Inclined slope with acceleration

In summary, the tension in the rope can be found by using the equations T = m(cosα)(tgα - a) when a < g/tgα and T = m√(a^2 + g^2) when a > g/tgα. This is because when a < g/tgα, N = 0 and the rope is at an angle greater than α, while when a > g/tgα, N is not equal to 0 and the rope is at an angle less than α. Additionally, to solve for T using the x-y axes, one can eliminate N by multiplying the first equation by cosα and the second by sinα and adding them together.
  • #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
Hi devanlevin! :smile:
devanlevin said:
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
but N is 90 degrees to the x-axis I am talking about, the axis parallel to T, look at the picture,
 
  • #4
Hi devanlevin! :smile:

(have an alpha: α and a square-root: √ :wink:)
devanlevin said:
but N is 90 degrees to the x-axis I am 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
thats the second case, the one i noted i don't know how to build the equation for,
when a>g/tg(alpha)
 
  • #6
how do i find this
 
  • #7
devanlevin said:
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
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 isn't 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
devanlevin said:
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
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
devanlevin said:
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
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
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
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
devanlevin said:
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
tiny-tim said:
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?? don't see it
 
  • #17
devanlevin said:
where should there be a minus?? don't see it

look at your own diagram … a is to the left, and n is to the right
 
  • #18
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 don't see it
 
  • #19
did what you said using my equations, without the - and it works, you sure there is a minus i haven't written
 
  • #20
thanks for the help, could you please look at this thread, desperately need help with it
https://www.physicsforums.com/showthread.php?p=1982691#post1982691
 
  • #21
oops!

devanlevin said:
did what you said using my equations, without the - and it works, you sure there is a minus i haven't written

Sorry … you're right … no extra minus. :redface:
 

1. What is an inclined slope with acceleration?

An inclined slope with acceleration is a type of motion in which an object is moving along a surface that is at an angle, or inclined, and is also experiencing a change in its velocity, or acceleration. This type of motion is often seen in physics experiments and can help us understand the effects of gravity and friction on an object's movement.

2. How is acceleration calculated on an inclined slope?

Acceleration on an inclined slope can be calculated using the formula a = gsinθ, where a is the acceleration, g is the acceleration due to gravity (usually 9.8 m/s²), and θ is the angle of the slope. This formula takes into account the effects of gravity and the slope angle on an object's acceleration.

3. What factors affect the acceleration of an object on an inclined slope?

The acceleration of an object on an inclined slope can be affected by several factors, including the angle of the slope, the mass of the object, the force of gravity, and the presence of friction. The steeper the slope and the greater the mass of the object, the greater the acceleration. However, friction can act to slow down or oppose the acceleration of the object.

4. How does an inclined slope with acceleration differ from a flat surface with acceleration?

An inclined slope with acceleration differs from a flat surface with acceleration in several ways. On an inclined slope, the acceleration is not constant as it is affected by the slope angle, whereas on a flat surface, the acceleration is constant. Additionally, the force of gravity plays a larger role in determining the acceleration on an inclined slope compared to a flat surface.

5. How is the motion of an object on an inclined slope with acceleration graphically represented?

The motion of an object on an inclined slope with acceleration can be graphically represented using a velocity-time graph. The slope of the graph represents the acceleration of the object, with a steeper slope indicating a higher acceleration. The area under the graph represents the displacement of the object. Alternatively, a position-time graph can also be used to show the changing position of the object over time on the inclined slope.

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