Static Equilibrium Beam/supports?

[0338jw]

Homework Statement

A man doing push-ups pauses in the position shown (in the figure I found on the net)
a= 40cm b= 95cm c= 30cm Tha mass of the person is 75kg. Determine the normal force exerted by the floor on each hand and each foot.

Homework Equations

\Sigma\tau=0
\SigmaF=0

The Attempt at a Solution

before I started I wanted to know if it would be okay to simply solve the problem supposing he were completely horizontal, but it might not be the case. Will i need to use trig, or can i just treat him like a beam with supports?

 

[0338jw]

Do I need to calculate the angles for the guy in this position, or can I just say he's horizontal?
I calculated the angle between his feet and the ground is 17.53 degrees. I'll get back to work and see where it takes me, but I'd like maybe some advice? Thanks!

 

[0338jw]

I have the sum of the force in the Y direction as Fn-mgsintheta+Fn=0 is this correct? The only forces acting on the guy are normal force and gravity, correct? Please someone help me soon :[

 

[rock.freak667]

Attatchments might take a few hours to be approved

upload it on http://imageshack.us and then past the link for forums.

 

[0338jw]

http://img95.imageshack.us/img95/714/picture1qn8.th.jpg

 

[0338jw]

Are the sum of the forces in the y direction FN-mg*sintheta*r+ FN -mgsintheta*r2? My axis is the center of gravity since i need both normal forces.

 

[rock.freak667]

R_a= Normal reaction of the arm
R_f= Normal reaction of the foot


if at that angle he is in equilibrium then


R_a+R_f-W=0



Now just take moments about any point and you can solve the two equations.

 

[0338jw]

So my sum of torques
\Sigma\tau= Fn-mgsinR +Fn -mgsinr

Should I have two sum of forces in the y direction to find each normal force? I'm sorry, but I'm not getting that sum of the torques. Would i be able tosolve for each individual one? Will I need to multiply each normal force by the angle, or would that just be the weight? Thanks for helping me!

 

[rock.freak667]

Take torques about the point where Rf acts and use \tau = Fr sin\theta

where r is the distance of the force to the point where Rf acts.

 

[0338jw]

I tried doing this, but the torques do not equal out. I have Tcw=mgsing72.47*.42m= 294.36
Tccw=mgsin17.5*.996=220.5

 

[rock.freak667]

Use the exact ratio of sine and then check again.

 

[0338jw]

I checked the sines but I keep getting the same thing. Should I be using the angles on either side formed by the FG pulling the center of gravity down? As you can se I'm using 17.53 (angled formed by his fet atthe ground) and 72.47 from the right side of the mg. Which angles should I use?

 

[rock.freak667]

Taking torques about Rf.

\tau_1=(R_a)(95+40)sin\theta


\tau_2=-95Wsin\theta

so now for equil. \tau_1=\tau_2

the sin\theta cancels and you can solve for Ra (Note W is weight)