How to Balance a See-Saw with Different Weights?

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To balance a see-saw with a 67-kg adult and a 23-kg child on a 10-m board, the pivot point should be placed at 2.56 meters from the adult. The calculation involves setting the moments equal, resulting in the equation 67d = (10-d)23. Solving this gives d = 2.56 m, indicating the correct placement for balance. It's important to clarify where the distance d is measured from for accuracy. The setup and calculations appear correct for achieving equilibrium.
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A 67-kg adult sits at one end of a 10-m board, on the other end of which sits his 23-kg child. Where should the pivot point be placed so the board (ignore its mass) is balanced?

67d=(10-d)23
67d=230-23d
90d=230
d=2.56 m

The fulcrum should be at 2.56 m

Have I set this up right
 
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Looks okay.
 
Looks okay to me, although you should state where your distance d is to me measured from.

Edit: Too slow again
 
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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