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

1. A 50-kg person in an elevator. Ubrzanje zbog gravitacije = 10 m / s²Odredite normalna sila exerted on the object by the elevator, if :

(a) the elevator is at rest

(b) the elevator is moving downward at a konstantna brzina

(c) elevator accelerated upward at a konstantno ubrzanje 5 /s²

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

(e) elevator in a slobodan pad

Riješenje

Poznato:

Person’s masa (m) = 50 kg

Ubrzanje zbog gravitacije (g) = 10 m/s²

Težina (w) = mg = (50)(10) = 500 Newtona

Htjela: The normal force (N)

rješenje:

(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

S – š = 0

N = w

N = 500 Newtona

(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

S – š = 0

N = w

N = 500 Newtona

(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 Newtona

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 = ma

N = w – ma

N = 500 – (50)(5)

N = 500 – 250

N = 250 Newtona

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 = ma

N = w – ma

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

Ubrzanje zbog gravitacije (g) = 10 m/s²

Riješenje

Poznato:

Elevator’s mass (m) = 2000 kg

Ubrzanje gravitacije (g) = 10 m/s²

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

Traži se: The tension force (T)

rješenje:

(a) elevator is at rest

dizalo 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 Newtona

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 Newtona

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 = mA

T = w + m a

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

T = 20,000 + 10,000

T = 30,000 Newtona

(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

[wpdm_package id='482']

1. Masa i težina
2. normalna snaga
3. Newtonov drugi zakon gibanja
4. Sila trenja
5. Gibanje po horizontalnoj površini bez sile trenja
6. The motion of two bodies with the same acceleration on rough horizontal surface with friction force
7. Gibanje po nagnutoj ravnini bez sile trenja
8. Gibanje po hrapavoj nagnutoj ravnini sa silom trenja
9. Kretanje u liftu
10. Gibanje tijela povezano je užetima i remenicama
11. Dva tijela s istom brzinom ubrzanja
12. Zaokruživanje ravne krivulje – dinamika kružnog gibanja
13. Zaokruživanje nagnute krivulje – dinamika kružnog gibanja
14. Jednoliko gibanje u horizontalnom krugu
15. Centripetalna sila u jednolikoj kružnoj kretnji