An elevator Physics Accelerating Frame

[kerimabdullah]

An elevator slows down with a 6 m/s² acceleration and an object m=3 kg is freely released as shown in figure There is no friction between object an semispherical shape.
The object reaches L point what is the force which the object exerts L point?
A)120 N
B)240 N
C)360 N
D)480 N
E)600 N

For elevator
a=6 m/s2
V2=Vo2-2ah
V=Vo-at
h=(a/2)t2

For object
inside the elevator gravitational constant g'=16m/s2 because
N-mg=ma
N=m(g+a)=mg'


I can't solve I need help

 

[tiny-tim]

welcome to pf!

hi kerimabdullah! welcome to pf!

For object
inside the elevator gravitational constant g'=16m/s2 because
N-mg=ma
N=m(g+a)=mg'

what about the centripetal acceleration?

(and if the elevator is going up but slowing down, isn't it 4 not 16?)

 

[kerimabdullah]

solution how to? g'=16m/s2 is correct
N-mg=ma+(m/r)Vfinal^2
Energy equation? m(g+a)Integral(rdr)? Because The height is changeabele for every second This is Turkish Physics Olympiad Question. I couldn't solve
Or (1/2)mV^2 +m(g+a)Integral(rdr)= (1/2)mVfinal^2

 

[tiny-tim]

what about the centripetal acceleration?

 

[Chestermiller]

This is a very interesting and challenging problem. In addition, as Tiny Tim pointed out, there is some ambiguity as to whether the elevator is moving upward initially, and slowing down, or whether the elevator is moving downward initially, and slowing down. So I guess it's best to work the problem both ways, and then to see whether either solution matches the choices offered.

Let's first concentrate on the motion of the elevator. Let an arbitrary origin be located on the centerline of the elevator shaft, at some point well below the elevator's starting location, and let y_o be the y coordinate of the top of the elevator at time t = 0. If the elevator is moving upward at t = 0 with an upward velocity of v_o, and is slowing down with a downward acceleration of 6m/s², what is its vertical coordinate y at time t? If the elevator is moving downward at t = 0 with a downward velocity of v_o, and is slowing down with an upward acceleration of 6 m/s², what is its vertical coordinate y at time t? If i_y represents a unit vector pointing in the upward y-direction, express the position vector from the arbitrary origin to the top center of the elevator in terms of y and i_y for each of these two cases.

After you answer this question, I will help you proceed further with the solution.

Chet

 

[kerimabdullah]

showthread.php?t=424269 BUT integral is difficult

 

[kerimabdullah]

Lets suppose the elevator moves with constant velocity downward and a man press the button to stop elevator then elevator makes slowing down motion downward with a=6m/s2 and angle and height for the object changes with the time.

 

[azizlwl]

If accelerating upward 6m/s²

v²=2aL
a=6+10
a_c=32m/s²

F=m(32+6+10)=3(48)N
None matches.

 

[kerimabdullah]

ELevator slows down ward with a constant acceleration of 6 m/s2 or (a = 6 N/kg )

 

[kerimabdullah]

Turkish Physics Olympic Questions (The original question 13) I am waiting solutions since 2 months but they didnt publish. I can't sleep nights because I didnt solve problem 13
www. fen.bilkent.edu.tr /~eurasia_pacific/ITAP_FOO/exams/xcj1p8id2cvtzu75ztrrk383b9hffup2wzucy3gq529vm7rbr8t12su3r29wnpp9eac33jkzbklnh2py19qurbn83mvg2bg6bj05.pdf

 

[Chestermiller]

If the mass were 30 kg, rather than 3 kg, then one of the answers would make sense. If the elevator were initially moving upward and decelerating at a rate of 6 m/s² and the mass were 30 kg, then the contact force would be 360 N. According to my analysis of this problem, the contact force on the mass at point L is given by:

N = 3m (g + a)

where a is the upward acceleration of the elevator. If a = -6 m/s², and m = 30 kg, then N = 360 N.

The main part of this problem is deriving the above equation. See if you can confirm my result.

Chet