Calculating Tension and Mass in Static Equilibrium: Are My Methods Correct?

[Peter G.]

1. A gymnast of weight 720 N is holding himself in the cross position on the high rings. He is quite still. A free-body force diagram for the gymnast shows the two upward pulls of the rings on the hands, each, size 380 N. Calculate the angle between the wires supporting the rings and the vertical:

So, the tension on both wires is the same and the angle too, so I assumed that the vertical component of the tension for both wires should be the same, and since the guy is in equilibrium: 720/2 = 360 N.

I then did Inverse Cos (360/380) which is equal to: 18.7 degrees

2. A very heavy sack is hung from a rope and pushed sideways. When the sideways push is 220N the rope supporting the sack is inclined at 18 degrees to the vertical:

Find the Tension and the Mass of the Sack:

So, I recognized that the tension was the hypothenuse of a triangle with angle 18 and opposite side 220N and the mass, the vertical part of the triangle (adjacent to the angle) divided by the pull of gravity:

I got 711.93 N for the tension and (209.23/10) for the Mass.

I am a bit insecure: I'm not sure if my methods/answers are correct.

Any help?

Thanks in advance,
Peter G.

 

[ashishsinghal]

Post the fbd of both the questions

 

[Peter G.]

That I've drawn? Because for the second question there's none, and for the first one there's the picture of the person

 

[ashishsinghal]

I asked to post, not draw. Post it on PF.

 

[Peter G.]

There you go.

 

[ashishsinghal]

first one is correct.
In second question, tension is correct but mass is wrong.

 

[ashishsinghal]

The method is correct. Check the calculations

 

[Peter G.]

Oh, for the mass I should've done: cos 18 x 711.9349551 = 677.0903782 / 10, right? And the other answers are correct? (I ask because I was a bit confused with the fact you posted twice)

Thanks,
Peter G

 

[ashishsinghal]

Yeah, correct