Physics Question

[shoehornlettuce]

I'm having trouble understanding how this is possible and was hoping somebody could explain it to me.

If you are lowering an object that is attached to a string and the object is accelerating upward, how can the tension force on the string be greater than mg?

I'm thinking this question is just poorly written or maybe I just have a fundamental misunderstanding of something. Wouldn't the object have to be accelerating downward for the tension force of the string to be greater than mg? Any help is greatly appreciated. Thanks

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[Erythropoietin]

The magnitude of mg always stays the same since you are at the same g, and mass isn't changing.
But when you are stretching the spring, you are increasing the restoring force (Hooke's law: F = -kx)
The negative sign in the formula indicates that it's a restorative force.
Anyways, since you are increasing the displacement x (stretching it), the upward force (tension) will be larger.

 

[shoehornlettuce]

Thanks for the quick replies. I guess I'm still a bit confused. Tension force will be equal to mass(9.8+acceleration). So if there is no acceleration the tension force just equals mg. If the tension force is greater than mg then the acceleration must be in the down direction correct? Assuming downward is the positive y direction.

 

[Onscene]

I always think about the different forces acting in the object. mg is acting down and the tension in the string is acting up. Because the object is being lowered and is accelerating upward (decelerating downward) the force in the string must be greater to overcome the effect of gravity. IF the two forces were equal, the object would either be at rest or moving at a constant velocity.

 

[Erythropoietin]

Thanks for the quick replies. I guess I'm still a bit confused. Tension force will be equal to mass(9.8+acceleration). So if there is no acceleration the tension force just equals mg. If the tension force is greater than mg then the acceleration must be in the down direction correct? Assuming downward is the positive y direction.

Where does this question come from, EK?

 

[shoehornlettuce]

Where does this question come from, EK?

Naw, just a question from a practice worksheet I got in my physics class. I think it was just poorly worded which is why I was getting confused. Anyway, thanks again for everyones help.

 

[Pisiform]

I'm having trouble understanding how this is possible and was hoping somebody could explain it to me.

If you are lowering an object that is attached to a string and the object is accelerating upward, how can the tension force on the string be greater than mg?

I'm thinking this question is just poorly written or maybe I just have a fundamental misunderstanding of something. Wouldn't the object have to be accelerating downward for the tension force of the string to be greater than mg? Any help is greatly appreciated. Thanks

No, lets see an example.

You have 100N object which is tied to a string and is being lowered. During this process, they say it is accelerating upwards at 5m/s^2. It also means that it is decelerating downwards with 5m/s^2 or its value is -5m/s^2 downwards.

Set up an equation:

Fnet = ma
T - 100N = 10 * a (Since it is accelerating upwards, then Tension would be greater than mg)
T - 100N = 10 * 5
T = 50 + 100
T = 150N

Thus Tension is greater than mg. 🙂

 

[Ssina]

No, lets see an example.

You have 100N object which is tied to a string and is being lowered. During this process, they say it is accelerating upwards at 5m/s^2. It also means that it is decelerating downwards with 5m/s^2 or its value is -5m/s^2 downwards.

Set up an equation:

Fnet = ma
T - 100N = 10 * a (Since it is accelerating upwards, then Tension would be greater than mg)
T - 100N = 10 * 5
T = 50 + 100
T = 150N

Thus Tension is greater than mg. 🙂

I think the important word in there is accelerating upwards= deaccelerating = slowing down... so for it to slow down some force has to slow it down so it falls at a slower rate each second otherwise g will act on it and pull down faster and faster.

 

[shoehornlettuce]

I think the important word in there is accelerating upwards= deaccelerating = slowing down... so for it to slow down some force has to slow it down so it falls at a slower rate each second otherwise g will act on it and pull down faster and faster.

So why wouldn't this cause tension force to be smaller than mg? If object is 10kg then you have a 98N object. Then another force going opposite mg (say 48N) so net force in the down direction of 50N. Wouldn't tension force then be 50N which is less than mg?

