How to calculate the force needed to get out of bed?

[marciokoko]

I know from physics 101 that f = ma. And if I want to lift an object straight up i have to multiply its mass times gravity (acceleration) and I get the force I need to require to overcome it.

But what about lifting your body out of bed? Just the part where you physically get up from the lying down position to the sitting up position. In other words, just for lifting your torso up from horizontal to a vertical?

I know it has something to with the the angle but I am not sure how to incorporate it into f=ma.

Thanks

 

[ash64449]

My initial guess is on center of mass. It is a point where all the external forces acting on a body can be considered acting at that point.

Well, initially, center of mass has one position and after the process, it has another which is higher than usual.

So, you need the same force mg to lift you off the bed.

I am not so sure about this treatment.

 

[mfb]

Your body is not a single point, so giving an answer as a single force value is problematic.
If you model the torso as rigid object that pivots around the hip, you can calculate a force. Initially the whole torso will move directly upwards, which allows to calculate the force as you described. That is not the same as forces inside your body to lift you up, as you have to include lever arms and other issues there.

 

[phion]

It might make more sense to calculate the energy required to get out of bed, considering the biological and chemical aspect of waking the **** up.

 

[marciokoko]

No I am not interested in Energy at all.

Lets imagine that you don't even use the arms. Let's just consider lifting the mass out of bed.

Even if its a rough calculation, how would I go about doing it?

 

[mfb]

Make a model of what moves where and how the energy changes as function of the position, force (in the direction of motion) is then the derivative of this.

 

[marciokoko]

Ok here are my drawings. 4 scenarios.

A: Simplistic F = ma of lifting an object up from the ground.

B: Force at an Angle = Not sure how but if the same object was being lifted up, but at an angle, the force would be different? Like in those examples of a horse pulling a cart vs the same horse pulling that same cart but from a sideroad, at an angle. I kinda remember we needed to consider the angle somehow.

What I really want to understand is the getting out of bed example.

C: Getting up out of bed requires you to lift a weight from 0 angle to 90 degrees (sitting up). This requires a certain force.

D: I want to udnerstand the basic C scenario in order to understand how that would change when inclining the bed, so the person needs to sit up, once again to 90 degrees from the bed, but this time the bed is inclined 20 degrees.

 

[CWatters]

C: Getting up out of bed requires you to lift a weight from 0 angle to 90 degrees (sitting up). This requires a certain force.

Actually it involves rotating a mass from the horizontal to the vertical position so it requires a torque rather than a force. The torque is at a maximum when the trunk is horizontal (just starting to rotate off the bed.)

 

[marciokoko]

Thanks CWatters.

So I ran into this online:

http://hyperphysics.phy-astr.gsu.edu/hbase/torq2.html

Im having trouble understanding what everything is in this diagram. The red line moves thru theta (dunno how to make that on the keyboard of my mac), to where the dotted black line is. I am guessing that's my body.

F is the Force moving that lever. Except that in my case, to that force I will have to add the force of gravity as well, right? because in that diagram I think its considered as being on a flat surface, like twisting a monkey wrench.

How would I enter the values?

 

[CWatters]

According to...
http://www.exrx.net/Kinesiology/Segments.html

Your legs and feet account for about 17% of your mass. Perhaps it would be reasonable to assume that the part being rotated (trunk, head, arms etc) accounts for the other 83%.

Lets say you weigh 100Kg then 83% is about 83kg. If the centre of mass of the moving part is say 0.4m from the pivot point then the required torque is 83 *9.8 * 0.4 = 325Nm.

If you don't like my assumptions you can easily change them.

PS: The above implies something needs to hold your feet down.

 

[CWatters]

Im having trouble understanding what everything is in this diagram. The red line moves thru theta (dunno how to make that on the keyboard of my mac), to where the dotted black line is. I am guessing that's my body.

No the red line doesn't move through theta. The diagram is showing how to calculate the length of the "lever arm" when it's not at 90 degrees to the applied force.

I'll make a diagram that shows what I meant in post #10. Back soon.

 

[marciokoko]

Ok but that website also has a calculator at the end for force, i think.

I did the following, although I like your assumptions better:

I weight 72 kg, so that's = 72 * 9.8 = 705N.

I considered 50% weight in the torso (yours seems more logical), so I went with a rough 350N.

Force Applied 350N
Distance: 1.7m (my height)
Theta: 90 degrees

I got stuck calculating the lever arm. For some reason the online calculator sets my lever arm automaticaly to 1.7m and gives me a value of 595 Nm.

 

[CWatters]

Ok so this shows how to calculate the max torque.

Torque = force * length of lever arm.

As you sit up the length of the lever arm reduces until the centre of mass is over the pivot point. The length of the lever arm and the torque is at a maximum when you are horizontal as shown.

