How Do Gymnasts Balance Forces During an Iron Cross?

[way2whack]

Gymnast pictured below performing an Iron Cross has a mass of 65.00 kg. Assume that the forces
in the cables are symmetric and that he is in static equilibrium.

A. Draw a complete free body diagram of the gymnast (just him – he is the system).

B. Calculate the vertical and horizontal components of the cable force acting on his left hand.
**answer: vertical component = 318.83 N horizontal component = -56.13 N**

C. Calculate the moment generated around his left shoulder by the forces in the cable. Make sure you include both components of the cable force.
**answer: -68.60 Nm**

D. Is this moment resisted by the shoulder abductors or adductors? Explain your choice using the results from part C. **answer: This moment is resisted by the shoulder adductors because shoulder abductors are responsible for moving muscles on the arms out to the side, which is a positive moment based on my assignment of rotation. However, my answer was a negative moment about the shoulder joint means the muscles responsible were moving the arms in (clockwise on the left arm).**

E. Would it be harder or easier to hold this move if the cable was vertical instead of tilted inwards? Explain your answer. **I have no idea**

F. If the shoulder muscles have an average moment arm around the shoulder joint of 40 mm, what is the total force required by those muscles to maintain this position? **I have no idea**

http://i.imgur.com/VNdfQ.png <-- picture of gymnast

 

[Mr.A.Gibson]

We should really consider his weight will be acting at this point. Hence the free body diagram will have the force of his weight, and the tension in the two ropes acting on his hands.

The components of the tension you seem to have done correctly.

Part C, moment = force x perpendicular distance from force to pivot
The total tension you work out from Pythagorus, using the two components you have calculated, The perpendicular distance is .62cos 20. You might need to draw out the diagram to convince yourself of this. You can then calculate the moment

D The tension is trying to pulls the arms up, he must pull them down in order to counter this.

E If the cable was vertical then the moment would be = 1/2mass x g perpendicular distance, so just calculate this moment and compare it to your previous answer.

Question F. 40mm is the distance from the pivot to where the force is acting in a shoulder joint, I assume it is referring to the position held in question E.
Hence moment calculated in E = Force x distance(m)
rearrange this for force.