How Do Forces Distribute in a Block and Tackle System?

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In a block and tackle system, the forces at various points can be calculated based on the weight being lifted. The user provided specific force values for points in the system, suggesting that point 1 has a force of 40, point 2 has 20, point 3 has 20, and point 4 has 10. There is a discussion about the mechanical advantage of the system, with clarification that a 10 unit weight can lift a 20 unit weight at point 2, indicating that mechanical advantage does exist. The conversation emphasizes the importance of force balance in analyzing the system's efficiency.
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Given a block and tackle system and a known weigt, give the magnitude of the forces at the numbered points. [see attached for figure]
https://www.physicsforums.com/attachment.php?attachmentid=7026&stc=1&d=1149034106
blocktackle.JPG

My answers:
Point 2: 20
Point 3: 20
Point 4: 10
Point 1: 40

I arrived at the answers simply by looking at the figure and do a force balance at each tackle. I think some (if not all) are wrong because the systems doesn't provide any mechanical advantage (according to my answers). Please Help.
 

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Looks OK to me (assuming all ropes are essentially vertical). Why do you say there's no mechanical advantage? A 10 unit weight (shown) lifts a 20 unit weight (that's point 2).
 
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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