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paulb203

Member

Registered: 2023-02-24

Posts: 405

Vectors, +/- signs

90kg man in a lift

Assigned signs; DOWN is MINUS; UP is POSITIVE

Acc. due to g = 10m/s/s DOWNWARDS
Acc. due to g = -10m/s/s

Fg=ma
Fg=90(-10)
Fg=-900

*

acc. due to lift’s motor =2m/s/s DOWNWARDS
a=-2m/s/s

F=ma
F=90(-2)
F=-180
Fnet=-180

*

What is the normal force?

Fn-Fg=Fnet
Fn-(-900)=Fnet
Fn+900=-180
Fn=-180-900
Fn=-1080?

But the normal force is always is in the UPWARDS direction, which given the assigned +/- in this example, is +; The Fn should be POSITIVE.
Where have I gone wrong?

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"The secret of getting ahead is getting started."
Mark Twain

Bob

Administrator

Registered: 2010-06-20

Posts: 10,753

Re: Vectors, +/- signs

I find 'man in lift' problems take some thinking about.  I'd like to chat about the general principles before taking on your problem.

If the lift isn't moving then 10 m/s/s is all there is and the reaction upwards opposes -10 x 90 so 900N up.

But reactions are not always up as the following will hopefully show.

Suppose the lift cable breaks and the lift tumbles down with only gravity acting. The man is now in free fall, effectively weightless and he could float off the floor so the reaction of the floor on him is zero.  If we compare with a parachute drop, the wind rush tells us we're falling. In our imaginary lift drop the air moves with the man so he'd have no way of telling that he wasn't able to float.

Now if we imagine a lift that is so powerful it can accelerate downwards at 20 m/s/s. The lift would drop faster than the man because he's already floating up, until he bangs his head on the ceiling. The reaction of the ceiling on him is definitely downwards.

But how much?  He is trying to fall at 10 m/s/s but the lift is pushing him at 20.  Effectively the force on him is 90 x (20-10) downwards.  So you need to subtract the lift's acceleration from gravity to determine the effective acceleration in the frame of reference.  It's like the moving train frames of reference.  If you 'bring the lift to rest' by applying an upwards acceleration of 2 m/s/s the gravity is similarly given that upwards acceleration so -10 + 2 = -8 m/s/s.

So I think the reaction (upwards) is ( 10 - 2 ) x 90 N

In really fast lifts you will notice the effect as your internal organs rise up inside your body.  Many years ago I stayed in the Luxor hotel in Vegas. The hotel is in the shape of a square based pyramid and the lift shafts follow the sloping line so not ony do you notice the stomach effect but also you tend to sway inwards because the motion is no longer just up/down but includes a sideways component. Really strange but then it was Vegas.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

paulb203

Member

Registered: 2023-02-24

Posts: 405

Re: Vectors, +/- signs

Suppose the lift cable breaks and the lift tumbles down with only gravity acting. The man is now in free fall, effectively weightless and he could float off the floor so the reaction of the floor on him is zero.

Thanks, Bob. When you say ‘effectively weightless’ are you referring to his apparent weight as opposed to his true weight (terms I’ve just come across)?
So if the cable breaks his apparent weight, the weight he ‘feels’(?), the weight he ‘seems’ to be(?); but his true weight his what it always is regardless of other factors, like acceleration, i.e, the gfs of 10N/kg times his mass of 90kg?

If we compare with a parachute drop, the wind rush tells us we're falling.

I’m told this isn’t actually freefall because of the force of friction involved; is that correct?

In our imaginary lift drop the air moves with the man so he'd have no way of telling that he wasn't able to float.

So the air, being matter, stuff with mass, accelerates too, at -10m/s/s?
You said he’d have no way of telling that he wasn’t able to float; so he isn’t able to float?

Now if we imagine a lift that is so powerful it can accelerate downwards at 20 m/s/s.

Do you mean 20m/s/s on top of the 10m/s/s due to gravity? Or 20m/s/s in total (10 due to gravity, 10 due to it’s motor/whatever)?

The lift would drop faster than the man because he's already floating up, until he bangs his head on the ceiling. The reaction of the ceiling on him is definitely downwards.

Ah, so he is floating, floating up. Why? Why isn’t he being pulled down at 10m/s/s due to the gfs?

But how much?  He is trying to fall at 10 m/s/s but the lift is pushing him at 20.

10 due to the gfs, 10 due to the motor/whatever?

Effectively the force on him is 90 x (20-10) downwards.  So you need to subtract the lift's acceleration from gravity to determine the effective acceleration in the frame of reference.

The force on him from the ceiling is 90(20-10)?

It's like the moving train frames of reference.  If you 'bring the lift to rest' by applying an upwards acceleration of 2 m/s/s the gravity is similarly given that upwards acceleration so -10 + 2 = -8 m/s/s.

I think I’m partly getting that, but I’ll need to try harder to grasp it better.

So I think the reaction (upwards) is ( 10 - 2 ) x 90 N

I’m not quite sure I’ve followed the above to get here but 10-2(90)=720 which I think was the correct answer to my original question.

In really fast lifts you will notice the effect as your internal organs rise up inside your body.

Rise up in the reference frame of your body? From the reference frame of a viewing platform would your organs remain stationary for a moment while you plummet, like a gory episode of The Road Runner?

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"The secret of getting ahead is getting started."
Mark Twain

Bob

Administrator

Registered: 2010-06-20

Posts: 10,753

Re: Vectors, +/- signs

I'd not come across this apparent/true weight dichotomy. I've always thought of it this way.  If you take a bathroom weighing device with you into the lift and stand on it then what does it record?  If you're in free fall, then the scale shows zero because your weight exerts no downward force on the scale.  It's worth googling "apparent and true weight" and the AI bot gives a good explanation ... that is until it gets to the bit about "fictitious force ".

What I meant was the bathroom scale he'd thoughtfully taken with him shows zero.

Parachute and 'free fall'.  Probably what you say is  correct .  Next time I jump out of an airplane I'll try to remember to take my bathroom scales with me.

The air in a lift falls the same as the man so it doesn't rush past his face.

I definitely meant 20 on top of 10 due to gravity.  If the lift shaft is open and quite tall too, and a stone is dropping next to it just under the effects of gravity (I'll disregard any air resistance here as it's small compared with the mass of the stone ... whoops I think I've just mixed units here). then the lift will overtake the stone.

He is still under the influence of gravity but the lift is accelerating faster than him which is why the ceiling catches up with him and gives him a bonk on the head.  Once the downwards reaction of the ceiling can exert a force on him he will start to accelerate at the same rate as the lift.

like a gory episode of The Road Runner?

  But it's well known that cartoon creatures don't obey Newton's laws at all.  When they fall off a cliff they spend a few seconds hanging in the air and only start to drop when they become aware of their predicament

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob