Force and motion in Newton's laws for elevator

In summary, when an elevator cab and its load, with a combined mass of 1600 kg, is brought to rest with constant acceleration in a distance of 42m while originally moving downward at 12 m/s, the tension in the supporting cable is found to be 18400 N. There may be variations in the solution, such as using the magnitude of the forces instead of their signs, but the final result should be the same.
  • #1
theunloved
43
1
An elevator cab and its load have a combined mass of 1600 kg. Find the tension in the supporting cable when the cab, originally moving downward at 12 m/s, is brought to rest with constant acceleration in a distance of 42m.

Taking downwards as positive
My solution:
F net = mg - T = ma
T = m(g - a) (1)

V^2 - Vi^2 = 2a (y - yi)
a = (-12)^2 / 2(42) = -1.7 m/s^2

(1) -----> T = m (g - a) = 1600 (9.8 + 1.7) = 18400 N

Did I do it right ? because my friend got the different answer from me, he basically got T = m (g + a)...
 
Physics news on Phys.org
  • #2
theunloved said:
An elevator cab and its load have a combined mass of 1600 kg. Find the tension in the supporting cable when the cab, originally moving downward at 12 m/s, is brought to rest with constant acceleration in a distance of 42m.

Taking downwards as positive
My solution:
F net = mg - T = ma
T = m(g - a) (1)

V^2 - Vi^2 = 2a (y - yi)
a = (-12)^2 / 2(42) = -1.7 m/s^2

(1) -----> T = m (g - a) = 1600 (9.8 + 1.7) = 18400 N

Did I do it right ? because my friend got the different answer from me, he basically got T = m (g + a)...

I have seen T = m(g + a) commonly used when the question is find the "magnitude of the force on the lift". In that case, the fact that g is down and a is up, so one of them should be negative, is sort of ignored because we expect that the two magnitudes are simply added - giving the same answer as yours.

Analysis of a ferris wheel is often written reaction force at top = mg - Fc and reaction force at bottom = mg + Fc.
This is an implication that we are talking magnitudes only.


Note that in your line of working that I coloured red, you have a - sign suddenly appearing on the right. Perhaps you just forgot to type is before the 42? ie distance traveled was -42 since the lift was traveling down.
 
  • #3
PeterO said:
I have seen T = m(g + a) commonly used when the question is find the "magnitude of the force on the lift". In that case, the fact that g is down and a is up, so one of them should be negative, is sort of ignored because we expect that the two magnitudes are simply added - giving the same answer as yours.

Analysis of a ferris wheel is often written reaction force at top = mg - Fc and reaction force at bottom = mg + Fc.
This is an implication that we are talking magnitudes only.


Note that in your line of working that I coloured red, you have a - sign suddenly appearing on the right. Perhaps you just forgot to type is before the 42? ie distance traveled was -42 since the lift was traveling down.

Thanks, it should be like this

V^2 - Vi^2 = 2a (y - yi)
sice V = 0, so we have
- Vi^2 = 2a (y - yi)
a = -(Vi^2) / 2(y - yi)
a = -(12^2) / 2(42) since am taking downwards as positive, so y > yi.
 

Related to Force and motion in Newton's laws for elevator

What are Newton's laws of motion?

Newton's laws of motion are three fundamental physical laws that describe the behavior of objects in motion. The first law states that an object will remain at rest or in motion at a constant velocity unless acted upon by an external force. The second law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The third law states that for every action, there is an equal and opposite reaction.

How do Newton's laws apply to elevators?

Newton's laws apply to elevators in the sense that they describe the forces acting on an elevator and how it responds to those forces. When the elevator is at rest, the first law states that it will remain at rest unless acted upon by an external force, such as the pull of gravity or the tension in the elevator cable. When the elevator is in motion, the second law applies as the net force on the elevator is equal to the mass of the elevator times its acceleration. The third law also applies as the elevator exerts an equal and opposite force on the building or support structure.

What role does gravity play in elevator motion?

Gravity plays a significant role in elevator motion as it is the force that pulls the elevator and its occupants towards the center of the Earth. Without gravity, there would be no weight or downward force on the elevator, and it would not be able to move. The force of gravity also affects the acceleration of the elevator, as stated in Newton's second law.

Why do elevators have maximum weight limits?

Elevators have maximum weight limits because of Newton's second law of motion. As the weight of the elevator and its occupants increases, so does the force required to accelerate it. If the weight exceeds the maximum limit, the elevator may not be able to accelerate properly, and it could potentially cause damage to the elevator or its supporting structure.

How do elevators use counterweights?

Elevators use counterweights to help balance the weight of the elevator car and its occupants. The counterweight is typically placed on the opposite side of the elevator car and is equal to the weight of the car and about 40-50% of its maximum capacity. This helps reduce the overall force needed to accelerate the elevator and makes it more energy-efficient. It also helps prevent the elevator from becoming imbalanced and potentially causing malfunctions or accidents.

Similar threads

  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
19
Views
865
  • Introductory Physics Homework Help
Replies
16
Views
469
  • Introductory Physics Homework Help
Replies
17
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top