Discussion
This one got me thinking: if you are hanging from a bar, the load is split into two point loads, totalling your weight.
Does the load increase if you then do a pull-up and if it does, by how much?
I say it doesn't, as there is no other reaction that you can add, as you are already hanging. A mate says it will increase, depending on how hard you pull yourself up (the faster you pull-up, the more the load will increase).
What say you all.
Does the load increase if you then do a pull-up and if it does, by how much?
I say it doesn't, as there is no other reaction that you can add, as you are already hanging. A mate says it will increase, depending on how hard you pull yourself up (the faster you pull-up, the more the load will increase).
What say you all.
The load does increase, as you have to overcome your inertia to move. If you are just hanging and not moving, everything is in balance and the downwards force on the bar is just as a result of your weight. In order to make you move upwards though, there has to be an acceleration. In order for there to be an acceleration, there has to be a net upwards force applied to your body. As you are creating this force by pulling on the bar with your arms, that force is applied to the bar and so the total force acting on it is your weight plus the force required to accelerate you upwards.
The faster your pull up the higher the additional force will be since force is directly proportional to acceleration, though it will act for a shorter period. The additional force only acts while your are actually accelerating yourself upwards and will return to just that caused by your weight if you stop at the top and hold the position.
The faster your pull up the higher the additional force will be since force is directly proportional to acceleration, though it will act for a shorter period. The additional force only acts while your are actually accelerating yourself upwards and will return to just that caused by your weight if you stop at the top and hold the position.
tank slapper said:
The load does increase, as you have to overcome your inertia to move. If you are just hanging and not moving, everything is in balance and the downwards force on the bar is just as a result of your weight. In order to make you move upwards though, there has to be an acceleration. In order for there to be an acceleration, there has to be a net upwards force applied to your body. As you are creating this force by pulling on the bar with your arms, that force is applied to the bar and so the total force acting on it is your weight plus the force required to accelerate you upwards.
The faster your pull up the higher the additional force will be since force is directly proportional to acceleration, though it will act for a shorter period. The additional force only acts while your are actually accelerating yourself upwards and will return to just that caused by your weight if you stop at the top and hold the position.
Makes sense - looks like I owe him a beer The faster your pull up the higher the additional force will be since force is directly proportional to acceleration, though it will act for a shorter period. The additional force only acts while your are actually accelerating yourself upwards and will return to just that caused by your weight if you stop at the top and hold the position.
Where me and physics parted ways was the concept that if you put a weight on a table, the table pushed back with an equal and opposite force. I thought - 'Fantastic, let's capture this force and use it to do stuff'. But the mysterious upwards force of an inanimate table cannot be captured it seems...
Simpo Two said:
Where me and physics parted ways was the concept that if you put a weight on a table, the table pushed back with an equal and opposite force. I thought - 'Fantastic, let's capture this force and use it to do stuff'. But the mysterious upwards force of an inanimate table cannot be captured it seems...
The upwards force is already being used to prevent the downward accelleration of the weight.Simpo Two said:
Take the weight off and the force is free...
Eh? what do you mean the "force is free" ? Take the weight away and the force supplied by the table disappears.The force from the table is just the reaction force to the weight. The forces are equal and opposite, hence zero acceleration.
Simpo Two said:
Or, if one object is exerting a force downwards and one is exerting it upwards, where are they getting the energy from to do it? Will both objects start getting colder...?
Confusion here between force, energy, and work. No energy is required to sustain the force, if the system is static the only energy to be considered is the gravitational *potential* energy. That is, a certain amount of energy can be recovered by allowing the weight to fall, eg tie a string to it and let it power a dynamo or similar.Same applies to a spring stretched between two posts. If nothing is moving then no energy needs to be supplied, or is being created.
I guess its this misperception of force as "energy" that leads people to thinking they can build a free energy generator using magnets. It aint so. Force is not energy, or work.
For movement, the propelling force needs to be greater than the opposing one so some energy transfer must take place. Therefore there must be an additional source of energy to be transferred (in this case the energy comes from the fuel in the tank).
It might help to realise that "energy" isn't actually an entity in itself, merely a way of describing the transferral of work from one system to another. If there is no movement, no work is being transferred, hence no net transfer in energy.
It might help to realise that "energy" isn't actually an entity in itself, merely a way of describing the transferral of work from one system to another. If there is no movement, no work is being transferred, hence no net transfer in energy.
Simpo, I think some situations create a misleading impression that energy is needed to exert a force.
For example, if I drive up a steep wet hill I could reach a point where the wheels spin and the car is stationary. I'm clearly pouring in energy to hold the car on the hill, but all the energy is being wasted as heat & noise etc in the slipping tires. None is going into the car as its not accelerating.
I could just pull the handbrake on and turn the engine off. Now the car will sit there with the same amount of force holding it up vs gravity but with no energy expended.
So a fat bloke, who weighs, say 100kg ( ) does a pull-up.
The initial load is 980N (kg x gravity) then the extra load is dependant on how quickly he goes from rest below the bar to rest above the bar (assuming the stop is instantaneous). Pull-up distance is, what, 30cm, lets say he takes 1 second to do it (he's unfit), so acceleration is 0.3 ms-2, making the additional load 100 x 0.3 = 30 kgms-2 (or 30N).
If he does it in half a second (as he gets fitter), then the load goes up to 60N.
Right?
The initial load is 980N (kg x gravity) then the extra load is dependant on how quickly he goes from rest below the bar to rest above the bar (assuming the stop is instantaneous). Pull-up distance is, what, 30cm, lets say he takes 1 second to do it (he's unfit), so acceleration is 0.3 ms-2, making the additional load 100 x 0.3 = 30 kgms-2 (or 30N).
If he does it in half a second (as he gets fitter), then the load goes up to 60N.
Right?
Not quite. Acceleration is not 0.3 ms^-2, but 0.6 ms^-2.
s=ut + 1/2 at^2
initial velocity u is 0, so that reduces to s= 1/2 at^2 in this case.
Solving for a gives a= 2s/t^2.
When the distance (s) is 0.3m, and time is 1 second, a is 0.6 ms^-2.
So that gives a force of 100kg x 0.6ms^-2 = 60N.
That doesn't reflect what really happens though. There would actually be a positive and negative acceleration, since the person isn't powering upwards to an instantaneous stop at the top. The total force on the beam would initially be higher and would then reduce as he slows down to the top. The peak force may also be significantly higher than the figure above depending on how violently the pull up is done.
s=ut + 1/2 at^2
initial velocity u is 0, so that reduces to s= 1/2 at^2 in this case.
Solving for a gives a= 2s/t^2.
When the distance (s) is 0.3m, and time is 1 second, a is 0.6 ms^-2.
So that gives a force of 100kg x 0.6ms^-2 = 60N.
That doesn't reflect what really happens though. There would actually be a positive and negative acceleration, since the person isn't powering upwards to an instantaneous stop at the top. The total force on the beam would initially be higher and would then reduce as he slows down to the top. The peak force may also be significantly higher than the figure above depending on how violently the pull up is done.
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