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Elevator kinematic

  • Thread starter ant284
  • Start date
  • #1
ant284
Hi,

here is the question:
there is a elevator, an man is in it, via a pulley he can raise the elevator.

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This is my picture so can't do better
z are empty space as this forums gets rid of them
there is a man in the middle of the box, pulling a pulley.
the elevator is 30kg the man is 70kg,
this man exert a force of 300N to rise himself, we need to find the acceleration.
the answer is 5m/s^2 but i can't find why.
What i do is that

the mass of the whole system is 100kg (man + elevator)
the tension in the string is 300N, and using the formula
Mg-T=ma
where g = 10
700-300=100a
4m/s^2
any feed back is appreciated
 
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Answers and Replies

  • #2
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
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I don't have a calculator handy, but can you check to see if the error comes from the approximation (g=10) you used? Try again with g=9.8 and see if it doesn't come out right.
 
  • #3
ant284
the acceleration is 10g.
But i can't figure out how to do it
I want to do
T-30g=30a
100g-T= 100 a (100 = 70+ 30)
if i add them together i get
700 = 130 a
a= 5.4m/s

But this is how close i get to the answer!!
and i'm sure i am wrong because the T = 300
so g = 10
in the first equation
T-30g=30a
300-30*10=30a
0=30a
so i can't go nowhere here.
 
  • #4
1,036
1
Hey, I posted an answer to this a few days ago (OK - it was slightly "camouflaged" with a few extra z's, but all you had to do was search/replace the z's with spaces) but that entire post somehow got erased.

Down with censorship!!

Anyway, am I getting the picture right? The guy is standing in an elevator, there is a rope attached to the top of the elevator, the rope goes over a pulley attached to the top of the elevator shaft, and the guy is pulling on the other end of the rope with 300 N force?

There's no way that elevator is going up!!

If the guy pulls on the rope with 300 N force, the rope pulls upward on him with 300 N force. And, assuming the tension everywhere in the rope is 300 N, the rope also pulls upward on the top of the elevator with a force of 300 N. So, total upward force = 600 N.

The downward force is (30 kg + 70 kg) x 10 m/s2 = 1000 N.

Net force (downward +) = 1000 - 600 = 400 N.

a = f/m = 400/100 = 4 m/s2 (downward)

If you use g=9.8, the acceleration is 3.8 m/s2 (still downward)
 
  • #5
ant284
Hi,

now the answer is 5m/s upwards,
because the man insinde the elevator is pully himself up with the rope.

imagine: there is a box (elevator with a hole just to let through a rope).
now this rope passes through a pulley and is attach to the top of the elevator. inside the elevator a man is hanging on the rope, and is pulley, exerting a force of the 300N so both the man and the elevator is going upwards, now we need to find this acceleration upwards.
 
  • #6
1,036
1
Impossible!

Forget about formulas & think about this logically. The man is not even exerting enough force to lift himself, without an elevator. How could he possibly be lifting himself AND the elevator?

To accelerate upwards at 5 m/s2, he would have to pull on the rope with a force of 750 N. (The tension would have to be 500 N just to stand still.)
 
  • #7
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,500
7
Originally posted by gnome
Hey, I posted an answer to this a few days ago (OK - it was slightly "camouflaged" with a few extra z's, but all you had to do was search/replace the z's with spaces) but that entire post somehow got erased.

Down with censorship!!
I thought you were just banging on the keyboard!
 
  • #8
schwarzchildradius
are you guys forgetting the force multiplication from the pulley?
BTW with first method I recieve:
mg - T = ma
a = g-T/m = (10 m/s2 - 300N )/100kg) = ( (10)-(3) )=7 m/s2
(T/m = (car+man kg)*10m/s2)
 
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  • #9
1,036
1
are you guys forgetting the force multiplication from the pulley?
HUH???
You are forgetting the force multiplication from the pulley.
Also, you wrote
a = g-T/m
but then you did
a = (g-T)/m

(Doesn't matter, though, since both are wrong.)

You have to do
mg - 2T = ma
a = (mg - 2T)/m = ((70+30)*10 - 2*300)/(70+30)
a = 4
and remember that if g = 10 m/s2 downward, then
a = 4 means 4 m/s2 downward
 
  • #10
schwarzchildradius
Aha!
very simple:
∑Fx = 0
∑Fy = T - m1g = m1a
∑Fy = m2g - T = m2a
add 2nd equation to 1st:
-m1g + m2g = m1a + m2a
--> a=((m2-m1)/(m1+m2))g = ((300 - 100)kg/(300+100)kg)9.8m/s2=4.9 m/s2

EDIT
with g = 10 m/s2 a = 5 m/s2, just like thread author reported.
 
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  • #11
1,036
1
[zz)]

I'da thought the guy would lose weight from all that exercise. Yet somehow he seems to have grown from 70 kilos to 300, and even stranger, the elevator put on some weight too, increasing from 30 kg. to 100. Where'd you get those numbers?

Going back to the original numbers:

M1 = 30 kg. (elevator)
M2 = 70 kg. (man)
T = 300N

With those numbers, using your method,
T - m1g = m1a
m2g - T = m2a
a = g*(m2-m1)/(m2+m1)
a = 10*(70-30)/(70+30)
a = 4 m/s2
This might look like it agrees with my answer, but actually it's completely wrong. Notice that the way you set up the equations, up is the positive direction (since in your ∑Fy you're subtracting m1g instead of adding). So this answer implies that a is 4 m/s upward.

Now let's put this result back into your original equations. Now we have
T - m1g = m1a
300 - 30*10 = 30*4 => 0 = 120
and
m2g - T = m2a
70*10 - 300 = 70*4 => 400 = 280

What went wrong? Doesn't it strike you as strange that you can completely eliminate T from the calculation? Meaning that the acceleration is the same no matter how hard the man pulls on the rope?

You can set it up that way only if the man is not standing in the elevator. Then the man accelerates down and the elevator accelerates up, and T is determined by their relative weights, not by how hard the man wants to pull.

But in this problem, the man is standing in the elevator. You omitted from your equations the force that the elevator exerts on the man, and the force the man exerts on the elevator. These are obviously equal and opposite, and we don't know the magnitude of that force, so call it R.

Now the equations are:
T - m1g - R = m1a for the elevator
T - m2g + R = m2a for the man
Add them together. Now it makes sense that R disappears, since it is an internal force within the man/elevator system:

2T - m1g - m2g = m1a + m2a
a = ((2T - (m1+m2)g)/(m1+m2) looks familiar?
a = (600 - (30+70)*10)/(30+70) = -4 m/sec2 as previously claimed. We set up the equations with up being positive, so this acceleration is downward.

Now you can plug that back into either of the equations and find that
R = T - m1a - m1g = 300 -(-120) - 300 = 120N
R = m2a + m2g - T = -280 +700 - 300 = 120N
 
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