If L1=.827m and L2=.363m andthe person's weight = 615N.Whatis thenormal force exerted by the floor oneach hand

Question

the man is showing doing pushups. His weight is 615N. The length from his feet to his center is .827m and from the center to to his hand is .363m. What I need to find out is how to calculate the normal force exerted by the floor on each hand, assuming that the person holds the position

Answer 1

F = (1/2)615(0.827)/(0.827 + 0.363)
F ≈ 213.7 N

Answer 2

First draw an free body force diagram:

Shows the axis of the man's body with a horizontal line.Show point F representing his feet at one end of the line. Show C representing his centre of gravity in the middle of the line and H representing his hands at the opposite end of the line.

Draw mg as the downward force of the man's weight = 615N.

Draw Nf as the normal reaction force of the floor pointing up through the man's feet.

Draw Nf as the normal reaction force of the floor up through the man's hands.

Take moments about the point F ( this gets rid of Nf and makes things simpler)

So taking moments about F...

mg X L1 = Nh X (L1 + L2)

Substitutung known values...

615 X 0.827 = Nh X (0.827 + 0.363)

Work out Nh.

Nh is the sum of the two Normal forces (one for each hand).

Make the assumption that an equal force acts through each hand and divide Nh by two.

Answer 3

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Answer 4

Ok, you need to use torques here, so let's consider a uniform body, the weight acts downward from the center of the boby, so we apply torques respect his hands.

615*((0.363 + 0.827)/2) = 2Fn*(0.363)

then, we can calculate the 2Fn :

Fn = 66.4 Newtons

That's the force exerted on each hand.

Hope that helps