Find the Tension in the cable when the lift is moving at constant speed

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Homework Help Overview

The discussion revolves around finding the tension in a cable when a lift is moving at constant speed, specifically focusing on the implications of acceleration and deceleration in different scenarios.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the relationship between tension, mass, and gravitational force, questioning how these dynamics change when the lift is moving downwards with deceleration compared to moving upwards with acceleration. There is a discussion about treating tension and gravity as vectors and the application of the equation T - mg = ma in different contexts.

Discussion Status

The discussion is active with participants sharing their reasoning and questioning the assumptions behind the equations used. Some participants express uncertainty about the application of the tension equation in different scenarios, while others emphasize the importance of reasoning through the problem independently.

Contextual Notes

There is a mention of homework rules that encourage showing workings and reasoning rather than simply seeking confirmation of answers. This suggests a focus on understanding the underlying principles rather than just obtaining a solution.

chwala
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Homework Statement
The maximum load that a lift of mass ##600## kg can hold is ##450##kg. Find the tension in the cable when the lift is holding a maximum load and is moving at a constant speed of ##3## m/s
Relevant Equations
Vertical motion - Mechanics.
aaaah just realized the solution after typing here...

... at constant speed, ##a=0##,

therefore

##T-mg = 0##

##T=(1050 × 10) =10, 500## Newtons

or any insight...welcome.



Maybe i should ask...when the lift is moving downwards and there is deceleration then would the Tension be treated in the same manner as moving upwards with acceleration?

Would the equation below apply to both scenario?

##T-mg = ma## ?

Cheers.
 
Physics news on Phys.org
With a reference direction of up is positive, then if it is moving down (negative) a positive acceleration will result in it slowing down (decelerate).
 
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Rather than saying T - mg, I would consider them as vectors, since g is down it is negative. Then you have: ma = T+ mg
 
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chwala said:
Maybe i should ask...when the lift is moving downwards and there is deceleration then would the Tension be treated in the same manner as moving upwards with acceleration?
What do you think? Why?

chwala said:
Would the equation below apply to both scenario?

##T-mg = ma## ?
What do you think? Why?

This approach to "learning" where you guess answers and seek confirmation from others is not learning at all: reasoning things out for yourself is an essential part of learning. That is why we ask you to show your workings.
 
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