Additional Force Required for Inclined Planes

[shobin]

I'm trying to calculate what additional force would be required if a load was to be moved up an inclined plane of 6.5 degrees instead of simply moving it laterally.

My question actually relates to "turning" an object up an inclined plane but I really just want to understand the % increase in required force if a gradient is introduced.

The load is 250kg. If there is a recognised formula to calculate this I would appreciate it if someone could let me know. Yours hopefully.

Stuart

 

[tiny-tim]

Welcome to PF!

Hi Stuart! Welcome to PF!

I'm trying to calculate what additional force would be required if a load was to be moved up an inclined plane of 6.5 degrees instead of simply moving it laterally.

My question actually relates to "turning" an object up an inclined plane but I really just want to understand the % increase in required force if a gradient is introduced.

The load is 250kg. If there is a recognised formula to calculate this I would appreciate it if someone could let me know. Yours hopefully.

Stuart

I don't really understand your question … moving a load laterally (horizontally) does not require any force, if there's no friction.

Anyway, to find the forces in any situation, find the components of all the forces in two perpendicular directions (horizontal and vertical, or parallel and normal to the slope), add them (for each direction separately), and set the two sums equal to zero (or to mass times acceleration if there is acceleration in either direction).

 

[shobin]

Ok, I'm not a Physics guy so I'm going to tell you my objective in the hope you may be able to suggest how I can calculate something.

If you have a person lying on their back on a bed and you turn them 90 degrees so that they now face you (side lying), how do you calculate the force (effort) required to complete that task if the person weighs 250kg ?

The second part of the question is how would you calculate the additional force (effort) required to repeat the same task up an inclined plane of 6.5 degrees.

Thanks for the help.

 

[tiny-tim]

oh, that's what you meant by "turning" …

ok, turning a patient on a horizontal bed (90º, from back to side) will raise his/her centre of gravity a few inches.

That requires energy, equal to mgh where m is the mass, g is "gravity", and h is the height difference (a few inches).

But turning a patient "uphill" on a sloping bed will raise his/her centre of gravity the same few inches, plus the extra height that the body has moved uphill, and so will require more energy.

The proportion of extra energy is the same as the proportion of the two heights.

(Conversely, turning a patient "downhill" on a sloping bed will raise his/her centre of gravity the same few inches, minus the height that the body has moved downhill, and so will require less energy.)

Are you a nurse or carer, asking about a particular actual situation?​

 

[Cleonis]

If you have a person lying on their back on a bed and you turn them 90 degrees so that they now face you (side lying), how do you calculate the force (effort) required to complete that task if the person weighs 250kg ?

The second part of the question is how would you calculate the additional force (effort) required to repeat the same task up an inclined plane of 6.5 degrees.

Again, in a real life situation the main factor will be friction, in one form or another, and I wouldn't know how to estimate the friction.


I will now say some things about lifting in general, I'm aware they will not be of practical value to you.

An elevator, such as the ones in high rise buildings, uses a counterweight. The counterweight is somewhat heavier than the empty elevator cabin, to get a good match when the cabin has the usual passenger load.
Imagine a situation where there is an exact match, and all of the mechanical parts operate without friction. Then a minute upward force upon the elevator cabin would suffice to make it go up.

Of course, a minute force would only impart minute velocity to the cabin. In actual elevators the force that the elevator motors exert is not so much for lifting the elevator cabin, but for accelerating the cabin/counterweight assembly.

If lifting a patient is the only requirement then the force that you need is pretty much equal to weight of the patient. Additional required force to set into motion is very small compared to that.

Same story for an inclined plane. If friction is zero, then you need to exert force just to prevent sliding down. Just a smidgeon of extra force would make the weight slide up.

All this brings me back to my first remark. The biggest problem is friction, in one form or another. That factor raises the required force significantly beyond the weight of the patient itself, but I wouldn't know by how much.

Cleonis

 

[shobin]

OK guys. Thanks for that. I work for a company that sells hospital beds and one of our competitors has launched a bed that instead of providing a flat surface for the patient, it curves down from the edge of the bed (hence the 6.5 degree difference between the edge of the bed and the bottom of the curve).

Therefore, if you were to turn the patient 90 degrees towards you (as though you were trying to remove a hospital sheet or inspect a dressing) then you would have to turn them up the 6.5 degrees slope.

The maximum permissable weight of the patient is 250kg.

It is widely accepted that carers experience musculo-skeletal disorders as a result of repetitive patient handling and I am trying to identify what the increased force requirements are if a 250kg patient has to be turned up an incline of 6.5 degrees as opposed to 0 degrees (their bed versus our bed).

I was hoping that their may be a physics formula that would easily explain the difference eg Force (Newtons) = mass * acceleration / inclination

I'm not a Physics guy as I said and I'm sure that the above probably sounds stupid but I need to try to explain to our customers how this inclined plane can increase the force required to handle patients and potentially increase handling risks.

Ideally, I would like to reach a conclusion where I could present a non-contestable equation which demonstrates that movement up a 6.5 degree slope represents an increased requirement in force of x%.

If you can help me out I'd be really grateful - let me know if you need any more detail.

Thx a million

Stu

 

[Cleonis]

I need to try to explain to our customers how this inclined plane can increase the force required to handle patients and potentially increase handling risks.

A very, very crude model is to compare with the case of rolling a barrel.

- Rolling a hard barrel over a hard horizontal surface requires a negligable amount of force.
- Rolling that same hard barrel up an hard surfaced incline of 6.5 degrees: then just to prevent the barrel from rolling back the force you need to exert is 11% of the barrel's weight.

If an object is located on an incline you can think of the force of gravity as decomposed in a component perpendicular to the inclined surface, and a component parallel to the surface.
The component parallel to the surface (the one that tends to make the barrel roll down) is given by the weight of the barrel, multiplied with the sine of the angle. The sine of 6.5 degrees is 0.11

Rolling an actual human over a mattress requires force in itself, but I do expect that the additional force requirement, to deal with the incline, will be pretty much the same in both cases. (Then again, if the patient can assist, by moving his/her arms/legs to prevent rolling back, then it's another story.)

Cleonis