Statics - Physics of push-ups

Problem Statement:

An athlete of mass m = 70 kg and height L = 1.80 m is doing push-ups on a horizontal surface without friction. Its center of mass lies at a distance LCM = 1.20 m from the point of support O (see figure). The distance between the hands and point B is LAB = 0.2 m. Calculate the reactions of the floor on the hands and feet of the athlete when his body forms an angle θ = 200 with the horizontal.

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Solution:

The athlete is in static equilibrium, so we will apply the two conditions that must be fulfilled to solve the problem:

At first we will draw the external forces that act on the athlete. As he leans on the ground, we will take into account the normal for the hands and feet. The weight applied in its center of mass also acts on him. The different external forces as well as the positive directions of the Cartesian axes are represented in the figure below:

The first condition of static equilibrium applied to the athlete is:

And since all the forces are vertical, we can project the previous equation on the y-axis, taking into account the sign of the projections:

The second condition of static equilibrium must be fulfilled for any point that we use as origin to calculate the torques of the forces. We will use the point O represented in the figure.

Recall that the torque (or moment) of any force is given by:

Where r is a vector that goes from the point we have taken as origin to the point of application of the force.

The direction of the torque is determined using the right-hand rule, and its magnitude is given by:

where θ is the angle formed by both vectors.

For the athlete the equation of moments with respect to the point O is:

The torque of the normal N1 is null because it is applied at the origin of moments. Therefore r = 0 for the normal.

Next we are going to calculate the two non null torques that appear in the previous equation.

Normal N2: The torque of N2 is:

In the following figure we have represented both vectors together with the right-hand rule.

As you can see in the figure, the moment has the direction of k.

On the other hand, the torque magnitude is given by:


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Weight P: The torque of P is:

We have represented both vectors in the following figure:

Using the right-hand rule as seen in the figure, we determine that the cross product has the direction of -k.

The torque magnitude is given by:

We have represented both vectors in the following figure:

As you can see, both vectors are on the z-axis, so we will project the equation of torques on that axis.

The torques equation written previously is:

And its projection on the z-axis is:

From equation (2) we get N2 by substituting the givens in the problem statement:

And from equation (1) we get N1:

Since the athlete has two feet and two hands, the value of the normal in each one is half what we have calculated.

In the problem we have used as g = 10 m/s2.

Do not forget to include the units in the results of the problems.

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