How to Calculate the force to push a wheelchair up a ramp?

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SUMMARY

To calculate the required push force for moving a wheelchair up a 5-degree ramp, one must consider the total weight of 96.8 kg and the rolling coefficient of friction of 0.1. The primary forces to overcome are the gravitational component parallel to the ramp and the rolling resistance of the wheels. Using the formula F = m * g * sin(θ), the estimated force required is approximately 82.7 N (or 20 pounds) when assuming constant velocity and negligible rolling resistance.

PREREQUISITES
  • Understanding of basic physics principles, specifically Newton's laws of motion
  • Familiarity with trigonometric functions, particularly sine for angle calculations
  • Knowledge of rolling resistance and its impact on motion
  • Ability to perform unit conversions between Newtons and kilograms-force (kgf)
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  • Research the calculation of forces on inclined planes using Newton's second law
  • Learn about rolling resistance and its effects on different types of wheels
  • Explore the impact of different ramp angles on required push force
  • Investigate wheelchair design features that minimize rolling resistance
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Engineers, healthcare professionals, and caregivers involved in wheelchair design and mobility assistance, as well as individuals interested in understanding the physics of inclined motion.

TommyH
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TL;DR
How do I calculate the required push force (in kgf) for pushing a wheelchair up a ramp?
- the ramp has a 5 degree incline
- the weight of the chair and person in it totals 96.8 kg
- the wheelchair has 4 wheels, all of equal diameter (15 cm)
- the wheels on the wheelchair have a rolling coefficient of friction with the floor of 0.1

I am really stuck on this. Any help would be greatly appreciated thank you
How do I calculate the required push force (in kgf) for pushing a wheelchair up a ramp?
- the ramp has a 5 degree incline
- the weight of the chair and person in it totals 96.8 kg
- the wheelchair has 4 wheels, all of equal diameter (15 cm)
- the wheels on the wheelchair have a rolling coefficient of friction with the floor of 0.1

I am really stuck on this. Any help would be greatly appreciated thank you
 
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We will try to work you through this.

Can you tell me what two forces need to be overcome before the wheel chair will begin moving up the ramp?
 
TommyH said:
I am really stuck on this.
Have you tried googling "forces on an incline"?
 
.Scott said:
We will try to work you through this.

Can you tell me what two forces need to be overcome before the wheel chair will begin moving up the ramp?
Normal force and frictional force?
 
The frictional force requires knowing the normal force - the force normal (perpendicular) to the surface of the ramp.

How do you calculate that normal force?

The first force I was considering was the frictional force. But there is a second force that is very important. It's the one that you will need to compute the normal force.

Also, by all means, take @A.T. 's advice and Google "forces on an incline". You will get to pick one of many sketches of the problem.
 
TommyH said:
Normal force and frictional force?
I don't think either of those is relevant. Friction between the wheels and the ramp doesn't contribute to a retarding force (unless the wheelchair's brakes are on, which would make no sense). The component of the weight of the wheelchair in the direction parallel to the ramp is one force, and the "rolling resistance" of the wheels is the other force that you need to overcome. The rolling resistance consists of the bearing friction for the wheels and any deformation of the wheels at the contact patches. Are the wheels pneumatic or solid rubber?

Are you familiar with trigonometry and algebra? We can show you how to calculate the force to overcome the component of the weight that is parallel to the ramp...

Also, will the pushing be done by a helper behind the wheelchair (hence a fairly continuous force at a constant speed), or by the person sitting in the wheelchair (so it will be in spurts)?
 
berkeman said:
Friction between the wheels and the ramp don't contribute to a retarding force.
The problem describes a "rolling coefficient of friction" with the floor.
So friction is (half of) a good answer.
Also, it is a unit-less value.
 
.Scott said:
The problem describes a "rolling coefficient of friction" with the floor.
So friction is (half of) a good answer.
Also, there are no units for that coefficient.
Is that meant to represent rolling resistance? The number did seem low for a coefficient of static friction...
 
berkeman said:
Is that meant to represent rolling resistance? The number did seem low for a coefficient of static friction...
That's how I took it.
 
  • #10
TommyH said:
Summary: How do I calculate the required push force (in kgf) for pushing a wheelchair up a ramp?
- the ramp has a 5 degree incline
- the weight of the chair and person in it totals 96.8 kg
- the wheelchair has 4 wheels, all of equal diameter (15 cm)
- the wheels on the wheelchair have a rolling coefficient of friction with the floor of 0.1

I am really stuck on this. Any help would be greatly appreciated thank you

Perhaps one reason you are stuck is that the problem conflates 'force' and 'velocity'. That is, since a resultant *acceleration* of the wheelchair is not provided, you are most likely assuming the wheelchair moves at a constant velocity. So, if the wheelchair moves at constant velocity, you can use a reduced form of Newton's second law (or two N2s in combination: one for linear motion and the other for rotational motion, connected by the assumption of rolling motion) to figure out what the applied force must be.
 
  • #11
Well, let's give the OP an initial ballpark estimate of the force. Hopefully that helps some.

All of the wheelchairs I've pushed in healthcare settings have solid rubber wheels, and very little rolling resistance. A 5 degree incline is pretty gentle, but let's assume that the rolling resistance is a negligible retarding force. Let's also assume that a helper is pushing the wheelchair at a constant velocity. Then the force required is just the component of the weight in the direction parallel to the incline:

F = m * g * sin(5 degrees) = 96.8kg * 9.8 \frac{m}{s^2} * sin(5 degrees) = 82.7N

82N is about 20 pounds of force. Does that help?

https://www.researchgate.net/profile/Catherine_Holloway/publication/267995686/figure/fig4/AS:669371231920138@1536601946564/5-Forces-acting-on-a-wheelchair-going-up-left-and-down-a-slope-W-is-the-weight-of.png

1566343254852.png
 
  • #12
BTW, slightly off topic -- If there is an Abilities Expo near you, try to stop by and check it out. I worked a recent one locally on EMS standby, and I was very impressed by the wonderful uses of technology to help folks with disabilities. I won't tell you which exhibit brought me to tears, but there are plenty of great examples of human creativity to help overcome the challenges of disabilities.

https://www.abilities.com/expos/
:smile:
 
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