Force and motion in Newton's laws for elevator

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SUMMARY

The discussion focuses on calculating the tension in the supporting cable of an elevator cab with a combined mass of 1600 kg, which is brought to rest from a downward speed of 12 m/s over a distance of 42 m. The correct formula for tension is derived as T = m(g - a), where 'g' is the acceleration due to gravity (9.8 m/s²) and 'a' is the deceleration (1.7 m/s²). The final calculation yields a tension of 18400 N. The confusion arises from differing interpretations of the acceleration's direction in relation to gravity.

PREREQUISITES
  • Understanding of Newton's Second Law (F = ma)
  • Knowledge of kinematic equations, specifically V² - Vi² = 2a(y - yi)
  • Familiarity with the concepts of tension and forces in vertical motion
  • Basic grasp of gravitational acceleration (9.8 m/s²)
NEXT STEPS
  • Study the application of Newton's laws in different contexts, such as free body diagrams
  • Learn more about kinematic equations and their applications in real-world scenarios
  • Explore the concept of forces in circular motion, particularly in systems like ferris wheels
  • Investigate the effects of acceleration on tension in various mechanical systems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, engineers working with elevator systems, and anyone interested in understanding forces and motion in practical applications.

theunloved
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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)...
 
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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.
 
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.
 

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