Calculating Normal Force Exerted by Floor on Feet/Hands

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SUMMARY

The discussion focuses on calculating the normal force exerted by the floor on a person's hands and feet while performing push-ups, given a weight of 587 N. The lengths from the center of gravity to the feet (0.817 m) and to the hands (0.414 m) are critical for the calculations. The normal force on each foot is determined to be 239.8 N, while the force on each hand is calculated at 121.5 N. The problem is approached using equilibrium principles and moments in two dimensions.

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Homework Statement


A person whose weight is W = 587 N doing push-ups.
Assume L1 = 0.817 m (Centre of Gravity to Feet) and L2 = 0.414 m. (Centre of Gravity to Hands) Calculate the normal force exerted by the floor on each hand (enter first) and each foot, assuming that the person holds this position

Weight = 587N
Length to Feet = 0.817m
Length to Hands = 0.414m

Homework Equations



I think I'm over thinking this one, but I've tried solving it as if it were a bridge problem. But to no prevail.

The Attempt at a Solution



F x d
587 x 0.817 = 479.6
479.6 / 2 = 239.8 by his feet

F x d
587 x 0.414 = 243.0
243.0 / 2 = 121.5 by his hands
 
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This sounds like a simple equilibrium problem - try using your equation for moments about either the hands or feet. Solve for the end reactions in 2-dimensions and then solve for each limb.
 

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