Calculating Normal Forces in a Push-Up Position

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Homework Help Overview

The discussion revolves around calculating the normal forces exerted by the floor on each hand and foot of a person performing push-ups, given their weight and distances related to their center of gravity.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationships between the weights and distances involved, with some attempting to express one variable in terms of another. Questions arise about the definitions of the variables L1, L2, X1, and X2, and the need for a diagram is mentioned. There is also exploration of how the center of gravity affects the calculations.

Discussion Status

Participants are actively engaging with the problem, raising questions about variable definitions and assumptions. Some guidance has been offered regarding the setup of equations and the interpretation of forces, but there is no explicit consensus on a solution.

Contextual Notes

There is a mention of the need to consider the distribution of weight across the hands and feet, as well as the implications of choosing different points for torque calculations. The original poster's assumptions about distances in relation to the center of gravity are also under discussion.

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Homework Statement


a person (weight W = 590 N, L1 = 0.830 m, L2 = 0.409 m) doing push-ups. Find the normal force exerted by the floor on each hand and each foot, assuming that the person holds this position.


Homework Equations


Xcg= (W1X1 + W2X2)/(W1+W2)
τ=Fl


The Attempt at a Solution



I tried solving for W1 in terms of W2 but that got me no where. I can't figure out how to resolve two varibles with one equation.
 
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What are L1, L2, X1, X2? A diagram would help.
 
pmd28 said:

Homework Statement


a person (weight W = 590 N, L1 = 0.830 m, L2 = 0.409 m) doing push-ups. Find the normal force exerted by the floor on each hand and each foot, assuming that the person holds this position.


Homework Equations


Xcg= (W1X1 + W2X2)/(W1+W2)
τ=Fl


The Attempt at a Solution



I tried solving for W1 in terms of W2 but that got me no where. I can't figure out how to resolve two varibles with one equation.
I am assuming that the distances L1 and L2 represent the distances from the persons cg and feet, and cg and hands, respectively??
 
PhanthomJay said:
I am assuming that the distances L1 and L2 represent the distances from the persons cg and feet, and cg and hands, respectively??

I had to draw out my own daigram. L1 is the distance from the feet to the cg, L2 is the distance from cg to the hand and X1 and X2 are just how I denoted distance in my notes, they're not given in the equations

So yes johnny you are correct.
 
W=W1+W2
Keep in mind that each hand also only caries half of W2.
 
Yea I caught on to the fact that it said per hand. And how does that play into the center of gravity formula.
 
Let the center of gravity be at 0 and have one L be positive and one be negative. You now have two equations with two unknowns. Have fun with the algebra!
 
would my F in torque be W=509?
 
pmd28 said:
would my F in torque be W=509?
Yes, that would be just one of the forces creating torque about your chosen point, provided that your chosen point is not the cg, in which case it creates no torque. You can choose any point you want to solve for torques about that point; the sum of the torques of all forces about that point, and the sum of all forces, must be zero for the equilibrium condition.
 
  • #10
OK I solved it. Thanks :)
 

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