How Does Acceleration Impact Stress and Extension in a Steel Rope?

  • Thread starter Thread starter sirhubert
  • Start date Start date
  • Tags Tags
    Extension Stress
AI Thread Summary
The discussion focuses on calculating the stress and extension in a steel rope supporting a 5-tonne load being lowered with an acceleration of 1.4 m/s². The initial calculations for stress used the weight of the load but did not account for the net force due to acceleration, leading to confusion. Participants clarified that the correct approach involves using the net force equation, where tension in the rope must balance the weight of the load minus the force due to acceleration. The correct tension is derived from the equation mg - T = ma, emphasizing the need to consider both gravitational and accelerating forces. Overall, understanding the role of net acceleration is crucial for accurate calculations in this scenario.
sirhubert
Messages
5
Reaction score
0

Homework Statement


A winding engine lowers a load of 5 tonne at the rate of 1.4m/s2 for a distance of 10m. Assuming that the load is suspended from a steel rope 12mm in diameter calculate the stress and the extension in the rope after traveling this distance. Assume Esteel=200 GN/m2 and g=10m/s2

Homework Equations



I've used the following equations to base my answer around:-
Stress=Force/Area
Strain=extension/original length
E=Stress/Strain

The Attempt at a Solution


My attempt to answer this is as follows:
A=3.142 * (6)^2
= 113.1mm2
= 113.1*10^-6 m2

F=5*1000*g where g=10m/s2
= 5*10^4

S=F/A = (5*10^4)/(113.1*10^-6)
Stress=442 MN/m2

using:- E=stress/strain
E= 200 GN/m2
strain=(442*10^6)/(200*10^9)
strain= 2.21*10^-3

My main query is where on Earth does the 1.4m/s2 come into the answer and am I anywhere near the correct solution? This has been bugging me for quite some time now so any help would be much appreciated :smile:
 
Last edited:
Physics news on Phys.org
sirhubert said:
My main query is where on Earth does the 1.4m/s2 come into the answer and am I anywhere near the correct solution?
Remember that 1.4m.s-2 is an acceleration.
 
Thank you for the advice what a complete idiot to miss out something so basic :-)

So I need to use Force=mass x acceleration to determine the actual 'downward pull', for want of a better word, of the load.

In that case if I have (5*10^4) * 1.4 the force would be 7 *10 ^4.

Would the rest of my calculations be correct if I plugged this value in?

Cheers
 
Last edited:
sirhubert said:
Thank you for the advice.

So I need to use Force=mass x acceleration to determine the actual 'downward pull', for want of a better word, of the load.

In that case if I have (5*10^4) * 1.4 the force would be 7 *10 ^4.

Would the rest of my calculations be correct if I plugged this value in?

Cheers
Careful! There are two forces acting on the mass, tension acting upwards and weight acting downwards, causing the net acceleration. Therefore;

mg - T =ma

Where the tension is the force acting on the steel cable. Does that make sense?

Also note that one tonne = 1x103kg
 
I do apologise for being such a numpty it's been far too many years since I last attempted anything like this so you'll have to bear with me.

Right, I presume what you're saying is mass x gravity - T = mass x acceleration
In that case, 5000x10 - T = 5000x8.6
T = 7000

That's presuming an upward acceleration of 8.6 m/s2 opposing gravity at 10m/s2 (net acceleration 1.4m/s2).

w.r.t conversion of tonnes I was trying to change the weight into Newtons, hence 1x10^4 and I think by doing so have completely confused myself. Sorry!
 
sirhubert said:
I do apologise for being such a numpty it's been far too many years since I last attempted anything like this so you'll have to bear with me.
No problem, that's why we're here :wink:
sirhubert said:
Right, I presume what you're saying is mass x gravity - T = mass x acceleration
Good
sirhubert said:
In that case, 5000x10 - T = 5000x8.6
T = 7000

That's presuming an upward acceleration of 8.6 m/s2 opposing gravity at 10m/s2 (net acceleration 1.4m/s2).
Your almost there, but the acceleration in the equation, is the net acceleration i.e 1.4m/s2. Do you understand why?
 
Hmmmm not sure! I may be on completely the wrong track here but these are my thoughts... if the object was accelerating at 10m/s2 then it would have no force against it so tension in the cable would be zero. However, if the object was stationary it would be working against the full gravitational pull so tension would be m*g. As our object is accelerating at 1.4m/s it has m*1.4 less amount of force needed to hold it (if that makes any sense) so the actual Tension would need to take that into account => mg - ma = Tension ?
 
So, you are proposing that the acceleration is the difference between the acceleration due to gravity and the actual acceleration? (just making sure I've read your post correctly)
 
Last edited:
Yes, I think :-) And your explanation is a hell of a lot simpler!
 
Last edited:
  • #10
why are you taking the acceleration to be 8.6 m/s^2.
Why is that so? That is wrong.

Using Newtons second law.
f=ma. f=net force acting on the body
the two forces acting on the object are its weight and tension in the rope.
so f=mg-T

hence, mg-T=ma
the net acceration given is 1.4 m?s^2

The tension force is there in the rope because the object is accelerating otherwise there would be no tension in the rope. i.e if a=0 m/s^2
 
Last edited:
Back
Top