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

1. A 50-kg person in an elevator. Gravitatsiya tufayli tezlanish = 10 m/s²Aniqlang normal kuch exerted on the object by the elevator, if :

(a) the elevator is at rest

(b) the elevator is moving downward at a doimiy tezlik

(c) elevator accelerated upward at a doimiy tezlanish 5 /s²

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

(e) elevator in a erkin tushish

qaror

Ma'lum:

Person’s massa (m) = 50 kg

Gravitatsiya tufayli tezlanish (g) = 10 m/s²

vazn (w) = mg = (50)(10) = 500 Nyuton

Kerakli: Normal kuch (N)

yechim:

(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 = ma

N – w = 0

N = w

N = 500 Nyuton

(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 = ma

N – w = 0

N = w

N = 500 Nyuton

(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

N = 750 Nyuton

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

N = 250 Nyuton

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

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

(d) elevator in a free fall

Gravitatsiya tufayli tezlanish (g) = 10 m/s²

qaror

Ma'lum:

Elevator’s mass (m) = 2000 kg

Og'irlik tezlanishi (g) = 10 m/s²

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

Qidirilgan: The tension force (T)

yechim:

(a) elevator is at rest

lift 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 = ma

T – w = 0

T = w

T = 20,000 Newton

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)

T = 20,000 – 10,000

T = 10,000 Newton

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)

T = 20,000 + 10,000

T = 30,000 Newton

(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)

T = 20,000 – 20,000

T = 0

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1. Massa va vazn
2. Oddiy kuch
3. Nyutonning ikkinchi harakat qonuni
4. Ishqalanish kuchi
5. Ishqalanish kuchisiz gorizontal sirt ustida harakat
6. The motion of two bodies with the same acceleration on rough horizontal surface with friction force
7. Ishqalanish kuchisiz qiyalik tekislikda harakat
8. Ishqalanish kuchi bilan qo'pol qiyalik tekislikda harakat
9. Liftda harakatlanish
10. Jismlarning harakati arqonlar va kasnaklar orqali bog'langan
11. Tezlanish kattaligi bir xil bo'lgan ikkita jism
12. Yassi egri chiziqni yaxlitlash – aylana harakatining dinamikasi
13. Egri chiziqni yaxlitlash – aylana harakatining dinamikasi
14. Gorizontal aylana bo'ylab bir tekis harakat
15. Bir tekis aylana harakatida markazdan qochiruvchi kuch