Inclined slope with acceleration

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  • #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
 
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  • #2
tiny-tim
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Hi devanlevin! :smile:
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.

what i did was:

F=T-mg[sin(alpha)]=ma[cos(alpha)]
No, you've left out N. :frown:
 
  • #3
devanlevin
but N is 90 degrees to the x axis im talking about, the axis parallel to T, look at the picture,
 
  • #4
tiny-tim
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Hi devanlevin! :smile:

(have an alpha: α and a square-root: √ :wink:)
but N is 90 degrees to the x axis im talking about, the axis parallel to T, look at the picture,
Sorry, I should have been more specific. :redface:

You didn't consider what hapens when N is zero …

then the box is hanging on for dear life (at an angle smaller than α). :biggrin:
 
  • #5
devanlevin
thats the second case, the one i noted i dont know how to build the equation for,
when a>g/tg(alpha)
 
  • #6
devanlevin
how do i find this
 
  • #7
tiny-tim
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how do i find this
The rope is now at an unknown angle.

Just write the horizontal and vertical equations, and solve by eliminating the angle. :smile:

(or just use a vector triangle :wink:)
 
  • #8
devanlevin
from what i understand,
i say, when a>g/tg(alpha) N=0, and so the total force is sqrt((T)^2+(mg)^2) and that is the size of the force on the box, now that is equal to ma???

so T=sqrt((ma)-(mg)^2)

but that isnt right, where am i going wrong,,, can you please show me how to solve this both ways, both with equation and with vector triangle
 
  • #9
tiny-tim
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from what i understand,
i say, when a>g/tg(alpha) N=0, and so the total force is sqrt((T)^2+(mg)^2) …
No, that is not the total force …

T and g are not perpendicular.
Just write the horizontal and vertical equations, and solve by eliminating the angle.

(or just use a vector triangle)
 
  • #10
devanlevin
right, so how would i divide my axises, parallel to a or parallel to t??
if parallel to a
F=Tcos(alpha)=ma

if parallel to a
F=F=T-mg[sin(alpha)]=ma[cos(alpha)]
which is what i did in the 1st place
 
  • #11
tiny-tim
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right, so how would i divide my axises, parallel to a or parallel to t??
if parallel to a
F=Tcos(alpha)=ma …
waaah! :cry:

it's not alpha!!

and you need two axes!
 
  • #12
devanlevin
so when a<g/tg(alpha) i divide the vector components parallel to T
and when a>g/tg(alpha) i divide them parallel to a

shouldnt i use the same x,y axises in both cases??? here once i look at it on the slopes axis once on the grounds axis?? how do i know how to divide and when
 
  • #13
tiny-tim
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No, you can use any pair of axes in both cases …

the only difference is that if a ≤ g/tanα, then the rope is at angle α, but if a > g/tanα, the rope is at an angle less than α.
 
  • #14
devanlevin
i can only see the 1st case with an axis parallel to t

F=T-mg[sin(alpha)]=ma[cos(alpha)]
T=m[(gsin(alpha))+(a(cos(alpha))]
T=m(cos(alpha))[tg(alpha)+a]

and with a regular axis, parallel to a for the second case

Ma(y)=0=Mg-T(y)
Ma(x)=ma=T(x)

T=sqrt[(mg)^2 + (ma)^2]

if i try the other way what are my equations? i get

for the 1st case
Ma(x)=ma=Tcos(alpha)-Nsin(alpha)
Tx=Nsin(alpha)+ma
Ma(y)=0=Tsin(alpha)+Ncos(aplha)-mg
Ty=mg-Ncos(alpha)

then to find T==>Tx^2+Ty^2
but i get an equation with N in it



could you please show me the equations you get for each axis
 
  • #15
tiny-tim
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if i try the other way what are my equations? i get

for the 1st case
Ma(x)=ma=Tcos(alpha)-Nsin(alpha)
Tx=Nsin(alpha)+ma
Ma(y)=0=Tsin(alpha)+Ncos(aplha)-mg
Ty=mg-Ncos(alpha)

then to find T==>Tx^2+Ty^2
but i get an equation with N in it

could you please show me the equations you get for each axis
Hi devanlevin! :smile:

Well that's pretty good :approve:

you've tried both methods (x-y axes and T-N axes), and you've got the right equations for them (except there should be minus signs in the Tx equation) …

you just haven't spotted the trick for solving the x-y one :wink:

you simply need to eliminate N …

so multiply the first equation by cosα, and the second by sinα, and add! :smile:
 
  • #16
devanlevin
Hi devanlevin! :smile:

Well that's pretty good :approve:

you've tried both methods (x-y axes and T-N axes), and you've got the right equations for them (except there should be minus signs in the Tx equation) …

you just haven't spotted the trick for solving the x-y one :wink:

you simply need to eliminate N …

so multiply the first equation by cosα, and the second by sinα, and add! :smile:


where should there be a minus?? dont see it
 
  • #17
tiny-tim
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where should there be a minus?? dont see it
look at your own diagram … a is to the left, and n is to the right
 
  • #18
devanlevin
X+ in the direction of a

Ma(x)=ma=Tcos(alpha)-Nsin(alpha) (here there is a minus)
if i wan Tcos(alpha) on one side of the =, i move -Nsin(alpha) to the side of ma,
becomes + Nsin(alpha)

Tx=Nsin(alpha)+ma

still dont see it
 
  • #19
devanlevin
did what you said using my equations, without the - and it works, you sure there is a minus i havent written
 
  • #21
tiny-tim
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oops!

did what you said using my equations, without the - and it works, you sure there is a minus i havent written
Sorry … you're right … no extra minus. :redface:
 

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