Calculating Normal Forces in a Push-Up Position

[pmd28]

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.

 

[frogjg2003]

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

 

[PhanthomJay]

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??

 

[pmd28]

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.

 

[frogjg2003]

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

 

[pmd28]

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

 

[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!

 

[pmd28]

would my F in torque be W=509?

 

[PhanthomJay]

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.

 

[pmd28]

OK I solved it. Thanks :)