How Far Can a Person Sit on a Table Without Tipping It Over?

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To determine how far a 66kg person can sit on a 20kg table without tipping it over, the balance of forces and the center of gravity must be analyzed. The table, measuring 2.2m in length with legs positioned 0.5m from each end and 1.2m apart, will tip when the person's weight shifts the center of gravity beyond the support points. The tipping occurs due to the disturbance in balance when the person sits near the edge. A diagram can help visualize the forces at play and clarify the tipping point. Understanding the relationship between the person's position and the table's center of gravity is crucial for solving this problem.
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Homework Statement



How Close to the edge of a 20kg table, can a 66kg person sit without tipping it over?
The table is 2.2m long and there are two legs placed 0.5 m from each end of the table with 1.2 meters between the two legs and the top of the table is 0.80 m tall


Homework Equations



I am totally at a loss here, do I use r/R=Fd/gravity?

The Attempt at a Solution

 
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The question you need to ask is why does the table tip up at all?
 
It tips because the guy sits on the edge and disturbs the balance (changes the center of gravity). I still have no idea how what equation I should be using.
 
The table is symmetric, and you can assume that its mass acts from the centre of the tabletop. If you draw yourself a diagram you should see what's going on. Let me know if you don't!
 
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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