Why Does the Scale Reading Change in an Elevator?

  • Thread starter Thread starter joe215
  • Start date Start date
  • Tags Tags
    Elevator Force
AI Thread Summary
The discussion focuses on the changing scale readings of a person in an elevator, analyzing the forces at play during different time intervals. The individual experiences varying weights due to acceleration: 0 m/s² from 0-5 seconds, 3.92 m/s² from 5-10 seconds, 0 m/s² from 10-15 seconds, and -3.92 m/s² from 15-20 seconds. The calculated velocities appear unrealistic, particularly the 39.2 m/s at 10 seconds, prompting a reevaluation of the time intervals used in the velocity formula. The correct approach emphasizes that the time in the velocity equation should reflect the duration of acceleration, not the total elapsed time. Understanding these dynamics is crucial for accurately interpreting the scale readings in relation to the elevator's motion.
joe215
Messages
25
Reaction score
0

Homework Statement



A person with a weight of 500N stands in an elevator on a scale.
from 0-5s his weight reads 500N.
from 5-10s his weight reads 700N.
from 10-15s his weight reads 500N.
from 15-20s his weight reads 300N.

Find his acceleration (in each time interval) and his velocity at 5s, 10s, 15s, and 20s.

The Attempt at a Solution



I think my accelerations are correct:
(Fn-mg) /m=a

0-5s: (500N-500N) / 51kg= 0m/s^2
5-10: (700-500)/51=3.92
10-15: (500-500)/51=0
15-20: 300-500/51= -3.92

Now, I think my velocities seem a little unreasonable.
vf=vi+at

at 5: v=0
10: v=0m/s + (3.92)(10)=39.2m/s <- that is really fast!
15: v=39.2m/s + (0)(15)= 39.2m/s
20: 39.2m/s+(-3.92)(20)= -39.2m/s

thanks for any help
 
Physics news on Phys.org
joe215 said:
Now, I think my velocities seem a little unreasonable.
vf=vi+at
That time is the time in the interval for which the acceleration is "a", not the time from the beginning. I'd write it as: Vf = Vi + aΔt.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top