Question

A person with mass m=80 kg is training for pushups. Assume that the distance between the shoulders and the feet is 1.5 m, and the center of mass is at 1 m from the feet. First, he is trying to do a normal pushup. Assuming only vertical forces, what is the force on the arms?After that proved to be too difficult, he tries a modified version, where his arms are on the kitchen counter instead (~92 cm from the ground), which causes his body to form a 450 angle with the ground. What is the force on the arms, and what happens as the person expands their arms?

Answer 1

given that ::

mass of the person, m = 80 kg

a person trying to do a normal pushup. assuming only vertical forces, force on the arms is given as ::

dW = F s cos                              { eq. 1 }

where, F = mg

s = center of mass = l cos d

inserting the values in above eq.

dW = mg l cos d                                    { eq. 2 }

where, l = distance between feet and center of mass = 1 m

work is equal to the force multiplied by the distance over which the force is exerted which is given as ::

dW = dF

mg l cos d = F L cos d                          { eq. 3 }

where, L = distance between the shoulder and feet = 1.5 m

or    F = mgl / L                               { eq. 4 }

inserting the values in above eq.

F = (80 kg) (9.8 m/s²) (1 m) / (1.5 m)

F = 522.6 N

and the person expands their arm which means that most of your arm weight does not count, since the lower arms do not go up and down, and the center of mass of the upper arms only moves half as far as your chest.