Push up weight distrubtion problem

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To solve the push-up weight distribution problem, the total weight of the individual must be balanced between the hands and feet using the principles of forces and torques. The center of mass is located 60 cm from the hands and 1 m from the feet, with the total distance between hands and feet being 1.6 m. By setting up two equations based on the equilibrium of forces and the cancellation of torques, the distribution of weight can be calculated. This involves treating the feet as the pivot point to find the forces exerted by both the hands and feet. Understanding these concepts is crucial for accurately determining how much weight is supported by each point.
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Homework Statement



While doing push ups part of the weight of an 80kg young man is supported by his hands and part by his feet. His center mass is located 60cm from his hands and 1m from his feet. (The hand ot feet distance is 1.6m) How much of the total weight is supported by his hands? How much of the weight is supported by his feet? Use the feet as the pivotal point.

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The Attempt at a Solution



I don't even know where to start...It's just completely escaped my mind. This was a question on a test we had last week and I'm quite anxious to figure out how to do it.
 
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You need to use the forces and the torques to figure this out.

Knowing that the forces of the hands and feet are equal to the weight and that the torques likewise cancel out, you should have 2 equations and 2 unknowns.
 
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