Suppose you throw the ball upwards at some speed $v$. Then the time it spends in the air is simply:

$$ t_{\text{air}} = 2 \frac{v}{g} $$

where $g$ is the acceleration due to gravity. When you catch the ball you have it in your hand for a time $t_{\text{hand}}$ and during this time you have to apply enough acceleration to it to slow the ball from it's descent velocity of $v$ downwards and throw it back up with a velocity $v$ upwards:

$$ t_{\text{hand}} = 2 \frac{v}{a - g} $$

Note that I've written the acceleration as $a - g$ because you have to apply at least an acceleration of $g$ to stop the ball accelerating downwards. The acceleration $a$ you have to apply is $g$ plus the extra acceleration to accelerate the ball upwards.

You want the time in the hand to be as long as possible so you can use as little acceleration as possible. However $t_{\text{hand}}$ can't be greater than $t_{\text{air}}$ otherwise there would be some time during which you were holding both balls. If you want to make sure you are only ever holding one ball at a time the best you can do is make $t_{\text{hand}}$ = $t_{\text{air}}$. If we substitute the expressions for $t_{\text{hand}}$ and $t_{\text{air}}$ from above and set them equal we get:

$$ 2 \frac{v}{g} = 2 \frac{v}{a - g} $$

which simplifies to:

$$ a = 2g $$

So while you are holding one 3kg ball you are applying an acceleration of $2g$ to it, and therefore the force you're applying to the ball is $2 \times 3 = 6$ kg.

In other words the force on the bridge when you're juggling the two balls (with the minimum possible force) is exactly the same as if you just walked across the bridge holding the two balls, and you're likely to get wet!

Just don't start writing equations without a complete Free-Body-Diagram(may be rough diagram) according to the frame of reference..

### 1)Ground Frame

Now . Newtons law. $$\sum \vec{F_{ext}}=\frac{dp}{dt}=ma(for\ constant \ mass)$$

So, $$N-mg=\sum \vec{F_{ext}}=ma$$ Where system is boy in ground's frame.

Here we equate the net external force to the acceleration of body . $ma$ is not a force it is the measure of net force which causes body to accelerate.

### 2)In a accelerating frame(non-inertial)

here we have to add a fictious force $-m\vec{a}$ where $\vec a$ is the acceleration of frame .

Here too $$N-mg-ma=\sum \vec{F_{ext}}=0(as \ in \ frame \ of \ elevator \ boy \ is \ at \ rest)$$
Again you get $$N-mg=ma$$

## Best Answer

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.