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

AI Thread Summary
When a lift moves at constant speed, the tension in the cable equals the weight of the lift, calculated as T = mg, resulting in a tension of 10,500 Newtons. The discussion raises a question about whether the same tension equation, T - mg = ma, applies when the lift is moving downwards with deceleration. It is noted that while considering the direction of forces, tension and gravitational force should be treated as vectors. The importance of reasoning through problems independently is emphasized, highlighting that simply seeking confirmation does not constitute effective learning. Overall, understanding the dynamics of tension in both upward and downward movements is crucial for accurate problem-solving.
chwala
Gold Member
Messages
2,825
Reaction score
413
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).
 
Rather than saying T - mg, I would consider them as vectors, since g is down it is negative. Then you have: ma = T+ mg
 
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.
 
  • Like
Likes scottdave, MatinSAR and PeroK
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top