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

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SUMMARY

The tension in the cable of a lift moving at constant speed is calculated using the equation T - mg = 0, resulting in T = 10,500 Newtons for a mass of 1050 kg. When the lift is moving downwards with deceleration, the same equation T - mg = ma applies, but the direction of acceleration must be considered. The discussion emphasizes the importance of understanding the reasoning behind the equations rather than merely seeking confirmation from others.

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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.
 
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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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