Question

The figure shows a person whose weight is W 700 N doing push-ups. Find the normal force exerted by the floor on (a) each hand and (b) each foot, assuming that the person holds this position. 0.410 m 0.840 m Units (a) Number Units (b) Number

Answer 1

Let the force exerted on each hand be x and on each foot be y.

Balancing the net force along the vertical direction:

2*x + 2*y = 700 [Multiplication by two as there are two hands and feet]

x + y = 350

The torque about the center of gravity of the man should be balanced:

2x*0.410 = 2*y*0.840

That is, x = (0.840/0.410) * y = 2.049y

Substituting the value of x in terms of y in the above equation, we get: 2.049y + y = 350

(b) => y = 114.8 N

(a) Hence x = 350 – 114.8 = 235.2 N

Kindly upvote :)