньютонов

Tension in cable (T) = elevator’s weight (w) = 20,000 Newton

(b) elevator accelerated downward at a constant 5 m/s²

The direction of the acceleration is downward, so we choose the positive direction as down.

w – T = m a

T = w – m a

T = 20,000 – (2000)(5)

Т = 20,000 – 10,000

T = 10,000 Ньютонов

c) elevator accelerated upward at a constant 5 m/s²

The direction of the acceleration is downward, so we choose the positive direction as up.

T – w = m a

T = w + m a

T = 20,000 + (2000)(5)

Т = 20,000 + 10,000

T = 30,000 Ньютонов

(d) elevator in a free fall

The direction of the acceleration is downward, so we choose the positive direction as down.

w – T = m a

T = w – m a

T = 20,000 – (2000)(10)

Т = 20,000 – 20,000

T = 0

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1. Масса и вес
2. Нормальная сила
3. Второй закон движения Ньютона
4. Сила трения
5. Движение по горизонтальной поверхности без силы трения.
6. The motion of two bodies with the same acceleration on rough horizontal surface with friction force
7. Движение по наклонной плоскости без силы трения.
8. Движение по шероховатой наклонной плоскости с учетом силы трения.
9. Движение в лифте
10. Движение тел обеспечивается с помощью тросов и блоков.
11. Два тела с одинаковой величиной ускорения
12. Округление пологой кривой – динамика кругового движения
13. Прохождение виражного поворота – динамика кругового движения
14. Равномерное движение по горизонтальной окружности
15. Центростремительная сила при равномерном круговом движении