Tension of a rope with a hanging mass and acceleration

In summary, the tension in a rope increases as the mass falls. However, when the mass falls at an accelerated rate, there is a force acting in a downward direction to oppose the acceleration. This force is called ma and is equal to mg-T.
  • #1
justadaftspark
17
0

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
 
Physics news on Phys.org
  • #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
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 I am not getting it
 
  • #7
justadaftspark said:
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
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
justadaftspark said:
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
justadaftspark said:
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

  • Untitled.jpg
    Untitled.jpg
    4.5 KB · Views: 1,189
  • #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
justadaftspark said:
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.
 
Last edited by a moderator:

1. What is the formula for calculating the tension of a rope with a hanging mass and acceleration?

The formula for calculating tension in a rope with a hanging mass and acceleration is T = m(g+a), where T is the tension, m is the mass of the object, g is the acceleration due to gravity, and a is the acceleration of the object.

2. How does the mass of the object affect the tension in the rope?

The tension in the rope is directly proportional to the mass of the object. This means that as the mass of the object increases, the tension in the rope will also increase.

3. What happens to the tension in the rope if the acceleration of the object changes?

If the acceleration of the object changes, the tension in the rope will also change. This is because the formula for calculating tension includes the acceleration of the object.

4. How does the angle at which the rope is suspended affect the tension?

The angle at which the rope is suspended has a direct effect on the tension. As the angle increases, the tension in the rope also increases. This is because a larger angle increases the horizontal component of the tension force.

5. Can the tension in a rope with a hanging mass and acceleration ever be zero?

No, the tension in a rope with a hanging mass and acceleration can never be zero. This is because the tension is created by the weight of the hanging mass and the acceleration of the object, both of which will always have some value.

Similar threads

  • Introductory Physics Homework Help
2
Replies
38
Views
972
  • Introductory Physics Homework Help
Replies
3
Views
516
  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
25
Views
3K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
2K
Back
Top