Weight distributions If a man is standing on two weighing machines (scales), with one foot on each, Will both machines show equal weight or his weight will be distributed in two machines? 
 A: If the person in concern is standing perfectly, such that his weight distribution is equal over both legs, and if both the weights are calibrated perfectly, then yes, both the weights will show equal readings. (The readings each being half the weight of the person.) 
In a realistic case (without 100% perfection), however, the weight would be unequally distributed over both the machines. The sum of these two readings would equal the weight of the person within a tiny error margin.
A: Remember Newton's first law $\sum F=0$. Vertically you have this:
$$\sum F_y=N_{scale1}+N_{scale2} - W=0$$
The scales will feel and display the $N_{scale}$ force. Solving for either scale force you will see that they share the total weight $W$. Together they must lift the whole weight $W$, not individually. 
A: 
$N_1$ the reading of force on the first weighing machine
$N_2$ the reading of force on the second weighing machine
$X_1$ the horizontal displacement of first leg from COM
$X_2$ the horizontal displacement of second leg from COM
we know that the man is in equilibrium. so

the weights shown on both meters will have their sum equal to the force of gravity on the man.
$$N_1 + N_2 = W$$
then the forces must satisfy that the total torque on the man is zero, that is
$$ N_1X_1 = N_2X_2 $$
by solving $$ N_1 = WX_2/(X_1+X_2) $$ $$ N_2 = WX_1/(X_1+X_2)$$
the thing is : it depends. it depends on the distances that your legs spread horizontally with the center of mass.if both distances are same then the weights are equal.
