In class, we discussed the conditions for static equilibrium in holding a plank

Question

In class, we discussed the conditions for static equilibrium in holding a plank pose. Let’s consider a man and woman of equal mass (75 kg) and equal height h = 5 feet, 8 inches. The man’s center of mass is located at 0.57h above the ground, and the woman’s is located at 0.54h.

(a) What are the forces the woman’s hands and feet have to exert on the ground (2 different numbers).

(b) Show that the choice of pivot point does not matter. (You probably chose the pivot point in part a) to be the COM. Now choose the feet, or any other point. Show you get the same results.)

(c) What are the forces the man’s hands and feet have to exert on the ground? Compare and contrast with the answer for part (a).

Explanation / Answer

(a) There two types of forces working here,

(1) Force to counter normal force,

Normal force exerted will be equal to four fources exteted by two legs and two hands.

F1+F2+F3+F4= 75*9.81 N = 735.75 N, Asumming all of them are equl then,

F1=F2=F3=F4= 735.75/4 =183.95 N

(2) Force to counter moment with respect to center of gravity/mass.

Above forces will counter moment at center of gravity/mass.

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