Tension of a rope with a hanging mass and acceleration

  • #1

Homework Statement


Find the tension of a rope with a hanging mass of 0.5kg that falls with an acceleration of 0.44ms^2


Homework Equations



T = ma
T = ma-mg



The Attempt at a Solution



T = mg = 0.5*9.81 = 4.905N, I would assume that this is correct for a static load on the rope ?

My question is. How does the relationship between gravity and acceleration act on the tension of the rope? Is the correct equation
T = mg-ma ?

Any help would be appreciated
 

Answers and Replies

  • #2
Yes, t = mg - ma. But, for clarity, draw a free-body diagram of the block and sum your forces.
 
  • #3
Thanks,

Tension acts in an upward direction
Gravity acts in an downward direction
Acceleration acts in a downward direction
so i get T-ma-mg = 0
transpose to get T = mg+ma
Or am i looking at acceleration going in the wrong direction ?
 
  • #4
I apologize, my previous statement was incorrect. Since the mass is moving downward, the tension has to support the accelerations of both gravity and of the mass, and thus T = m(g+a).
Your second statement is correct.
 
  • #5
266
2
From an inertial frame of reference ( say, a person standing on the ground) assuming the downward direction to be +ve, ma = mg - T.
 
  • #6
For a person standing on the ground ma would equal m, correct? no acceleration

Transpose ma=mg-T for T
T=mg-ma ?

can someone explain why it is either g-a or g+a. This may be simple but im not getting it
 
  • #7
For a person standing on the ground ma would equal m, correct? no acceleration

Sorry i mean to say that the person on the ground ma would equal mg
 
  • #8
266
2
No I think according to the question, the person on the ground sees it coming down with an acceleration of a=0.44m/s2. So ma=mg -T.
 
  • #9
So we are saying that
ma=mg-T
T= m(g-a)
T= 0.5(9.81-0.44) = 4.685N in an upward direction

Sunil Simha, where did i go wrong in post #3?
 
  • #10
266
2
So we are saying that
ma=mg-T
T= m(g-a)
T= 0.5(9.81-0.44) = 4.685N in an upward direction

Sunil Simha, where did i go wrong in post #3?

I think you took the frame of reference of the falling block. Is that right?

If yes, ma would be a pseudo force in upward direction and thus the equation would be T+ma=mg as the block is in rest in its own frame.

If not, then maybe you took the frame of reference of the ground in which case the force equation would still be the same i.e. ma=mg-T.
 
  • #11
I maybe should have mentioned that the rope is on a flywheel

would it be that acceleration (0.44) is working against gravity as it is slower?
 
  • #12
266
2
I maybe should have mentioned that the rope is on a flywheel

would it be that acceleration (0.44) is working against gravity as it is slower?

Check out my attachment... Is it the correct diagram?

If yes then you can see the force equations.
If no, I'm sorry for the inconvenience:redface:
 

Attachments

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  • #13
Not an inconvenience at all mate, thank you for all your patience and help.
I am trying to understand why the equation is the way it is.

Yes my diagram is similar but without the rope on the left hand side being pulled
to re cap
T= up
mg= down
ma=down
thus T-mg-ma = 0 correct?
 
  • #14
The mass is allowed to fall at a rate of 0.44ms^-2
 
  • #15
266
2
Not an inconvenience at all mate, thank you for all your patience and help.
I am trying to understand why the equation is the way it is.

Yes my diagram is similar but without the rope on the left hand side being pulled
to re cap
T= up
mg= down
ma=down
thus T-mg-ma = 0 correct?

You see, ma is actually the resultant of mg and T, right? Because (in the frame of reference of the ground) they are the only two forces acting on the block. So ma=mg-T.
 
  • #16
Ahh.... big help with

t = mg

Not sure where you are in the world but you may have heard the penny drop.

Thanks again Sunil Simha.
 
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