# A person doing a pushup.

1. Oct 27, 2012

### pmd28

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Xcg= (W1X1 + W2X2)/(W1+W2)
τ=Fl

3. 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.

2. Oct 27, 2012

### frogjg2003

What are L1, L2, X1, X2? A diagram would help.

3. Oct 27, 2012

### PhanthomJay

I am assuming that the distances L1 and L2 represent the distances from the persons cg and feet, and cg and hands, respectively??

4. Oct 27, 2012

### pmd28

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.

5. Oct 27, 2012

### frogjg2003

W=W1+W2
Keep in mind that each hand also only caries half of W2.

6. Oct 27, 2012

### pmd28

Yea I caught on to the fact that it said per hand. And how does that play into the center of gravity forumla.

7. Oct 27, 2012

### frogjg2003

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!

8. Oct 27, 2012

### pmd28

would my F in torque be W=509?

9. Oct 28, 2012

### PhanthomJay

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. Oct 28, 2012

### pmd28

OK I solved it. Thanks :)