Whats wrong with this torque set up

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The discussion centers on calculating the forces on a man's hands and feet while doing push-ups, given his weight of 70 kg and the distances from his center of gravity. The torque equation used indicates that the sum of the torques should equal zero, leading to the calculation of forces on the feet and hands. The initial calculation resulted in 175 N for the feet, but it was clarified that the hands should bear significantly more weight, calculated at 511 N. The confusion arose from rounding in the reference material, rendering some provided measurements, like the 30 cm height, unnecessary for the calculation. Ultimately, the correct understanding of the torque setup confirms that the hands support more weight than the feet during push-ups.
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The problem involves a man doing push ups, he weighs 70 KG. It wants to know the force on each hand and each foot. His center of gravity is 25 cm from his two hands, and 73 cm from his feet. The center of gravity is also 30 cm above both is hands and feet.

Using the hands as the rotation point, weight bearing down negative, force of the ground on the feet upwards

Sum of the torque equals 0 = (-70)(9.8)(.25) + (.98)(Force of Feet)

This comes out to be 175 N, which I know is incorrect, but do not see what is wrong with the set up. Obviously the hands should be bearing a decent amount more than the feet.
 
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175 N on the feet and 511 N for the Hands 511N>175N your work is right
 
Ah, stupid book must have rounded a lot, that's why I thought I was off, thanks, so the 30 cm was useless info
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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