Equilibrium of Two Rods: Finding the Point of Toppling

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The discussion centers on determining the distance from point A at which rod AB will topple when a boy weighing 600 N moves along it. The participants agree that the rod will topple when the torque from rod CD cannot balance the torque from rod AB, specifically when the normal force at point C becomes zero. The suggested pivot point for calculations is at point E of rod CD. After analyzing the forces and torques, the consensus is that the correct distance for the boy to be from point A for the rod to topple is 0.5 m. The conversation concludes with expressions of gratitude for the assistance provided.
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Homework Statement


Weight of AB = CD = 200 N
Length of AB = 3 m
Length of CD = 2 m
A boy moves from point A toward B. At what distance from point A the rod will topple?
a. 0.4 m b. 0.5 m c. 0.6 m d. 1 m e. 2 m
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Homework Equations


Newton's law
torque


The Attempt at a Solution


When rod AB topples, the normal force at B = normal force at C = 0 right?

Will there be normal force at point A? Where should I take the pivot?

Thanks
 
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What is the boy's weight?
The rod AB topples only when DC does it. It happens when the torque of the weight of DC around E can not balance the torque arising from the other rod. In that case the support C does not exert force. Put the pivot of rod DC at E.

ehild
 
ehild said:
What is the boy's weight?
The rod AB topples only when DC does it. It happens when the torque of the weight of DC around E can not balance the torque arising from the other rod. In that case the support C does not exert force. Put the pivot of rod DC at E.

ehild

Oh I'm sorry. The boy's weight is 600 N. Is 0.5 m the correct answer? Thanks
 
songoku said:
Oh I'm sorry. The boy's weight is 600 N. Is 0.5 m the correct answer? Thanks

I got the same. :smile:

ehild
 
ehild said:
I got the same. :smile:

ehild

Thanks for the help :)
 
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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