- #1
devanlevin
in the following question, a box with a mass of m is tied to a pole at the top of a slope with an incline of (alpha), the whole slope accelerated at "a", find the tension "T" in the rope.
http://picasaweb.google.com/devanlevin/DropBox?authkey=sbH95pBl_D8#5273613370813875458 [Broken]
what i did was:
F=T-mg[sin(alpha)]=ma[cos(alpha)]
T=m[(gsin(alpha))+(a(cos(alpha))]
T=m(cos(alpha))[tg(alpha)-a]
which according to the answer sheet is only a part of the correct answer,
it says this is correct when-- a<g/tg(alpha)
but when a>g/tg(alpha)
then
T=m*sqrt(a^2+g^2)
why specifically a>g/tg(alpha) or a<g/tg(alpha), what happens physically at that point? is that the critical acceleration where N=0??
how do i see this from the equations?, the second T, seems to me like the sum of the 2 vectors ma and mg, what is the logic in doing that
http://picasaweb.google.com/devanlevin/DropBox?authkey=sbH95pBl_D8#5273613370813875458 [Broken]
what i did was:
F=T-mg[sin(alpha)]=ma[cos(alpha)]
T=m[(gsin(alpha))+(a(cos(alpha))]
T=m(cos(alpha))[tg(alpha)-a]
which according to the answer sheet is only a part of the correct answer,
it says this is correct when-- a<g/tg(alpha)
but when a>g/tg(alpha)
then
T=m*sqrt(a^2+g^2)
why specifically a>g/tg(alpha) or a<g/tg(alpha), what happens physically at that point? is that the critical acceleration where N=0??
how do i see this from the equations?, the second T, seems to me like the sum of the 2 vectors ma and mg, what is the logic in doing that
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