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Finding force using moments problem 
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#1
Apr1612, 07:37 AM

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Hello
I am having trouble understanding a concept of what force i have found and how i can maniplate this data. I have attached image of a triangular frame which has a 10Kg mass hanging from it at one point. I know how to calculate the Force Fh using moments. i am unsure what this force actually is. Is it a static force or what? For my application some person will push the triangle at point where Fh is applied. To find the power (force x velocity) do i need to convert the force i have back to equivelant mass and multiply by my nex acceleration of the person or am i wrong? 


#2
Apr1612, 07:52 AM

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What force have you found? Fh?
I think you need to restate your question so that there is no doubt about what we are discussing. Is there some 'scenario' which has yielded the question? 


#3
Apr1612, 08:21 AM

P: 14

Sory for being unclear.
I know Fh as i have calculated this for various positions. The scenario is from a piece of gym epuipment where a person pushes a mass around a fixed pivot point. As the persn pushes they need to apply more force. I an interested in finding out what is the fh i have found? I think i have static force? Is it still unclear? 


#4
Apr1612, 12:18 PM

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Finding force using moments problem
A static force is one where there is no acceleration (in most cases, no movement in an inertial frame) i.e. equilibrium.
In the diagram, it looks to me as if the person is causing the 10kg mass to accelerate more than it would naturally accelerate due to gravity (it would move in a circle about the pivot  like a pendulum if left to itself). It must be nonstatic. 


#5
Apr1612, 12:26 PM

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Do you want to calculate how much the mass will accelerate?



#6
Apr1612, 04:10 PM

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I want to know the force the person has to apply which is Fh. I have attached a link with the machine:
http://www.google.ie/imgres?hl=en&sa...9,r:7,s:0,i:81 


#7
Apr1612, 05:17 PM

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It is still not clear what you want to know, exactly but the picture on the link makes the basic setup more clear.
Do you want to know the force that will BALANCE the mass at a certain angle? You wrote that you can calculate that ("static force") Do you want to know what force will accelerate the mass at some rate? If you do, then you need to specify what rate. If there is not movement then there is no work done (strictly). I guess what I can tell you is that if you want to work out the total force that the guy will have to use, it will be the static (equilibrium) force plus whatever he needs to accelerate the mass. That's a bit of an open question, of course but, as you will realise, the further he pushes, the greater the force will be as the mass goes further out from the pivot. The Power at any point will be force X velocity, as you wrote. Perhaps that's enough for you to get on with it now? A similar (weightlifting) thread has just run to hundreds of posts when someone seemed to think that there is a direct relationship between the Work Done, in the strict sense, and the energy used up by the muscles. They are doing their crust even when you aren't moving at all  just keeping the mass up in the air, so it's a bit of a nonsense really. I reckon if it hurts then it must be doing you some good. Grrrrrrr . The thread was put out of its misery by being locked, thank God. When I was at School in the early 60s, they had a scrum practice frame with a set of pads and enormous springs. Very uncomfortable thing to get involved with. Worse than the real thing! 


#8
Apr1612, 06:14 PM

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Let the triangle have sides x (horizontal) and y (vertical).
The force you calculate by equating the moments (W.x/y) is the static force to maintain equilibrium. Exceeding that by a small amount will result in movement. But as the movement progresses the moments will change, so more force may be needed. How much force is eventually needed will depend on how far you need to push it. E.g. as the rotation of the apparatus approaches 90° the horizontal force required approaches infinity. You'd need to put in some upward force instead. The least force is required (at any point of the turn) when the force is kept at right angles to (what is initially) the upright strut  i.e. tangential to the rotation. In this case the max force required is W.√(x^2+y^2)/y. 


#9
Apr1712, 03:24 AM

P: 14

Thanks, yes by static force i meant the force to balance the system at certain angles.
I have a sensor on the equipment that will tell me acceleration and velocity throughout the movement. By multiplying the force mentioned above by velocity i get a power reading. This is not the true power reading i imagine as a result of the initial acceleration of player to go from static to his peak velocity is not taken into account. I don't mean to sound silly but is there a way of incorporating initial acceleration into my calculation or how to find the force created by the acceleration of mass? 


#10
Apr1712, 04:51 PM

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It all depends what you actually want.
You could work out the force at any point in the rotation. You could work out (in steps, rather than the full integration) the work done over the full stroke by adding the force times distance values. If you are actually measuring the force of the shoulder against the bar, then this would be a 'good' answer. If you have timing information as well, then you could also work out the power being delivered: P=work done/ time interval I appears to be a very dynamic exercise and there must be more and less efficient ways of getting the mass to a given height. Getting it right could give you greater height for a given amount of (unspecified units) 'effort' on the part of the person. But is this representative of what is needed in a real rugby scrum? The real load is not at all like the mass / pendulum arrangement. It strikes me that what you need is some form of damping on this machine to stop you from 'cheating' by taking a run up at the start of the push. 


#11
Apr1712, 05:51 PM

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As Sophiecentaur says, it would help if you would go into more details of what you're trying to achieve. Are you trying to measure peak force, peak power, average power or total effort?
Maybe all four? Bear in mind when multiplying force by distance or velocity that these are vectors, and it's the dot product that you want. I.e. if the force is horizontal but the movement at angle theta to horizontal you have to multiply by cos(theta). Fortunately that's equivalent to assuming the force applied is tangential to the swing. When the apparatus has swung through angle theta, the (minimum) force applied will be W.(sin(theta) + x.cos(theta)/y) As you say, the actual force applied may be sometimes more, sometimes less, producing accelerations. In fact, to get the maximum swing the user would probably apply a larger force than given above in the early part of the swing to get a bit of speed up. The momentum gained will then help as the equilibrium force increases. Do you care about this? I.e. do you want to penalise the user by only counting the peak power, or do you regard this as just good technique? 


#12
Apr1812, 03:47 AM

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I want a means of comparing players against each other. So i think by finding their power at certain positions using the static force and velocity at that point will do for comparing players as the force will increase if a player chooses to put a heavier weight on the machine and also it will take into account their velocity. Using my position sensor i can find acceleration and from this i can find peak acceleration. I can then compare players who were lifting the same mass to see where there peak acceleration is.
Thanks for your help, your explanations have helped me come to a decision that it would just be too difficult to find exact power as there are too many variables to mingle in a computer programme for a college project. Thanks again lads 


#13
Apr1812, 04:22 AM

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