 

[CWatters]

Distance: 1.7m (my height)

That would imply a pivot point at your feet.

 

[russ_watters]

PS: The above implies something needs to hold your feet down.

Edit: deleted, think I missed your point. You were referring to the torque being great enough to lift your feet instead of your torso. Agreed.

 

[marciokoko]

Yes CWatters, so changing 1,7 to 0.85 I get 297.5Nm.

I know I am leaving out the "someone holding your feet down" counter force and the fact that the lever arm changes.

So but is that correct? I guess it makes sense that you require less force to lift 1/2 your body up to a sitting position that to lift your entire body.

Now I have 2 lingering questions:

A/ If the bed was at an incline, as in scenario D from my image. What would change is the angle, from 90 to 90+20, correct?

B/ If I wanted to consider the changing length of the lever arm as I sit up, I would need to use calculus?

 

[CWatters]

Minor issue, but no it doesn't. That's both the torque required to hold your body rigid and horizontal or to rotate at constant speed. The torque to start the rotation is an arbitrarily small amount larger.

Sure. What I meant is... If 83% of your mass is above the pivot point (your hips) then your feet will lift up rather than your head.

 

[CWatters]

Yes CWatters, so changing 1,7 to 0.85...

0.85 is on the high side. It should be the distance from the pivot to the centre of mass of the moving part. If the pivot point is at your hips (as in doing sit ups) then I would expect the centre of mass of the top half of you body to be around chest height (hence my estimate of 0.4).

 

[CWatters]

Now I have 2 lingering questions:

A/ If the bed was at an incline, as in scenario D from my image. What would change is the angle, from 90 to 90+20, correct?

Yes. However since gravity acts vertically the torque will still be at a maximum when the trunk is horizontal. eg raising the legs 20 degrees won't affect the max torque required.

B/ If I wanted to consider the changing length of the lever arm as I sit up, I would need to use calculus?

Basic trig might be all you need. For example if you lift up to the 45 degree position the lever arm is 0.4 Cos 45 = 0.28 and the torque falls to 83 * 9.8 * 0.28 = 227 Nm.

 

[marciokoko]

Ok I didnt understand either of your answers. I am slow. :-)

A/ I didnt mean raise just the legs. I meant that the entire bed, instead of being flat (0 degrees), the bed is now inclined 20 degrees. Its like making situps in a gym, where you have those boards that you can incline to make the situp workout harder.

How does that factor in? I thought it just meant the torque angle was greater by 20 degrees?

 

[sophiecentaur]

No I am not interested in Energy at all.

Lets imagine that you don't even use the arms. Let's just consider lifting the mass out of bed.

Even if its a rough calculation, how would I go about doing it?

If you had a rope and a pulley, the only force needed would be your weight force - that's eeeaasy!

 

[sgphysics]

Early in the morning it appears to require more force to get out of bed than later in the day, an observation independently confirmed by several people. A possible explanation is a temporal increase in gravity during night/morning, which then is restored to normal level in the day. Hard to calculate.

 

[Drakkith]

Early in the morning it appears to require more force to get out of bed than later in the day, an observation independently confirmed by several people. A possible explanation is a temporal increase in gravity during night/morning, which then is restored to normal level in the day. Hard to calculate.

I expect that caffeine must have certain 'antigravity' properties then.
I wonder what @OmCheeto has to say about this peculiar effect...

 

[CWatters]

A/ I didnt mean raise just the legs. I meant that the entire bed, instead of being flat (0 degrees), the bed is now inclined 20 degrees. Its like making situps in a gym, where you have those boards that you can incline to make the situp workout harder.

Yes I understood that.

The peak torque required is still the same and it still occurs when the trunk is horizontal (see new diagram).

I agree it is harder to do sit ups that way. That's because now you start with your head lower and have to do work just to get to the horizontal position.

 

[OmCheeto]

I expect that caffeine must have certain 'antigravity' properties then.
I wonder what @OmCheeto has to say about this peculiar effect...

A. I'm no longer allowed to get out of bed like a young person. My physical therapists spent 15 minutes explaining how old people are supposed to get out of bed, to prevent applying excess torque on the lower back.
B. I've doubled my upper body mass, and halved my lower body mass, over the last 366 days, so this problem has no solution, in my case.
C. See below: "Proper method for old people to get out of bed"

But in all seriousness, mfb implied the correct solution in post #3: The force is continuously variable.
Expanded later by CWatter in post #8.
So, IMHO, it is a cantilever problem.
And since "Statics" was the most boring college course ever invented: "The study of things just sitting there", and caused me to fall asleep within the first 5 minutes of class, causing me to fail the class 3 times, at which point I purchased some toothpicks, which allowed me to finally pass the class, allowing me to progress to "Dynamics", which I aced by the way, I would fall asleep if I actually tried to solve this problem.

