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Fun elevator problem, find accel

  • Thread starter bigman8424
  • Start date
25
0
This is a really fun probl, i've been trying to get for a while:
A 600 N man stands on a bathroom scale on elevator. After, it starts moving the scale reads 900 N.
Find Acceleration (magnitude/direction)
EF = ma
weight - T = ma
wt - T = ma
(600 N)(9.8 m/sec2) - T = ma
i noticed i didn't have accleration,
Ft - meg = meae = -mea

900-600/900+600 = (1/5)g(9.8)
i got 1.96 m/sec2

does this seem right?
 

Answers and Replies

91
0
First thing you see is that the man's mass goes from 600 to 900N. Therefore the elevator is accelerating upward.

Taking the case where the elevator is stationary.

F = mg
600 = m(9.8)
m = 61.2kg This is the mass of the man.

Where you are getting into some troubles is in setting up the exact situation.

Think about what you know and what you don't.

The first case you have is that the man's weight is 600N when the elevator isn't moving. The second thing you know is that when the elevator is moving, his weight increases to 900N. So you need to set up an equation to determine this.
 
391
2
I have not thought this through completely, but if the final scale reads 1.5 times greater, can you just assume that he is expirencing an acceleration 1.5 times greater and subtract [tex]a_g[/tex] from that number to find the elevator's acceleration?
 
99
1
Yes Candyman that is corect.
 
538
2
Actually what the wighing machine reads is the Normal force on the person standing on it.

When the lift was stationary , it must be reading 600 N . because N=Mg=600 N
But as per the problem when the lift starts accelerating , The reading increases, this means than N should increase , therefore a force has emerged in direction of Mg , this is possible when the lift accelerates upwards with "acceleration 'a' , then the pseudo force 'ma' adds to Mg.

Therefore final equation becomes:

Mg + Ma = 800
 
25
0
Oh i see, if the scale read zero, then he's free falling. ?
 
538
2
yep, then he must be falling down with acceleration 'g'!

B.J
 

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