Calculating Normal Force Exerted by Floor on Feet/Hands

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To calculate the normal force exerted by the floor on each hand and foot during push-ups, the total weight of the person is 587 N, with distances from the center of gravity to the feet and hands being 0.817 m and 0.414 m, respectively. The initial calculations suggest that the force on each foot is approximately 239.8 N and on each hand is about 121.5 N. However, a more effective approach involves applying equilibrium principles and calculating moments about either the hands or feet. This method allows for a clearer understanding of the distribution of forces in two dimensions. Ultimately, using equilibrium equations will yield accurate reactions for each limb.
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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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