Proper method for old people to get out of bed:
1. Roll to the edge of the bed.
2. Face the edge of the bed, with your body in a sitting position. Knees, calves, and feet, just at the edge.
3. Prepare your lower arm to lift your torso, as things are going to happen quickly.
4. Move those parts of your body below the knees across the ledge of the bed, which will cause gravity to pull them down, simultaneously pushing with your arm, elevating your torso to a vertical position, thus minimizing torque on your lumbar discs.
5. Get caffeine.

ps.

According to...
http://www.exrx.net/Kinesiology/Segments.html

Your legs and feet account for about 17% of your mass.

Even though you provided a grand reference, I think you inappropriately used the term "legs and feet", as your reference says they only account for 6.43% of total body mass.
I learned that today, a while back.

Today I learned that we have two kinds of "legs".

 

[marciokoko]

Ok so wait. Does it require more force to get out of a 20 degree inclined bed? Why is it harder, actually impossible to do so?

 

[CWatters]

I believe the peak force or torque is the same. It just takes more work.

 

[OmCheeto]

Ok so wait. Does it require more force to get out of a 20 degree inclined bed? Why is it harder, actually impossible to do so?

As has been implied several times before, the torque, and therefore force, change constantly.
This is why I despise wordy solutions to wordy problems. It takes forever to describe, what can be explained with a simple diagram, and an equation.

Torque = Force on the center of gravity on my upper body(constant = 160 lbf) * lever arm length(constant = 1 foot) * cos(θ)
The lever arm length being the distance from my hip joints to my center of gravity.
θ = 0° when laying down
θ = 90° when sitting up
The force you are looking for, to generate the torque, to counterbalance your upper body weight, changes in both magnitude, and direction, as you move.

Given my weight is 160 lb, and my legs are now, a negligible mass:

@ -20°, the cantilever solution for this angle indicates that the Om-extracto-crane attached to my center of gravity is applying 150 lbf.
continuing:
@ 0°: 160 lbf
@ 20°: 150 lbf
@ 40°: 123 lbf
@ 60°: 80 lbf
@ 80°: 27.8 lbf
@ 90°: 0 lbf

So, it is not impossible. It simply requires a crane.

Of course, as you mentioned in your original question: f=ma
So there will be an extra force, at the beginning, to get your butt out of bed.
Very minor, in my case.

 

[sophiecentaur]

This puts me in mind of the Wallace and Grommit film "The Wrong Trousers" in which Wallace's bed tips up and he slides down through the floor into the ground floor and is poured into his trousers on the way. Zero work needed. Just using his GPE from being upstairs.

 

[marciokoko]

Wait, so is it that the force is the same but the work done is greater?

Or is the force greater?

 

[CWatters]

Think that was what I said in #24.

The peak torque or force is the same. The work done is greater.

 

[marciokoko]

Ok and why is the work done greater? Is it simply because the "distance" covered is now 90 degrees + 20 = 110 degrees total?

Or does the inclination or something else have anything to do with it?

IOW, does it take the same amount of work from degree -20 to degree 0 as it does from degree 0 to degree 20?

 

[sophiecentaur]

Ok and why is the work done greater? Is it simply because the "distance" covered is now 90 degrees + 20 = 110 degrees total?

Or does the inclination or something else have anything to do with it?

IOW, does it take the same amount of work from degree -20 to degree 0 as it does from degree 0 to degree 20?

Why do you want to treat this in such an arm waving way? If you tried to treat your finances in this way you would soon end up with no money. It's fairly elementary mechanics and you should just do the proper sums. It's not a matter of 'opinion' and 'the way its put'; it's just a straightforward calculation.

 

[marciokoko]

Sophie,

If its a straightforward calculation to calculate the difference between getting up from an incline of -20 vs from a 0 degree flat board, then what is that simple calculation? Can you show me?

 

[sophiecentaur]

A question that's well formed will often almost answer itself (that's a comment frequently made on PF).
If you draw a diagram of exactly the situation that you are interested in and mark the forces involved, you can use the few basic 'statics' equations (such as the principal of moments and the definition of Work) to tell you what you want.
In the case of getting up out of bed, the model needs to be drastically simplified if you want to get anywhere. Say exactly what the before and after situations are and you can determine the Work done. (mgh would be a start, where h is the height by which your cm is raised).
Whatever you do and to whatever level you want the analysis, you have to specify the problem. Only the questioner can do that (with a bit of guidance, possibly). So start at the very beginning and build ( that means draw, probably) a model to describe the question. . . . . . . . .