Static Equilibrium Beam/supports?

1. Jul 2, 2008

0338jw

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

2. Relevant equations
$$\Sigma\tau$$=0
$$\SigmaF=0$$

3. 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?

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2. Jul 2, 2008

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!

Last edited: Jul 2, 2008
3. Jul 2, 2008

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 :[

4. Jul 2, 2008

rock.freak667

Attatchments might take a few hours to be approved

5. Jul 2, 2008

6. Jul 2, 2008

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.

7. Jul 2, 2008

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.

8. Jul 2, 2008

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!

9. Jul 2, 2008

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.

10. Jul 2, 2008

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

11. Jul 2, 2008

rock.freak667

Use the exact ratio of sine and then check again.

12. Jul 2, 2008

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?

13. Jul 2, 2008

rock.freak667

$$\tau_1=(R_a)(95+40)sin\theta$$
$$\tau_2=-95Wsin\theta$$
so now for equil. $\tau_1=\tau_2$
the $sin\theta$ cancels and you can solve for Ra (Note W is weight)