AP Physics - Static Equilibrium
posted by Phillip .
A man doing push-ups pauses in the position shown in the figure. His mass is 75-kg. Determine the normal force exerted by the floor on each hand and on each foot.
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a = 40 cm
b = 95 cm
c = 30 cm
W = mg = 735 N
I can't figure out how to find the amount of force being exerted at the hands/feet if you don't know the mass at each location. Help would be very much appreciated.
Assume the forces on each foot are equal, and the forces on each hand are equal. The total on all four appendages must equal 735 N.
Set the total moment about a line through the feet/floor contact points equal to zero. This will let you solve for the sum of the forces on the hands. You will need to know where the center of mass of the body is. This is shown as "b" in your figure
I thought W was the center of mass?
Also, are you saying to use moment of inertia I to solve, and if so, how would I set up the equations? I'm not good at equilibrium problems.
W is the center of mass. The moment of inertia does not play a role in statics problems.
I still don't know get to set up how to solve for the forces of the hands/feet if you don't know their distinct masses and only have the center of mass...
You don't need to know the distinct masses of hands and feet... just the total mass and that it acts as if it were at the Center of Mass.
The height of the CM above the floor, although shown in the figure, will not matter in this problem.