- #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

Last edited by a moderator: