Question

A drawing shows a person (weight W = 583 N, L₁ = 0.828 m, L₂ = 0.406 m) doing push-ups. Find the normal force exerted by the floor on each hand and eachfoot, assuming that the person holds this position.

force on each hand: ? N

force on each foot: ? N

Answer 1

Here, consider the vertical and rotational equilibrium.

Since, vertical equilibrium tells us all vertical forces are balanced. So,

2F+2H = 583

where F is he force of the floor on each foot and H is the force of the floor on each hand; dividing by 2:

=> F+H = 291.5

the sum of torques is also zero; if we sum torques around the feet, we have

sum of torques = 583 L1 - 2H (L1+L2) =0

the force F exerts no torque around the feet since F has no lever arm around the feet

substituting numbers, we have

583x0.828 - 2H(0.828 + 0.406)=0

=> H = 482.72 / 2.468 = 195.6 N

Now, from the above,

H+F = 291.5 N,

=> F = 291.5 - 195.6 = 95.9 N Which approximates to F = 96 N

So, force on each hand = 195.6 N

force on each foot = 96 N