Force exerted on the floor by gymnast

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A 74 kg gymnast hangs motionless from rings, with both arms at a 26° angle from the vertical, exerting a force of 290 N on each arm. To find the force exerted by the floor on his feet, one must first calculate the gymnast's weight using W = mg. The forces from the rings can be broken into components: F(sub x) = Fcos(26°) and F(sub y) = Fsin(26°). The total weight supported by the rings is determined by summing the vertical components of the forces from both arms. The remaining weight is then supported by the floor, allowing for the calculation of the force exerted by the floor.
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Can anyone help me with this ?

Before practicing his routine on the rings, a 74 kg gymnast hangs motionless, with one hand grasping each ring and his feet touching the ground. Both arms make an angle of 26° with the vertical.


- If the force exerted by the rings on each arm has a magnitude of 290 N, what is the magnitude of the force exerted by the floor on his feet?
 
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What are you having trouble with? What have you tried so far?
 
Everytime you see a force at a particular angle...you should immediately realize that it can be broken down into components.

Try thinking about the force in components and see what you come up with.
 
Re

thats just it I am not sure how or what to break the components into can anyone help ?
 
Find the guy's weight: W = mg

Break both forces into F(sub x) and F(sub y): Fcos(26), Fsin(26)

Find how much of his weight the rings support: W - (F(sub y)rope1 + F(sub y)rope2)

Remaining weight is supported by floor, thus, F(floor) = (whatever answer you got for the section immediately above)

Hope that helps

Pat
 

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