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

#### steph_parker

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.

#### studiot

It isn't as simple as that.

Maximal forces are impulsive, but of very short duration

#### Woody

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.

studiot

#### benit13

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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