Physics of punching a heavy bag

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SUMMARY

The discussion focuses on calculating the forces acting on a heavy punching bag suspended from a roof structure, specifically the tension in the rope and the effects of an impulsive force from a punch. Key equations include the impulse equation F = Δmv/Δt and the tension formula T = mg + mv²/l, where T is tension, m is mass, g is gravitational acceleration, v is velocity, and l is the length of the rope. The maximum tension occurs at the bottom of the swing, while the tension at the top of the arc is zero when the bag's velocity is also zero. Understanding these dynamics is crucial for designing the roof members to support the punching bags.

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I am working on a project where my client wants to hang several heavy punching bags from the roof structure of his building. As my physics is quite rusty, I need help in determining the reactions a struck heavy bag will have on a roof member.

I have attached a diagram with a brief description of the problem. Any assistance on this matter would be greatly appreciated.
 

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Hi pkumor! Welcome to PF :smile:

Start by observing all the forces acting on the punching bag. In which position would the tension be maximum?
 
Infinitum,
Am I missing a force?

I am guessing the max tension would be when the bag is vertical, but I also need to know what the tension at the top of the swing would be to determine the maximum horizontal force being applied at the support...
 
You don't seem to have the tension labelled on the bag. I missed it while looking earlier.

The maximum tension is at the bottom most point, when the bag is given a velocity to freely move in circular motion. But what is the question exactly, the person applies a force F, continuously, or is it impulsive?

If it is impulsive(as a punch is :-p), you need to include the centripetal force into your force equations.
 
For the sake of this question the force is an impulse, the boxer hits the bag and it moves away from him.

And that is where I need the help, as I stated my physics is rusty. I do not know how to correctly applying the equations to this problem and need some detailed guidance (or just the answer) so I can move onto designing the roof members.

Also (T) was included in the diagram, but not the centripetal force...
 
pkumor said:
For the sake of this question the force is an impulse, the boxer hits the bag and it moves away from him.

And that is where I need the help, as I stated my physics is rusty. I do not know how to correctly applying the equations to this problem and need some detailed guidance (or just the answer) so I can move onto designing the roof members.

Also (T) was included in the diagram, but not the centripetal force...

So if it is an impulsive force, you need to calculate the velocity the bag attains when it was punched, from the impulse equation,

F = \frac{\Delta mv}{\Delta t}

Now, getting this velocity the of the bag, the maximum tension(which is at the bottommost point) will be given as,

T = mg + \frac{mv^2}{l}
 
ok, but isn't v dependent on how far up the bag moved through the arc, which would be determined by the initial F?

How do I calculate \varphi from the initial punch?

Also how do I calculate the tension in the rope at the top of the arc when the velocity equals zero?

If it helps we can say F=500lb, r=8ft, w=100lb...
 
pkumor said:
ok, but isn't v dependent on how far up the bag moved through the arc, which would be determined by the initial F?

Of course it is dependent on initial F :smile:


How do I calculate \varphi from the initial punch?

At the highest point the bag reaches, its velocity is zero. So, apply conservation of energy at this point to determine what height it reaches. Now, you know that the bag is moving in a circle, so what is the angle displaced by it, if you know how high it has gone? :wink:

Also how do I calculate the tension in the rope at the top of the arc when the velocity equals zero?

When the velocity is zero, the bag is at the most extreme position of the circular arc it is travelling(somewhat like a simple pendulum) So which in which direction will it have no net acceleration. What can you say about the forces in this direction?
 

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