Torque required to lift a given mass around a point.

[Daz50]

Homework Statement

Dear Physics Forums, I was hoping you may be able to help with a problem I am having in my final postgraduate project.

A short background of the project to put the problem into context - I am designing and manufacturing a form of powered orthosis to help a person move from a sitting position to a standing position.

I am currently trying to calculate the forces involved in the sit-to-stand tranition. For simplicity at this stage, my image shows a very basic model I created of a mass on the end of a moment arm.

The attached file shows an image of this simple model.

The mass, M (90kg) is designed to represent a person's bodyweight, the length L (65.8cm) is the buttock-knee length of a person and the pivot point O, represents the knee joint. Anything below the knee is disregarded at this stage. The mass M is required to travel 90degrees in 1.5 seconds, finishing at M1 (the green dashed line) assuming a constant angular velocity. It is also on a vertical plain as this is supposed to model someone standing up.

Homework Equations

Torque = I Alpha (The equation I think I should be using but I am not sure how)

Torque = mass x Alpha x L² (The equation I actually used but I think is wrong)

The Attempt at a Solution

My working so far is as follows:

1) convert 90degrees into radians 90 / (pi/2) = 1.575 radians

2) Work out how many radians per second 1.575 radians / 1.5 seconds = 1.05r/s

3) Factor in the mass M, and the length, L into the equation.

90kg x 1.05rad/s x 0.658m² = 40.9Newton Metres.

I have a calculated value of 40.9Newton metres to move the 90kg mass through 90degrees, however I am not sure how to factor the effects of gravity into this equation as this would certainly have an impact on the torque required to move M to M1.

Apologies if this post is a bit all over the place - I do not have a physics background because I came from a different undergrad degree discipline so am trying my best!

Any assistance would be appreciated!

Kind regards,

Daz

 

[Villyer]

A torque causes angular acceleration (alpha).
Hence if you want constant angular velocity, your torque would be zero.

A person sitting has no angular velocity, so in order to stand they have to use their muscles to produce a torque on their leg that accelerates it to some angular velocity, and then decelerates it back to stationary when the person is upright. I guess this can be done either through the natural torque produced by gravity, or by a torque produced my your muscles on the other side of your leg. Having muscles stop you is a more precise way of doing it, since the torque produced by them isn't constant and can be adjusted.

 

[Daz50]

Hi Villyer,

Thanks for clarifying a few things. Obviously I don't want the person to have no angular velocity - otherwise they're staying sitting!

I agree with you though about the stopping method. There is some counter activation by the opposing muscles (the hamstrings) according to my research so it makes sense to suggest that they play a role in bringing the legs to a stop. As well as the natural joint range of the knee.

Could you suggest a rudimentary way to model this?

Thanks,

Daz

 

[Villyer]

The first step would be to figure out how you want the leg to move.
I would map out over what angles you want to be accelerating, when angles you want to use constant angular velocity, and when you plan on decelerating.

 

[asteriatic]

Hello Daz,

First of all, congratulations on your postgraduate project and thank you for reaching out to the Physics Forums for help.

To answer your question, the equation you should be using is indeed Torque = I*Alpha, where I is the moment of inertia and Alpha is the angular acceleration. In your case, the moment of inertia would be that of a point mass, which is simply the mass (M) times the distance from the pivot point (O) squared, or I = ML^2. This means that the torque required to lift the given mass around the pivot point would be Torque = ML^2*Alpha.

To factor in the effects of gravity, you would need to consider the weight of the mass (M) and the distance from the pivot point (O) to the center of mass of the mass (M). This can be represented by a new variable, d, which is the distance from the pivot point (O) to the center of mass. So the new equation would be Torque = Mgd*Alpha, where g is the acceleration due to gravity.

In your case, since the mass is being lifted vertically, the distance (d) would be equal to the length (L) of the moment arm. So the final equation would be Torque = MgL*Alpha.

I hope this helps and clarifies any confusion you had. Good luck with your project!

Best regards,
Physicist

 

[asteriatic]

I would like to commend you on your thorough and well-thought-out approach to your project. It is clear that you have put a lot of effort into understanding the forces involved in the sit-to-stand transition.

To address your question about factoring in the effects of gravity, you are correct in thinking that it would have an impact on the torque required to move the mass. In order to calculate the total torque required, you would need to include the torque due to gravity as well as the torque due to the angular acceleration. The equation you used, Torque = mass x Alpha x L2, only accounts for the torque due to angular acceleration.

To incorporate the torque due to gravity, you would need to use the equation Torque = mgLsin(theta), where m is the mass, g is the acceleration due to gravity, L is the length of the moment arm, and theta is the angle between the moment arm and the direction of gravity. In your case, theta would be 90 degrees since the mass is being lifted vertically.

To calculate the total torque required, you would add the torque due to gravity to the torque due to angular acceleration. So, the equation would be:

Total Torque = mgLsin(theta) + mass x Alpha x L2

I hope this helps clarify your calculations. Best of luck with your project!