You are Here: Home >< Maths

# Why does the maximum tension in the cable occur when it is accelerating? watch

1. Just for part ii) Why does the maximum tension in the cable occur when it is accelerating and not in any of the other phases? It says that in the mark scheme for ii. Many thanks

Attachment 725396725398
Attached Images

2. Well, given that all forces acting on the object are vectors with directions, you know that the forces upwards will counteract the forces downwards. Hence the equation:
T - mg = ma (where mg is the weight of the object and ma is the resultant force in the upwards direction)

notice that this is the equation used in the answer

Rearranging this gives:

T = mg + ma

This means that if the object wasn't accelerating upwards, travelling at a constant speed, the tension would become:

T = mg + m(0) = mg

And if the object was decelerating, the tension would become:

T = mg - deceleration

Since tension is amplified by any force in the same direction, the phase at which it has the maximum tension will be when it has maximum acceleration.
3. (Original post by Maivnek)
Well, given that all forces acting on the object are vectors with directions, you know that the forces upwards will counteract the forces downwards. Hence the equation:
T - mg = ma (where mg is the weight of the object and ma is the resultant force in the upwards direction)

notice that this is the equation used in the answer

Rearranging this gives:

T = mg + ma

This means that if the object wasn't accelerating upwards, travelling at a constant speed, the tension would become:

T = mg + m(0) = mg

And if the object was decelerating, the tension would become:

T = mg - deceleration

Since tension is amplified by any force in the same direction, the phase at which it has the maximum tension will be when it has maximum acceleration.
Thank you very much But what do you mean by T = mg - deceleration ?
4. (Original post by h26)
Thank you very much But what do you mean by T = mg - deceleration ?
Maybe I wrote it in a confusing way, sorry about that. Deceleration is actually just negative acceleration (or acceleration in the opposite direction). So if you have the equation:

T = mg + ma (where a = 0.05 ms^-2)

and replace acceleration with deceleration, you get:

T = mg + m(-a) = mg - ma (where a = 0.1 ms^-2)

What I meant was that for when the object is decelerating, tension is equal its weight minus the value of its deceleration times its mass.
5. (Original post by Maivnek)
Maybe I wrote it in a confusing way, sorry about that. Deceleration is actually just negative acceleration (or acceleration in the opposite direction). So if you have the equation:

T = mg + ma (where a = 0.05 ms^-2)

and replace acceleration with deceleration, you get:

T = mg + m(-a) = mg - ma (where a = 0.1 ms^-2)

What I meant was that for when the object is decelerating, tension is equal its weight minus the value of its deceleration times its mass.
I understand it now Thank you very much for the help! Really appreciate it So Tension is greatest when accelerating, then in the middle when stationary and lowest when decelerating, right?
6. When the acceleration is stationary and its travelling at a constant speed, then yes it will be in the middle out of all three phases.

You got it!
7. (Original post by Maivnek)
When the acceleration is stationary and its travelling at a constant speed, then yes it will be in the middle out of all three phases.

You got it!
Thanks ever so much!
8. (Original post by h26)
Thanks ever so much!
I now realise that whether the object is at rest (0 velocity) or moving at a constant speed, its tension will be the same. So it really didn't matter whether or not you meant stationary motion or stationary acceleration.

All the best!

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: February 16, 2018
The home of Results and Clearing

### 2,920

people online now

### 1,567,000

students helped last year
Today on TSR

### University open days

1. Sheffield Hallam University
Tue, 21 Aug '18
2. Bournemouth University
Wed, 22 Aug '18
3. University of Buckingham
Thu, 23 Aug '18
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams