How do I work out the maximum amount of force someone could exert to propel themselves backwards on a chair whilst sitting on it

Feb 2020
1
0
Glasgow
I would like to find the maximum force that someone could exert backwards whilst sitting on a chair with their feet on the ground. I have values for the maximum force that someone can exert to jump whilst standing but I would like to know this from when sitting down.
 
Apr 2015
1,227
356
Somerset, England
It isn't as simple as that.

Maximal forces are impulsive, but of very short duration
 
Jun 2016
1,358
689
England
It is not entirely clear from the Original Post,
but it seems that you are wanting to estimate the force that can be exerted by a persons legs from a sitting position
given knowledge of the force they can exert while jumping (from a standing position).

Similar to estimating (for example) how good someone might be at rowing given their performance at long jump.

While there will be correlations, I don't think they will be direct or simple.
I don't think that you will be able to work out an equation based on the purely geometrical considerations.
There will be too many individual circumstances to make a simple comparison.
 
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Oct 2017
676
348
Glasgow
Mechanics problems involving the human body are usually very difficult because the ability for muscles to lift or push is often not constant as a function of time or as a function of leverage angle (for things like flexing). Such considerations have a big impact on the problem. For example, the optimal "throw-in" angle in football is closer to 30 degrees rather than usual 45 degrees obtained from standard ballistics calculations.

However, in the hopes of getting a back-of-the envelope estimate, you can instead make some rather severe assumptions, such as:

1. The pushing force on the floor is vertical and instantaneous, resulting in an initial upwards velocity, \(\displaystyle u\)
2. The rear legs of the chair are fixed to the ground (i.e. are a pivot point for rotation)

You can then solve a moments problem to figure out the magnitude of u required to get the man + chair to topple over rather than fall back to the sitting position.

Example:

Set origin to be the position of the pivot point (where the rear chair leg touches the ground) = (0, 0)
Distance between rear and front legs of chair = W
Initial position of front chair leg where it touches the ground = (W, 0)
Weight of person + chair = -Mg (upwards is positive)
Initial centre of mass position of person + chair = \(\displaystyle (x, y) = (x_0, y_0)\) with \(\displaystyle 0 < x_0 < W\)
Radial distance between centre of mass and pivot point = R (constant)
So initial position of centre of mass in polar coordinates is \(\displaystyle (r,\theta) = (R, \theta_0)\) with \(\displaystyle \theta_0\) in quadrant 1.

Differential equation is the same as a pendulum:
\(\displaystyle \frac{d^2 \theta}{dt^2} + \frac{g}{R} \sin \theta = 0\)

This is difficult to solve analytically. Using the small angle approximation (which is not really true in our example, but hey ho...), we instead solve:
\(\displaystyle \frac{d^2 \theta}{dt^2} + \frac{g}{R} \theta = 0\)

This is a homogeneous second order ODE. The characteristic equation is:
\(\displaystyle \lambda^2 + \frac{g}{R} = 0\)

Solve this to give:
\(\displaystyle \lambda = \pm \sqrt{\frac{g}{R}} i\)

Consequently, we use the complex roots trial solution (with a = 0 and b = \(\displaystyle \sqrt{\frac{g}{R}}\)):
\(\displaystyle \theta(t) = c_1 e^{at}\cos(bt) + c_2 e^{at} \sin (bt)\)

Substituting for a and b:
\(\displaystyle \theta(t) = c_1 \cos\left(\sqrt{\frac{g}{R}} t\right) + c_2 \sin\left(\sqrt{\frac{g}{R}} t\right)\)

Now to apply boundary conditions. First up we have \(\displaystyle \theta(0) = \theta_0\). The sine term evaluates to zero and the cosine term evaluates to 1, so
\(\displaystyle \theta(0) = c_1\)

Next up we have \(\displaystyle \omega(0) = uW\). Differentiate the trial solution with respect to t to get the angular velocity as a function of time:
\(\displaystyle \omega(t) = - c_1 \sqrt{\frac{g}{R}} \sin\left(\sqrt{\frac{g}{R}} t\right) + c_2 \sqrt{\frac{g}{R}} \cos\left(\sqrt{\frac{g}{R}} t\right)\)

The sine term evaluates to zero and the cosine term evaluates to 1, so
\(\displaystyle \omega(0) = c_2 \sqrt{\frac{g}{R}} = uW\)
\(\displaystyle c_2 = uW \sqrt{\frac{R}{g}}\)

We know \(\displaystyle c_1\) and \(\displaystyle c_2\) now, so the solution is
\(\displaystyle \theta(t) = \theta_0 \cos\left(\sqrt{\frac{g}{R}} t\right) + uW \sqrt{\frac{R}{g}} \sin\left(\sqrt{\frac{g}{R}} t\right)\)

You can then use this formula to solve some problems, like a constraint on u that allows the centre of mass to rotate beyond an angle of \(\displaystyle \pi/2\) before the angular speed drops to zero.

There's probably easier methods of estimating what you're after...
 
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