I don't know why I'm having so much trouble with this. It seems like a simple enough problem.

 

[Ssina]

So why wouldn't this cause tension force to be smaller than mg? If object is 10kg then you have a 98N object. Then another force going opposite mg (say 48N) so net force in the down direction of 50N. Wouldn't tension force then be 50N which is less than mg?

I don't know why I'm having so much trouble with this. It seems like a simple enough problem.

if I understand your example correctly you have an mg= 98N pointing down and you say another force, tension, in opposite direction 48N (which then it has to be pointing up) and correctly you will have a NET force that will be 50N and pointing DOWN, since down is bigger than up.... so you will accelerate downwards

the way I think about these questions is that anytime you have a NET force in any direction, you will accelerate in that direction (regardless of gravity (g) and other forces) because this is our Fnet that takes into account all the forces acting...

now going back to your example when you have a net force pointing down, then you will be accelerating downwards (not upwards/decelerating as the original questions says)... and that is because your tension force is less than mg (50N vs 98N)

what I would do is to think of Fnet, Fgravity, and Tension all separately

Imagine we drop a box 10kg (attached to a string) from some hight:
at first tension<mg, like your example, T=49N and Fgrav=98N.... so the box will accelerate downwards because Fgravity, which is the bigger force, is pointing down:
F(net)=mg-T
Fnet=ma=98-49= 50N down
so now you can use the Fnet=m(a)=10a, so a=5m/s^2 down
here the box is falling downwards at a=5 not a=g=9.8 (because there is a tension force acting on the box in the opposing direction)

now lets assume T>mg, the box is still falling/being lowered, but someone is pulling on the string creating the tension force upwards to slow down the falling, e.g. T=149N and Fgrav=mg=98N
so now Fnet=T-mg
Fnet=ma=149-98=50N up
so Fnet=50=10(a), so a=5m/s^2 up
here the box is still falling down (or being lowered) but at a=5 pointing up... all it means it's that the box is slowing down... if at t=0s is going v=20m/s then at t=1s is going v=15m/s... t=2s then v=10m/s... basically changing it's velocity by 5 m/s every second

I hope this helps a bit and it's not confusing....

 

[shoehornlettuce]

if I understand your example correctly you have an mg= 98N pointing down and you say another force, tension, in opposite direction 48N (which then it has to be pointing up) and correctly you will have a NET force that will be 50N and pointing DOWN, since down is bigger than up.... so you will accelerate downwards

the way I think about these questions is that anytime you have a NET force in any direction, you will accelerate in that direction (regardless of gravity (g) and other forces) because this is our Fnet that takes into account all the forces acting...

now going back to your example when you have a net force pointing down, then you will be accelerating downwards (not upwards/decelerating as the original questions says)... and that is because your tension force is less than mg (50N vs 98N)

what I would do is to think of Fnet, Fgravity, and Tension all separately

Imagine we drop a box 10kg (attached to a string) from some hight:
at first tension<mg, like your example, T=49N and Fgrav=98N.... so the box will accelerate downwards because Fgravity, which is the bigger force, is pointing down:
F(net)=mg-T
Fnet=ma=98-49= 50N down
so now you can use the Fnet=m(a)=10a, so a=5m/s^2 down
here the box is falling downwards at a=5 not a=g=9.8 (because there is a tension force acting on the box in the opposing direction)

now lets assume T>mg, the box is still falling/being lowered, but someone is pulling on the string creating the tension force upwards to slow down the falling, e.g. T=149N and Fgrav=mg=98N
so now Fnet=T-mg
Fnet=ma=149-98=50N up
so Fnet=50=10(a), so a=5m/s^2 up
here the box is still falling down (or being lowered) but at a=5 pointing up... all it means it's that the box is slowing down... if at t=0s is going v=20m/s then at t=1s is going v=15m/s... t=2s then v=10m/s... basically changing it's velocity by 5 m/s every second

I hope this helps a bit and it's not confusing....

Thanks, I appreciate you taking the time to help me out. I think I have a pretty good understanding of this now.