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A steel chain, whose mass measured m per unit , is placed on ground. Now grab one end of the chain and lift it vertically ,making it rise at a constant speed v , when the end of the chain is at height x, determine the grabing force F at this moment.
I did F=m(x*g+1/2*v^2) but the correct answer is F=m(x*g+v^2)
Here is what I did
there must be a acceleration process , so assume the end of the chain went through a distant L (limL--->0) before it reaches the speed v . when at the height x , it will require the force of it's weight to keep the speed v , so the F=xmg+gL
We go back to the acceleration process , the work the force does equal to TL=1/2*L*m*v^2
T=gL , after tranformation gL=1/2*m*v^2
substitute the gL , F=m(x*g+1/2*v^2) , but the real answer doesn't contain that 1/2, so I wanna know what had gone wrong?
Numbers are the essence of the Universe
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This is a calculus problem, unless you know some tricks and short-cuts for this type of Physics problem.
Also, what is "grabing" force? What if we assume the chain starts off moving upward at terminal speed since it weighs nothing
at first, so can accelerate infinitely to proper speed in first billionth of a second, perhaps?
Also don't forget ilovephysics.com, I haven't been there for a long time.
igloo myrtilles fourmis
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